## KSEEB Solutions for Class 10 Maths Chapter 5 Areas Related to Circles Additional Questions

Students can Download Class 10 Maths Chapter 5 Areas Related to Circles Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 5 Areas Related to Circles Additional Questions

I. Multiple Choice Questions:

Question 1.
The radius of a circle is 5cm, then its area is
a) 154 cm3
b) 1386 cm3
c) 169 cm3
d) 308 cm3
Answer:
a) 154 cm3

Question 2.
The diameter of a semicircle is 4cm Its area is.
a) π cm2
b) 2π cm2
c) 4π cm2
d) 8π cm2
Answer:
b) 2π cm2

Question 3.
The radii of two concentric circles are 3cm and 2cm. the ratio of their areas.
a) 3 : 4
b) 9 : 2
c) 9 : 4
d) 4 : 3
Answer:
c) 9 : 4

Question 4.
The diameter of a circle whose area is equal to the areas of two circles of radii 24cm and 7cm is
a) 25 cm
b) 31 cm
c) 28 cm
d) 50 cm
Answer:
d) 50 cm

Question 5.
The radius of a circle is 21cm then its circumference is
a) 66 cm
b) 33 cm
c) 132 cm
d) 99 cm
Answer:
c) 132 cm

Question 6.
Formula to find area of Quadrant is
a) $\frac{\pi r^{2}}{2}$
b) πr2
c) $\frac{\pi r^{2}}{3}$
d) $\frac{\pi r^{2}}{4}$
Answer:
d) $\frac{\pi r^{2}}{4}$

Question 7.
Find the perimeter of a Quadrant of a circle of radius 7cm.
a) 25 cm
b) 49 cm
c) 14 cm
d) 66 cm.
Answer:
a) 25 cm

Question 8.
If the perimeter and the area of a circle are equal numerically, then the diameter of the circle is.
a) 8 units.
b) n units
c) 4 units
d) 2 units.
Answer:
c) 4 units

Question 9.
The central angle of a Quadrant of a circle measures.
a) 60°
b) 90°
c) 120
d) 30°
Answer:
b) 90°

Question 10.
The measure of the central angle of a circle is.
a) 360°
b) 180°
c) 270°
d) 90°
Answer:
a) 360°

Question 11.
If the diameter of a semi-circular protractor is 7cm, then its perimeter is
a) 7 cm
b) 36 cm
c) 21 cm
d) 18 cm
Answer:
d) 18 cm

Question 12.
The external and internal radius of a circular path are 5m and 3m respectively. The area of the circular path is.
a) 87π m2
b) 167π m2
c) 157π m2
d) 327π m2
Answer:
b) 167π m2

Question 13.
Find the area of a ring-shaped region enclosed between two concentric circles of diameter 8cm and 6cm.
a) 22 cm2
b) 44 cm2
c) 66 cm2
d) 88 cm2
Answer:
a) 22 cm2

Question 14.
If the perimeter of a circle is equal to that of a square, then the ratio of their areas is.
a) 11 : 4
b) 22 : 7
c) 14 : 1
d) 7 : 22
Answer:
c) 14 : 1

Question 15.
The area of circle whose circumference is 44cm is
a) 77 cm2
b) 308 cm2
c) 168 cm2
d) 154 cm2
Answer:
d) 154 cm2

Question 16.
A bicycle wheel makes 2500 revolutions in moving. 11km. Find the circumference of the wheel.
a) 220 cm
b) 55 cm
c) 110 cm
b) 100 cm.
Answer:
c) 110 cm

Question 17.
The minute hand of a clock is of length 4cm. then the angle swept by the minute hand in 15 minutes.
a) 45°
b) 60°
c) 120
d) 90°
Answer:
d) 90°

Question 18.
The perimeter of the sector with radius 10.5cm and sector angle 60° is
a) 23 cm
b) 32 cm
c) 35 cm
d) 28 cm
Answer:
b) 32 cm

Question 19.
The radius of a semicircular protractor is 35m. then its perimeter.
a) 90 m
b) 45 m
c) 180 m
d) I20 m.
Answer:
c) 180 m

Question 20.
The area swept out by a horse tied in a rectangular grass field with a rope 8m long is
a) 647π m2
b) 327π m2
c) 128π m2
a) 48π m2
Answer:
a) 647π m2

Question 21.
The area (in cm2) of the circle that can be inscribed in a square of side 8cm is.
a) 64π
b) 16π
c) 8π
d) 32π
Answer:
b) 16π

Question 22.
The area of a circle that can be inscribed in a square, of side 10cm is
a) 407π cm2
b) 307π cm2
c) 1007π cm2
d) 257π cm2
Answer:
d) 257π cm2

Question 23.
The area of the largest triangle that can be inscribed in a semicircle of radius r is
a) r2
b) 2r2
c) r2
d) 2r3
Answer:
a) r2

Question 24.
The area of the largest square that can be inscribed in a circle of radius 12cm is.
a) 24 cm2
b) 249 cm2
c) 288 cm2
d) 196 $\sqrt{2}$ cm2
Answer:
c) 288 cm2

Question 25.
The perimeter of a square circumscribing a circle of radius a unit is
a) 2a units
b) 4a units
c) 8a units
d) 16a units.
Answer:
c) 8 a units

II. Short Answer Questions:

Question 1.
Find the area of the circle inscribed in a square of side b cm.

Answer:
Diameter of a circle = b ; r = b/2

Question 2.
Find the area of a square inscribed in a circle of radius a.

Answer:
radius = a & diameter = 2a
Diagonal of square = 2a
AC2 = AB2 + BC2
(2a)2 = (side)2 + (side)2
4a2 = 2side2

Area of square = side2 = 2a2.

Question 3.
If the diameter of semicircular protractor is 14 cm then find its perimeter.
Answer:
Perimeter of semicircular protractor
= πr + diameter
= $\frac{22}{7}$ × 7 + 14 = 22 + 14 = 36 cm.

Question 4.
Find the number of revolutions if circumference of a wheel is 22cm and travells a distance 110km.
Answer:
Number of revolutions

Question 5.
If circumference and the area of a circle are numerically equal, find the diameter of the circle
Answer:
2πr = πr2
2r = r2
r = 2
d = 2r = 2 × 2 = 4 units.

Question 6.
Difference between the circumference and radius of a circle is 37cm. Find the area of circle.
Answer:
2πr – r = 37
r(2π – 1) = 37
pppppppp

Question 7.
Find the radius the area of a circular playground is 22176m2
Answer:
Area of circular playground = 22176 m2
πr2 = 22176
$\frac{22}{7}$ r2 = 22176
r = 84 m

III. Long Answer Questions.

Question 1.
The area of a circular playground is 2217m2. Find the cost of fencing this ground at the rate of? 50 per/m.
Answer:
Area of circular playground

= 44 x 12
= 528 m
Cost of fencing this ground
= 528 x 50
= ₹ 26400

Question 2.
The circumference of a circle exceeds the diameter by 16.8cm. Find the radius of the circle.
Answer:
Let the radius of circle be r cm.
∴ Diameter = 2r cm
circumference = 2πr cm
circumference = 16.8 + Diameter
2πr = 16.8 + 2r

Hence radius is 3.92 cm

Question 3.
A race track is in the form of a ring whose inner circumference is 352m, and the outer circumference is 396 in. Find the width of the track.
Answer:
Inner circumference = 352m
27πr = 352m

∴ Width of track = R – r = 63 – 56 = 9m

Question 4.
The minute hand of a clock is 10cm long. Find the area of the face of the clock described by the minute hand between 10 AM and 10-30 AM.
Answer:
Angle described by the minute hand in one minute = 6°
∴ Angle describes by minute hand between 10 AM and 10.30 Am = 30 minutes = 30 x 6 = 180°
∴ Area swept by the minute hand in 30 minutes
= Area of sector of circle radius 10cm

Question 5.
A circle of radius 6cm is inscribed in a square as shown in the given figure. Find the shaded region.

Answer:
From the figure, side of a square = Diameter of a circle = 2r
= 2 x 6 = 12cm
∴ Area of square = (side)2
= (12)2 = 144cm2

∴ Area of the shaded region
= Area of square-Area of the circle
= 144 – 133.04
= 30.96 cm2

Question 6.
Find the length of the arc that subtends an angle of 72° at the centre of a circle of radius 5cm.
Answer:

Question 7.
In the given figure, arcs are drawn by taking vertices A, B and C of an equilateral triangle of side 10cm as centers, to intersect the sides BC, CA and AB at their respective midpoints D,E and F. Find the area of the shaded region.

Answer:
= $\frac{1}{2}$ (Side of equilateral ∆le)
= $\frac{1}{2}$ × 10 = 5 cm
Area of shaded region
= Sum of areas of there equal sectors

Question 8.
A paper is in the form of a rectangle ABCD in which AB = 18 cm and BC = 14cm. A semicircle with BC as diameter is cut off. Find the area of the remaining portion.

Answer:
∴ Area of rectangle ABCD = l × b
= AB × BC
= 18 × 14 = 252 cm2
Radius of semicircle (r)

Area of the remaining portion
= 252 – 77
= 175cm2

## KSEEB Solutions for Class 10 Maths Chapter 6 Constructions Additional Questions

Students can Download Class 10 Maths Chapter 6 Constructions Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 6 Constructions Additional Questions

Question 1.
Construct a triangle similar to a given triangle ABC with its sides equal to 5/3 of the corresponding sides of triangle ABC.
Answer:

BDE is the corresponding similar triangle to ABC.

Question 2.
Draw a line segment PQ 7 cm and divide it in the ratio 3 : 4
Answer:
Ratio 3 : 4 = 3 + 4 = 7

The point A divides PQ
∴ PA : AQ = 3 : 4

Question 3.
Draw a circle of radius 4 cm and chord of length is 6 cm. Construct tangents at the ends of chord.
Answer:

∴ PQ and RS are two required tangents.

Question 4.

Construct another ∆ ADE similar to ∆ ABC. Such that AD = $\frac{4}{5}$AB

∴ ADE is the required similar triangle

Question 5.
Draw a circle of radius 3.5 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

∴ PA and PB are two required tangents from an external point P.
∴ QC and QD are two required from an external point Q.

## KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Additional Questions

I. Multiple-choice Questions:

Question 1.
The co-ordinates of origin are
a. 0,0.
b. 0, 1.
c. 1, 0.
d. 1, 1.
Answer:
a. 0,0.

Question 2.
The distance of the point (3, 4) x – co-ordinate is equal to
a. 3
b. 4
c. 1
d. 7
Answer:
b. 4

Question 3.
The distance of the point (3, 4) from y-axis is
a. 3
b. 4
c. 5
d. 1
Answer:
a. 3

Question 4.
The angle between x-axis and y-axis is
a. 0°
b. 45°
c. 60°
d. 90°
Answer:
d. 90°

Question 5.
The distance of the point (α, β) from origin is
a. α.β
b. α2. β2
c. $\sqrt{\alpha^{2}-\beta^{2}}$
d. $\sqrt{\alpha^{2}+\beta^{2}}$
Answer:
d. $\sqrt{\alpha^{2}+\beta^{2}}$

Question 6.
The distance between the point (x1,y1) and (x2, y2) is

Answer:

Question 7.
Which of the following is distance formula?
Answer:
a. The distance between 2 points (x1,y1) and (x2, y2) is

b. The mid point of the lines segment join the points (x, y) and (x2, y2)

c. The co-ordinates of the centroid of the ∆le ABC where A, B, C are (x1,y1)

d.

Question 8.
The distance b/w the point (0, 3) and (- 2, 0) is
a. $\sqrt{14}$
b. $\sqrt{15}$
c. $\sqrt{13}$
d. $\sqrt{5}$
Answer:
c. $\sqrt{13}$

Question 9.
The ∆le with vertices (- 2, 1) (2, – 2) and (5, 2) is
a. Scalene
b. Equilateral
c. Isosceles
d. right angled isosceles
Answer:
d. right angled isosceles

Question 10.
The Co-ordinates of the point which divides the join of (x, y) and (x2, y2) in the ratio m1 : m2 internally are

Answer:

Question 11.
If the distance b/w the point (3, a) and (4,1) is $\sqrt{10}$, then, find the values of a
a. 3,- 1
b. 4, – 2
c. 2, – 2
d. 5, – 3
Answer:
c. 2, – 2

Question 12.
The distance of the mid – point of the line -segment joining the points 6, 8. and 2, 4. from the point (1, 2) is
a. 3
b. 4
c. 5
d. 6
Answer:
c. 5

Question 13.
The coordinates of the mid – point of the line – segment joining (- 8, 13) and (x, – 7)
a. 16
b. 10
c. 4
d. 8
Answer:
a. 16

Question 14.
Find the value of k if the distance b/w (k, 3) and (2, 3) is 5
a. 5
b. 6
c. 7
d. 8
Answer:
c. 7

Question 15.
The distance b/w the point (Sin θ, Sin θ) & (cos θ, – cos θ) is
a. $\sqrt{2}$
b. 2
c. $\sqrt{3}$
d. 1
Answer:
a. $\sqrt{2}$

II. Short Answers Questions:

Question 1.
Write the coordinate of the origin.
Answer:
(0, 0)

Question 2.
What is the ordinate of all the point on x-axis
Answer:
0(zero)

Question 3.
Find the perpendicular distance of the point (- 3, 5) form the x – axis.
Answer:
5

Question 4.
Find the distance of B(3 + $\sqrt{3}$, 3 – $\sqrt{3}$) form origin.
Answer:

d = $\sqrt{24}$
d = $\sqrt{4 \times 6}$
d = 2$\sqrt{6}$

Question 5.
Write the formula to find distance b/w two points
Answer:

Question 6.
Write the formula to find distance b/W point and origin.
Answer:
d = $\sqrt{x^{2}+y^{2}}$

Question 7.
Write the formula to find area of triangle.
Answer:
area of triangle

Question 8.
Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.
Answer:
Since (3, a) lies on the line 2x – 3y = 5
Then 2(3) – 3(a) = 5
6 – 3a = 5
– 3a = 5 – 6
– 3a = – 1
a = 1/3

Question 9.
Write the section formula when it divide in ration p : q
Answer:
Co-ordinates

III. Long Answer Questions:

Question 1.
Find the distance between the following pairs of points.
i) (6, 5) and (4, 4)
Answer:

Question 2.
The distance between the point (3, 1) and (0, x) in 5 unit. Find x.
Answer:
(x1, y1) = (3, 1); (x2, y2) = (o, x)
d = 5

(∵ squaring on both the sides)
25 = (3)2 + (x – 1)2
x2 + 1 – 2x + 9 = 25
x2 + 1 – 2x + 9 – 25 = 0
x2 + 1 – 2x + 9 – 25 = 0
x2 – 2x – 15 = 0
x2 – 5x + 3x – 15 = 0
x(x – 5) + 3(x – 5) = 0
(x – 5)(x + 3) = 0
x – 5 = 0, x + 3 = 0
x = 5, x = – 3

Question 3.
Find a point on y-axis which is equidistant from the points (5, 2) and (- 4, 3)
Answer:
A Point lie on y – axis = (0, y) the points are equidistant from the point on y – axis.

6y – 4y = 25 – 29
2y = – 4
y = $\frac{-4}{2}$ ⇒ y = – 2
∴ point on y – axis is y = – 2

Question 4.
Find the radius of a circle whose centre is (- 5, 4) and which passes through the point (- 7,1)

Question 5.
Prove that each of the set of co¬ordinates are the vertices of parallelograms.
i. (- 5, – 3), (1, – 11), (7, – 6), (1, 2).
Answer:

∴ AD = BC & AB = CD
∴ Given vertices from a parallelogram.

Question 6.
In what ratio does the point (- 2, 3) divide the line segment joining the point (- 3, 5) and (4, – a)?
Answer:
(x1, y1) = (3, 5)
(x2, y2) = (4, – a)
p(x, y) = (- 2, 3)

4m – 3n = – 2(m + n)
4m – 3n = – 2m – 2n
4m + 2m = – 2n + 3n
6m = n
$\frac{m}{n}=\frac{1}{6}$
m : n = 1 : 6 (or)
$\frac{-9 m+5 n}{m+n}$ = 3
– 9m + 5n = 3(m + n)
– 9m + 5n = 3m + 3n
– 9m – 3m = 3n – 5n
– 12m = – 2n
$\frac{m}{n}=\frac{-2}{-12}=\frac{1}{6}$
m : n = 1 : 6

Question 7.
In the point C(1, 1) divides the line segment joining A(- 2, 7) and B in the ration 3 : 2 find the co-ordinates of B.
Answer:
Point C(1, 1)
(x1, y1) = A(- 2, 7) m : n = 3 : 2
B(x2, y2) =?

∴ Co-ordinate of B = (3, – 3)

Question 8.
Find the ratio in which the point (- 1, k) divides the line joining the points (- 3, 10) and (6, – 8) and also find ‘k’.
Answer:
P(x, y) = (- 1, k); (x1, y1) = (- 3,10)
(x2, y2) = (6,- 8)

6m – 3n = – (m + n)
6m – 3n = – m – n
6m + m = 3n – n, $\frac{-8 m+10 n}{m+n}$ = k

Question 9.
Three Consecutive vertices of a parallelogram are A(1, 2), B(2, 3) and (8, 5). Find the fourth vertex.
Answer:

‘O’ is the mid point of AC and by using mid point formulas AC is a line and ‘O’ is the point
A(x1, y1) =(1, 2);
c(x2, y2) = (8, 5)
Co-ordinates of

Co-ordinates of ‘O’ = $\frac{9}{2}, \frac{7}{2}$
BD is a line and ‘O’ is midpoint
B(2, 3) = (x1, y1); D(x, y) = (x2, y2)
$\left(\frac{9}{2} \frac{7}{2}\right)$ = (x, y)
By midpoint formula
Co-ordinates of

2 + x = 9 3 + y = 7
x = 9 – 2 y = 7 – 3
x = 7 y = 4
the fourth vertex is (7, 4)

Question 10.
Find the relation between x and y, if the points (x, y) (1, 2) and (7, 0) are colliner.
Answer:
Given point are A(x, y), B(1, 2) and C(7, 0)
These points will be collinear if the area of the triangle formed by them is zero.
Now, ar(ABC)

⇒ 2x + 6y – 14 = 0
⇒ x + 3y = 7
which is the required relation between x & y

Question 11.
Find the ratio in which the line segment joining the points A(3, – 3) and B(- 2,7) is divides by x-axis. Also find the Co-ordinates of the point of division.
Answer:

Here, point Q is an x-axis its ordinate & 0
Let ratio be k : 1 and Co-ordinate of point Q be (x, 0)

Question 12.
If A(4, 2), B = (7, 6) and C(1, 4) are the vertices of ∆ ABC and AD is its median, P.T that median AD divides ∆ ABC into two triangle of equal areas.D is the median on BC
⇒ The Co-ordinates of midpoint D are given by

Co-ordinates of D are (4, 5)
Now, Area of triangle
ABD = $\frac{1}{2}$[x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)
= $\frac{1}{2}$[4(6 – 5) + 7(52) + 4(2 – 6)
= $\frac{1}{2}$[4 + 21 – 16] = $\frac{1}{2}$ sq.units
Area of
∆ ACD = $\frac{1}{2}$[4[4 – 5] + 1[5 – 2] + 4[2 – 4]]
= $\frac{1}{2}$[- 4 + 3 – 8] = $\frac{-9}{2}=\frac{9}{2}$ sq.units

Question 13.
In what ratio does the y – axis divide the line segment joining the point p(- 4, 5) and Q(3, 7). Also, find the co-ordinates of the intersection.
Answer:

Suppose y – axis divides PQ in the ratio k : 1. Then the co-ordinates of the point of division are

Since, R lies on y-axis and X Co-ordinate of every point on y-axis is zero.
$\frac{3 \mathrm{K}-4}{\mathrm{K}+1}$ = 0
⇒ 3k – 4 = 0 ⇒ K = $\frac{4}{3}$
Hence, the required ratio is 4/3 : 1 i.e, 4 : 3
Putting K = 4/3 in the Co-ordinates of R, we find that its Co-ordinates are (0, – 13/7)

Question 14.
Show that ∆ ABC with vertices A(- 2, 0), B(2, 0) and C(0, 2) is similar to ∆ DEF with vertices D(- 4, 0), E(4, 0) and F(0, 4)
Answer:
Given vertices of ∆ ABC and ∆ DEF are

A(- 2, 0), B(2, 0), C(0, 2), D(- 4, 0), E(4, 0) and F(0, 4)

Here, we see that sides of ∆ DEF are twice the sides of a ∆ ABC.
Hence, both triangle are similar.

## KSEEB Solutions for Class 10 Maths Chapter 4 Circles Additional Questions

Students can Download Class 10 Maths Chapter 4 Circles Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 4 Circles Additional Questions

I. Multiple Choice Questions:

Question 1.
Line segment joining the centre and a point on the circle is called
(a) radius
(b) diameter
(c) Chord
(d) Arc
Answer:
(a) radius

Question 2.
Part of a circle is called
(a) Chord
(b) diameter
(c) Segment
(d) Arc
Answer:
(d) Arc

Question 3.
The biggest chord in a circle is called
(a) radius
(b) diameter
(c) chord
(d) Arc
Answer:
(b) diameter

Question 4.
The region, bounded by a major arc and a chord is called
(a) Segment
(b) major segment
(c) minor segment
(d) major arc
Answer:
(b) major segment

Question 5.
The length of the biggest chord is 8 cm then the value of radius is
(a) 8 cm
(b) 4 cm
(c) 3 cm
(d) 5 cm
Answer:
(b) 4 cm

Question 6.
How many radius can be drawn in circle
(a) 1
(b) 2
(c) only 3
(d) many
Answer:
(d) many

Question 7.
An angle in a semicircle is.
(a) 60°
(b) 30°
(c) 90°
(d) 180°
Answer:
(c) 90°

Question 8.
Equal chords of a circle are.
(a) Equidistant from the centre.
(b) Equal
(c) Unequal
(d) Not equidistant from the centre
Answer:
(b) Equal

Question 9.
If the length of the chord increases its perpendicular distance from the centre.
(a) Increases
(b) Decreases
(c) Equal
(d) Constant
Answer:
(b) Decreases

Question 10.
The perpendicular distance between the biggest chord and the centre is.
(a) zero
(b) Equal
(c) 9 cm
(d) 10cm
Answer:
(a) zero

Question 11.
In a circle angles in the major segment are called.
(a) Obtuse angles
(b) Acute angles.
(c) Right angles
(d) Complete angle
Answer:
(b) Acute angles.

Question 12.
In a circle angles in the minor segment are called.
(a) Obtuse angles
(b) Acute angles.
(c) Right angles
(d) zero angle
Answer:
(a) Obtuse angles

Question 13.
In a circle angles in the same segment are
(a) Not equal
(b) Right angles
(c) Equal
(d) zero angle.
Answer:
(c) Equal

Question 14.
Circles having the same centre but different radii are called.
(a) Congruent circles
(b) Concentric circles
(c) Equal circles
(d) None of these
Answer:
(b) Concentric circles

Question 15.
Circles having same radii but different centres are called
(a) Congruent circles
(b) Concentric circles
(c) Equal circles.
(d) Intersecting circles
Answer:
(a) Congruent circles

Question 16.
The number of circles are drawn through three non-collinear points in a plane is.
(a) 1
(b) 2
(c)3
(d) 4
Answer:
(a) 1

Question 17.
A line which intersects a circle in two points is called
(a) A secant
(b) A chord
(c) An arc
(d) A tangent
Answer:
(a) A secant

Question 18.
A line which intersects a circle in only one point is called
(a) A secant
(b) A tangent
(c) A chord
(d) A diameter
Answer:
(b) A tangent

Question 19.
A tangent to a circle intersects the circle is
(a) one point only
(b) Two points
(c) No point
(d) Three points
Answer:
(a) one point only

Question 20.
A secant of a circle intersects the circle in
(a) only one point
(b) Two points
(c) Three points
(d) No point
Answer:
(b) Two points

Question 21.
The point where a tangent line intersects a circle is called the
(a) centre
(b) point of contact
(c) End-point
(d) None of these
Answer:
(b) point of contact

Question 22.
The angle between the tangent at any point of a circle and the radius through the point of contact is
(a) 60°
(b) 90°
(c) 45°
(d) 30°
Answer:
(b) 90°

Question 23.
How many tangents can be drawn to a circle at any point of it?
(a) 1
(b) 2
(c) 3
(d) None of these
Answer:
(a) 1

Question 24.
How many parallel tangents can a circle have at the most?
(a) 1
(b) 2
(c) 4
(d) 3
Answer:
(b) 2

Question 25.
Two circles of radii 5 cm and 3 cm touch each other externally. The distance between their centres is
(a) 5 cm
(b) 3 cm
(c) 2 cm
(d) 8 cm
Answer:
(d) 8 cm

Question 26.
Two circles of radii 5 cm and 3 cm touch each other internally. The distance between their centres is
(a) 5 cm
(b) 3 cm
(c) 2 cm
(d) 8 cm
Answer:
(c) 2 cm

Question 27.
The tangents at the endpoints of a diameter of circle are
(a) perpendicular
(b) parallel
(c) intersecting
(d) inclined at 60°
Answer:
(b) parallel

Question 28.
The length of the tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. The radius of the circle is
(a) 3 cm
(b) 2 cm
(c) 5 cm
(d) 4 cm
Answer:
(a) 3 cm

Question 29.
Two concentric circles of radii 5 cm, and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
(a) 8 cm
(b) 6 cm
(c) 4 cm
(d) 10 cm
Answer:
(a) 8 cm

Question 30.
How many tangent lines can be drawn to a circle from a point outside the circle?
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 31.
How many tangents can be drawn to a circle have
(a) 2
(b) infinitely many
(c) one
(d) no
Answer:
(b) infinitely many

Question 32.
In the following figure, |PBA is
(a) 60°
(b) 30°
(d) 45°
(d) none of these

Answer:
(a) 60°

Question 33.
In the following figure, Find AP if AB = 5 cm
(a) 5 cm
(b) 4 cm
(c) 3 ccm
(d) 2 cm

Answer:
(a) 5 cm

Question 34.
In the following figure, find the perimeter of A APQ if AB = 6 cm.
(a) 10 cm
(b)12cm
(c) 15 cm
(d) 6 cm

Answer:
(b)12cm

Question 35.
In the following Figure, find the length of the chord AB if PA = 4 cm and OP = 3 cm.
(a) 2 cm
(b) 4 cm
(c) 10cm
(d) 5 cm

Answer:
(c) 10cm

II. Short Answer Questions:

Question 1.
Define concentric circles.
Answer:
Circles which are having same centre and different radii are called concentric circles.

Question 2.
Define congruent circles.
Answer:
Circles which are having same radii & different centres are called congruent circles.

Question 3.
Define the sector of a circle.
Answer:
The region bounded by an arc of a circle and its two bounding radii is called sector.

Question 4.
Name the biggest chord of a circle.
Answer:
Diameter

Question 5.
Write the formula to find the perimeter of a circle.
Answer:
Circumference = C = 2nx.

Question 6.
Name the angle formed in a semi-circle is
Answer:
Right angle

Question 7.
Name the angle formed in a major segment.
Answer:
Acute angle

Question 8.
Name the angle formed in a minor segment.
Answer:
Obtuse angle.

Question 9.
The length of the tangent to a circle from a point P, which is 25 cm away from the centre, is 24 cm. What is the radius of the circle?
Answer:

Question 10.
In a quadrilateral, ABCD circumscribes a circle with centre O. If ∠AOB = 110°, then find ∠COD.
Answer:

Question 11.
Write the relationship between radius and diameter.
Answer:
d = 2r (or) r = d/2

Question 12.
Angles formed in a same segment are ________
Answer:
Equal

III. Long Answer Questions:

Question 1.
In the following figure if ∠DAB = 60°
and ∠ACB = 70°, find the measure of ∠DBA.
Answer:

Question 2.
Prove that the perpendicular from the centre of a circle to a chord, bisect the chord.
Answer:

AB is the chord of a circle with centre O and OD ⊥ AB. We have to prove that AD = DB
In ∆ ODA and ∆ ODB.
OA = OB (radii)
OD = OD (common)
∠ODA = ∠ODB = 90° (OD ⊥ AB)
∆ ODA = ∆ ODB (R H S)
AD = DB [C.P.C.T].

Question 3.
If two equal chords of a circle intersect within the circle, prove that the segments of an chord are equal to corresponding segment of the other chord.
Answer:

OL ⊥ AB and OM ⊥ CD are drawn and OP is joined.
In ∆ OLP and ∆ OMP, since equal chords are at equal distance from centres.
OL = OM (OP common)
∠OLP = ∠OMP =90°
∆ OPL ≅ ∆ OPM. (R H S)
=> PL = PM [C.P.C.T]
But, AL = CM
=> AL – PL = CM – PM
=> AP = CP
Also, AB – AP = CD – CP
∴BP = DP
Thus, corresponding segments are equals.

Question 4.
If two interesting chords of a circle make equal angel with that: diameter passing through their point of intersecting. Prove that the chords are equal.
Answer:

OM ⊥ AB and ON⊥CD
In ∆ OME and ∆ ONE

OE = OE (common)
∴ By A AS congruence
∆ OME ≅ ∆ ONE
OM = ON (C.P.C.T)
∴ AB = GD [Distance between the center to the chords equal.]
∴ Length of the chords are equal.

Question 5.
Two concentric circles of radii 13 cm and 5 cm are drawn. Find the length of the chord of the outer circle which touches the inner circle.
Answer:

In A OPB [P = 90°
OB2 = OP2 + BP2
(13)2 = (5)2 + BP2
BP2 = 165 – 25 = 144

AB = 2BP = 2 x 12 = 24cm

Question 6.
Tangents AP and AQ are drawn to circle with centre O, from an external point A. Prove that ∠PAQ = 2∠OPQ
Answer:

‘O’ is the center of a circle.
AP & AQ are tangents.
In A POQ, OP = OQ [radius]

Question 7.
Circles C1 and C2 touch internally at a point A and AB is a chord at the circle C1 intersecting C2 at P. Prove that AP = PB.
Answer:

Join OA, OP and OB.

HYP OA = HYP OB [radii]
OP = OP [common]
∆ OAP ≅ ∆ OPB [by RHS theorem]
AP = PB [C.P.C.T]

Question 8.
A circle is touching the side BC of ∆ ABC at P. AB and AC when produced are touching the circle at Q and R respectively. Prove that AQ = V2 (Perimeter of ∆ ABC)
Answer:

AQ = AR → (1) [tangents drawn from A]
BP = BQ → (2) [tangents drawn from B]
CQ = CR → (3) [tangents drawn from C]
Perimeter of ∆ ABC = AB + BC + AC
= AB + BP + PC + AC
= AB + BQ + CR + AC
[From (1), (2) & (3)]
= AQ + AR
= AQ + AQ [∴ using(1)]
Perimeter of ∆ ABC = 2 AQ
AQ = 1/2 (Perimeter of ∆ ABC)

Question 9.
A straight line drawn through the point of contact of two circles with centres A and B intersect the circles at P and Q respectively. Show that AP and BQ are parallel.
Answer:

In ∆ APC
AP = AC [radii of same circle]

Question 10.
Two circles with centres X and Y touch each other externally at P. Two diameters AB and CD are drawn one in each circle parallel to other. Prove that B, P and C are coliinear.
Answer:

∴ APD and BPC are two intersecting lines
∴ BPC is a straight line
∴ B, P, C are coliinear.

## KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Additional Questions

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Additional Questions

I. Multiple Choice Questions:

Question 1.
Which of the following is not a linear equation
(a) x + 3y = 8
(b) x = 2y
(c) x2 + 5x + 8 = 0
(d) y = 0.
Answer:
(c) x2 + 5x + 8 = 0

Question 2.
x = 1 and y = 3 is a solution of a linear equation.
(a) x – y = 4
(b) x + y = 4
(c) 2x + y = 8
(d) 2y + x = 9
Answer:
(b) x + y = 4

Question 3.
The pair of co-ordinates satisfying 2x + y = 6 is
(a) (2, 2)
(b) (2, 3)
(c) (4, 1)
(d) (5, 1)
Answer:
(a) (2, 2)

Question 4.
The pair of equations x = 0 and y = 0 represents.
(a) Parallel lines
(b) Coincident lines
(c) non-intersecting lines
(d) Perpendicular intersecting lines
Answer:
(d) Perpendicular intersecting lines

Question 5.
If the ten’s and unit’s digits of a two digit number are y and x respectively, then the number will be.
(a) 10y + x
(b) 10x + y
(c) 10xy
(d) y + x
Answer:
(a) 10y + x

Question 6.
If the line y = px – 2 passes through the point (2, 2), then the value of P is
(a) 3
(b) 2
(c) 5
(d) 4
Answer:
(b) 2

Question 7.
The coordinates of the origin are.
(a) (0, 1)
(b) (1, 0)
(c) (2, 2)
(d) (0, 0)
Answer:
(d) (0, 0)

Question 8.
The point (-3, -4) lies in the Quadrant.
(a) I
(b) II
(c) III
(d) IV
Answer:
(c) III

Question 9.
The point of intersection of the lines x – 2 – 0 and y + 6 = 0 is
(a) (0, 0)
(b) (0, -6)
(c) (0, 2)
(d) (2, -6)
Answer:
(d) (2, -6)

Question 10.
The pair of equations x = 4 and y = 3 graphically represents lines which are
(a) Parallel
(b) intersecting at (3, 4)
(c) Coincident
(d) intersecting at (4, 3)
Answer:
(d) intersecting at (4, 3)

Question 11.
Two lines are given to be parallel the equation of one of the lines is 4x + 3y = 14. The equation of the second line can be.
(a) 3x + 4y =14
(b) 8x + 6y = 28
(c) – 12x = 9y
(d) 12x + 9y = 42
Answer:
(c) – 12x = 9y

Question 12.
The pair of equations x + y = 0 and x + y = – 7 has
(a) one solution
(b) no solution
(c) two solutions
(d) infinitely many solutions
Answer:
(b) no solution

Question 13.
The pair of linear equations 3x + 5y = 3 and 6x + Ky = 8 does not have a solution if.
(a) K = 5
(b) K ≠ 10
(c) K = 10
(d) K ≠ 5.
Answer:
(c) K = 10

Question 14.
If y = 2x – 3 and y = 5 then the value of x is.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Question 15.
The sum of two numbers is 8 and their difference is 2. Find the numbers.
(a) 5 and 3
(b) 6 and 4
(c) 4 and 2
(d) 4 and 4
Answer:
(a) 5 and 3

Question 16.
The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(a) 98°, 82°
(b) 99°, 81°
(c) 118, 100°
(d) 80°, 100°
Answer:
(b) 99°, 81°

Question 17.
The value of a so that the point (3, a) lies on the line represented by 2x – 3y = 5 is
(a) 1/3
(b) – 1/3
(c) 1
(d) 2/3
Answer:
(b) – 1/3

Question 18.
If the pair of linear equations a1 x + b, y + c, = 0 and a2x + b2y + c2 = 0 has a unique solution, then.

Answer:

Question 19.
The solution of the pair of equations x – y = 0, 2x – y = 0 is
(a) x = 3, y = 5
(b) x = 3, y = 2
(c) x = 2, y = 2
(d) x = 5, y = 5
Answer:
(c) x = 2, y = 2

Question 20.
The value of x if y = 1/2x and 3x + 4y = 20 is
(a) 3
(b) 5
(c) 6
(d) 4
Answer:
(d) 4

II. Very Short Answer Questions:

Question 1.
Find the coordinate where the line x – y = 8 will intersect the y-axis.
Answer:
The given line will intersect y-axis when x = 0 ∴ 0 – y = 8
y = – 8

Question 2.
If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then find value of k.
Answer:
Since given lines are parallel

Question 3.
Write the number of solutions of the following pair of linear equations.
Answer:

∴ The given pair of linear equations has infinitely many solutions.

Question 4.
For all real values of c, the pair of equations x – 2y = 8, 5x + 10y = c have a unique solution. Justify true (or) false.
Answer:

∴ Given pair of equations have unique solution.
∴ Given statement is true.

## KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is tw ice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Answer:
Let the ages of Ani and Biju be ‘x’ years and ‘y’ years respectively.
x – y = 3 (x > y) (or) y – x = 3 (y > x) → (1)
Age of Ani’s father Dharm = 2x years
Age of Biju’s sister = $\frac{y}{2}$ years.
According to the Question
2x – $\frac{y}{2}$ = 30
4x – y = 60 → (2)
case (i) when x – y = 3
x – y= 3 → (1)
4x – y = 60 → (2)
Subtract equation (1) and (2)

x = 19 years
Put x = 19 in equation (1)
x – y = 3
19 – y = 3
– y = 3 – 19
y = 16
Ani’s age = x = 19 years.
Biju’s age = y = 16 years.
case (ii) y – x = 3
– x + y = 3 → (1)
4x – y = 60 → (2)
Adding equation (1) and (2)

x = 21
Put x = 21 in equation (1)
– 21 + y = 3
y = 3 + 21
y = 24
Ani’s age is 21 years and Biju age is 24 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
Answer:
Let the amount of their respective capitals be ₹x and ₹y
x + 100 = 2(y – 100)
x+ 100 = 2y – 200
x – 2y = – 300 → (1)
and y + 10 = 6 (x – 10)
y + 10 = 6x – 60
– 6x + y = – 70 multiply by – 2
12x – 2y = 140 → (2)
Subtract equation (1) and (2)

x = 40
Put x = 40 in equation (1)
40 – 2y = – 300
– 2y = – 300 – 40
– 2y = – 340
y = $\frac{340}{2}$
y = 170
∴ amounts of their respective capitals are ₹40 and ₹170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Answer:
Let the actual speed of the train be x km/hr and the actual time taken be ‘y’ hours.
Distance = speed x time = x × y
= xy km
xy = (x + 10) (y – 2)
xy = xy – 2x + 10y – 20
2x- 10y = – 20
xy = (x – 10) (y + 3)
xy = xy + 3x – 10y – 30
3x – 10y = 30 → (2)
Subtract equation (1) and (2)

Put x = 50 in equation (1)
2(50) – 10y = – 20
– 10y = – 20 – 100
– 10y = – 120
y = $\frac{-120}{-10}$
y = 12
distance covered by train = xy
= 50 × 12 = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer:
Let the number of students in the class be x and the number of rows be y. Then the number of students in each row $\frac{x}{y}$ .
If 3 students are extra in row, then there would be 1 row less i,e when each row
has
Students then the number of rows is (y – 1)
∴ Total number of students
= number of rows x number of students in each row

and, if 3 students are less in a row, then there would be 2 rows more i.e, when each row has $\left(\frac{x}{y}-3\right)$ students, then the number of rows is (y + 2).
∴ Total number of students = Number of rows x number of students in each row.

Then equation (1) and (2) can be written as
a – 3y = – 3 → (3)
2a – 3y = 6 → (4)
Subtract equation (3) and (4)

a = 9
Put a = 9 in equation (3)
9 – 3y = – 3
– 3y = – 3 – 9
– 3 y = – 12
x = 9 × 4
x = 36
Hence number of students in class is 36 and number of rows is 4.

Question 5.
In a ∆ ABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.
Answer:
Given

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
Answer:
5x – y = 5 → (1)
3x – y = 3 → (2)
From equation (1)
y = 5x – 5

From equation (2)
y = 3x – 3

In the graph, we observe that the coordinates of the vertices’s of the triangle ABC formed by the two lines represented, by the A(0, -3) and B(0, -5).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q
Answer:
px + qy = p – q → (1)
qx-py = p + q → (2)
Multiply equation (1) p and equation (2) by q
p2x + pqy = p2 – pq → (3)
q2x – pqy = pq + q2 → (4)
Adding equation (3) and (4)

Put x = 1 in equation (1).
p(1) + qy = p – q
qy = p – q – p
qy = – q
y = $\frac{-q}{q}$
y = – 1
Hence, the solution of the given pair of linear equation is x = 1, y = – 1

(ii) ax + by – c
bx + ay = 1 + c
Answer:
ax + by = c
bx + ay = 1 + c
then ax + by – c = 0 → (1)
bx + ay – (1 + c) = 0 → (2)
Solve by cross multiplication method.

Hence, the solution of the given pair of linear equation is

ax + by = a2 + b2
Answer:

Hence, the solution of the given pair of equations x = a, y = b.

(iv) (a – b)x + (a + b)y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2
Answer:
(a – b)x + (a + b)y = a2 – 2ab – b2 → (1)
(a + b) (x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 → (2)
Subtract equation (1) and (2),

(a – b)x – (a + b)x = a2 – 2ab – b2 – a2 – b2 ax – bx – ax – bx = – 2ab – 2b2.

Put x = a + b in equation (1)
(a – b) (a + b) + (a + b)y = a2 – 2ab – b2
a2 – b2 + (a + b)y = a2 – 2ab – b2
(a + b)y = a2 – 2ab – b2 – a2 + b2
(a + b)y = – 2ab

Hence, solution of the pair of linear equation is x = a + b,

(v) 152x – 378y = – 74
– 378x + 152y = – 604
Answer:
152x – 378y = – 74 → (1)
– 378x + 152y = – 604 → (2)
Adding equation (1) and (2)

226x – 226y = – 678 divide by – 226
x + y = 3 → (3)
Subtract equation (1) and (2)

Put x = 2 in equation (3)
2 + y = 3
y = 3 – 2
y = 1
Hence, the solution of the given pair of linear equations is x = 2, y = 1

Question 8.
ABCD is a cyclic quadrilateral (see Fig.3.7). Find the angles of the cyclic quadrilateral.

Answer:
We know that the opposite angles of a cyclic Quadrilateral are supplementary,

4y + 20 (4x) = 180°
4y – 4x = 180 – 20
4y – 4x = 160 divide by 4
y – x = 40 → (1)

3y – 5 + (- 7x + 5 ) = 180°
3y – 5 – 7x + 5 = 180°
3y – 7x = 180° → (2)
From equation (1)
y = 40 + x → (3)
Put equation (3) in equation (2)
3(40 + x) – 7x = 180°
120 + 3x – 7x = 180°
– 4x = 60°
x = $\frac{60}{-4}$
x = – 15
Put x = -15 in equation (1)
y – x = 40
y = 40 4 – x
y = 40 – 15
y = 25°

## KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Additional Questions

Students can Download Class 10 Maths Chapter 2 Triangles Additional Questions Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

## Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Additional Questions

I Multiple choice questions:

Question 1.
Two triangles are said to be similar if their corresponding sides are
a) Proportional
b) Equal
c) Paralled
d) Perpendicular
Answer:
a) Proportional

Question 2.
Congruent triangles are always,
a) Equal
b) similar
c) Not similar
d) Not equal
Answer:
a) Equal

Question 3.
Which of the following is can form similar triangles i) 3, 6, 9 ii) 9,18, 27 iii) 2,4, 6 iv) 5, 6, 7
a) i and ii
b) ii and iii
c) i and iii
d) i and iv
Answer:
a) i and ii

Question 4.
Basic proportional theorem is proposed by
a) Pythagoras
b) Thales
c) aryabhata
d) newton
Answer:
b) Thales

Question 5.
If straight line is drawn parallel to one side of a triangle then it divides the other two sides proportionally is called.
a) thales theorem
b) converse of thales theorem
c) Corollary of thales
d) Pythagoras theorem
Answer:
a) thales theorem

Question 6.
If straight-line divides two sides of a triangle proportionally then the straight line is parallel to the third side is called.
a) converse of thales theorem
b) pythogoral theorem
c) thales theorem
d) converse of Pythagoras theorem
Answer:
a) converse of thales theorem

Question 7.
In the given fig. which of following is

Answer:

Question 8.
In the given fig. which of the following Is true

Answer:

Question 9.
In the following figure DE || BC. AD = 4 cm, DB = 6 cm and AE =5 cm, then the value of EC is

a) 6.5 cm
c) 7.5 cm
b) 7 cm
d) 8.0 cm
Answer:
c) 7.5 cm

Question 10.
In the following figure DE || BC then the value of ‘x’ is

Answer:

Question 11.
In a trapezium the line joining the midpoints of non-parallel sides is.
a) Parallel to the parallel sides.
b) Half of the sum of the parallel side
c) both a and b
d) perpendicular to each other
Answer:
c) both a and b

Question 12.
If two triangles are equiangular then their corresponding sides are
a) proportional
b) equal
e) non-parallel
d) perpendicular
Answer:
a) proportional

Question 13.
If the three sides of a triangle are proportional to the corresponding three sides of another triangle then their corresponding angles are
a) equal
b) proportional
e) not equal
d) not proportional
Answer:
a) equal

Question 14.
In the given figure ∆ ADB ~ ∆ BDC then BD2 is

a) AD . AC
e) CD . AC
b) AD . CD
d) AB . BC
Answer:
b) AD . CD

Question 15.
In the given figure ∆ BDC ~ ∆ ABC then BC2 is

a) AC . CD
b) AC . AD
c) AD . CD
d) A13 . BC
Answer:
a) AC . CD

Question 16.
In the given figure ∆ ADB ~ ∆ BDC then B2 is

a) AC . AD
c) AD . CD
b) AC . CD
d) AB . BC
Answer:
a) AC . AD

Question 17.
If the area of two similar triangles are equal then they are
a) similar
b) congruent
e) proportional
d) equal
Answer:
d) equal

Question 18.
Area of similar triangles are proportional to
a) squares on their corresponding sides
b) squares on their corresponding altitude
e) squares on their corresponding medians
d) all the above
Answer:
d) all the above

Question 19.
The corresponding altitudes of two similar triangles are 9 cm and 15cm respectively then the ratio between their areas.
a) 9 : 15
b) 100 : 25
c) 81 : 225
d) 3 : 15
Answer:
c) 81 : 225

Question 20.
In a trapezium ABCD AB || CD and Its diagonals intersect at 0. if AB = 6 cm and DC = 3 cm then the ratio of the area of AOB and COD
a) 4 : 1
b) 1 : 2
c) 2 : 1
d) 1 : 4
Answer:
a) 4 : 1

II. Short Answer Questions:

Question 1.
State thales theorem (Basic proportionality theorem).
Answer:
A straight drawn parallel to one side of triangle then other two sides divides proportionally.

Question 2.
State converse of thales theorem.
Answer:
If two sides divides proportionally then straight line drawn parallel to given side.

Question 3.
What is the ratio of areas of two similar triangles whose sides are in the ratio 3 : 4
Answer:

∴ Ratio of areas are 9 : 16

Question 4.
If two triangles are similar such that the ratio of their areas 36 : 121, then what is the ratio of their corresponding medians.
Answer:

∴ ratio of their corresponding medians 6 : 11

Question 5.
Is the triangle with sides 12 cm, 16 cm and 18 cm a right triangle? give reason.
Answer:
122 + 162 = 182
144 + 256 = 324
400 ≠ 324
LHS ≠ RHS
∴ Given sides not form right angle ∆le.

Question 6.
What is the name given to the longest side of a right-angled triangle
Answer:
hypotenuse

Question 7.
State pythagoras theorem
Answer:
In a right-angled triangle square on the hypotenuse is equal to sum of the squares of other sides.

Question 8.
State converse of pythagoras theorem.
Answer:
“If the square on the longest side of a triangle is equal to the sum of the squares on the other two sides, then those two sides contain a right angle”.

Question 9.
Verify m2 – n2, 2mn, m2 + n2 form the sides of a right-angled triangle.
Answer:
(m2 + n2)2 = (m2 – n2)2 + (2mn)2
m4 + 2m2n2 + n4 = m4 – 2m2n2 + n4 + 4m2n2
m4 + n4 + 2m2n2 m – n4 + 2m2n2
LHS = RHS
given sides form right angle ∆le.

Question 10.
In figure ∠ABC = 90°, BD ⊥ AC if BD = 8 cm, AD = 4 cm, find CD

Answer:
BD2 = AD × CD
(8)2 = 4 × CD
64 = 4 CD

III. Long answer questions:

Question 1.
In Fig. $\frac{A O}{O C}=\frac{B O}{O D}=\frac{1}{2}$ and AB = 5 cm. Find the value of DC.

Answer:

Question 2.
ABC is an equilateral triangle of side 4a. Find each of its altitude.
Answer:

Let ABC be an equilateral triangle of side 4a units, We draw AD ⊥ BC. Then D is the mid-point of BC.

Now ABD is a right ∆le right angle at D.
∴ AB2 = AD2 + (2a)2. [By Pythagoras theorem]
= (4a)2 = AD2 + 4a2
= AD2 = 16a2 – 4a2 = 12a2

Question 3.
If the diagonals of a quadrilateral divide each other proportionally prove that is a trapezium?
Answer:

Question 4.
ABC is a triangle in which AB = AC and D is a point on AC such that BC2 = AC × CD Prove that BD = BC.
Answer:

Given: ∆ ABC in which AB = AC and D is a point on the side AC such that BC2 = AC x CD.
To prove: BD = BC
Construction: Join BD
Proof: We have,

Question 5.
A door of width 6 meter has an arc above it having a height of 2 meters find the radius of the arc?

Answer:

Let the radius of the semi – circle
OA = OC = OB
= (x + 2)m
=> OA = OB. In ∆ OAB
∆ OAB is an isosceles ∆le
∴ OD ⊥ AB.
=> OD bisects AB
∴ AD = BD = 3 cm.
In ∆le OAD, ∠PDA = 90°
AO2 = AD2 + OD2
(x+2)2 = 32 + x2
x2 + 4x + 4 = 9 + x2
4x = 9 – 4 = 5
x = $\frac{5}{4}$
Hence, radius = x + 2

Question 6.
In an equilateral ∆le ABC, AD is drawn ⊥r to BC meeting BC in D, prove that AD2 = 3BD2.
Answer:
In ∆ ABD, ∠ADB = 90°
Then by pythagoras theorem
=> AB2 = AD2 + BD2
=> BC2 = AD2 + BD2, [BC = 2BD]
∴ AB = BC = CA
=> (2BD)2 = AD2 + BD2
∴ (⊥ is the median in an equilateral ∆le)
3BD2 = AD2.

Question 7.
In the given figure, AD ⊥ BC and BD = $\frac{1}{3}$ CD prove that 2AC2 = 2AB2 + BC2.
Answer:
In right ∆ ABD, by pythogoras theorem.
∴ AB2 = AD2 + BD2
In right ∆ ABC, by pythagoras theorem.
∴ AC2 = AD2 + DC2
Subtracting (ii) from (i) we get,
∴ AB2 – AC2 = BD2 – DC2

Question 8.
Triangle ABC is right angles at B and D is the mid-point of BC? Prove that AC2 = 4AD2 – 3AB2.

Answer:
Given: ABC is right angled at B and D is the mid-point of B C
BD = DC = $\frac{1}{2}$ BC
In ∆ ABD, AD2 = AB2 + BD2
[pythogoras theorem] – (1)
In ∆ ABC, AC2 = AB2 + BC2
[pythogoras theorem] – (2)
From eqn (1) and (2)

=> 4AD2 = 4AB2 + BC2
=> BC2 = 4AD2 – 4AB2 – (3)
Using this eqn in (2)
AC2 = AB2 + 4AD2 – 4AB2
AC2 = 4AD2 – 3AB2.

Question 9.
Through the midpoint M of the sides of parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced to E. Prove that EL = 2BL.
Answer:

Question 10.
Prove that any two medians of a ∆le divide each other in the ratio 2:1.
Answer:

In ∆le ABC, BD, CE are two medians intersecting at G.
∴ AE = EB, AD = DC [medians divides equally]

Question 11.
A ladder of length 2.6 m is lended against a wall. When it is at a distance of 2.4 m from the foot of the wall, the top of the ladder touches the bottom edge of the window in the wall, if the foot of the ladder is moved 1.4 m towards the wall it touches the top edge of the window, find the height of the window.
Answer:

Length of ladder AC = DE = 2.6 m
Height of window AD =?
BC = 2.4 m
EC = 1.4 m
BE = 1 m
In ∆le ABC
AC2 = AB2 + BC2
AB2 = AC2 – BC2
AB2 = (2.6)2 – (2.4)2
= (2.6 + 2.4) x (2.6 – 2.4)
= 5 x 0.2 = lm
AB = 1m
In ∆le DBE
DE2 = BD2 + BE2
(2.6)2 – 12 = BD2
BD2 = (2.6 + 1) (2.6 – 1)
= 3.6 x 1.6
BD =
BD = 2.4 m
∴ Height of window
AD = BD – AB
= 2.4 – 1 = 1.4
Height of window = 1.4m

Question 12.
In a ∆le ABC, D and E be two points on side AB such that AD = BE : IF DE || BC and EQ || AC, then prove that PQ || AB.
Answer:

From fig, given BE = AD
EA = AD + DE
= BE + ED
= BD
Then (2) becomes

∴ PQ || AB [converse of BPT]

Question 13.
A tree 32 m tall broke due to gale and Its top fell at a distance of 16m from its foot at what height above the ground did the tree break?
Answer:

Height of the tree AB = 32 m
AD is broken part of tree = x
BC is distance between foot to the top oftreeBC = 16m
BD is the remaining height of tree = 32 – x
In Ale BDC ∠B = 90°
(CD)2 = BD2 + BC2
x2 = (32 -x)2 + (16)2
x2 = 1024 + x2 – 64x + 256
x2 – x2 = 1280 – 64x
64x = 1280
x = $\frac{1280}{64}$
x = 20m
Remaining part of a tree
BD = (32 – x) = 32 – 20 = 12m

Question 14.
In ∆le ABC, i) ∠ACB = 60°, ii) AD ⊥ar BC. Derive an expression for AB in terms of AC and BC.
Answer:

In ∆le ABC
i) ∠ACB = 60° and
ii) AD ⊥r BC
Construction:
Draw DE such that ∠EDC = 60°
∴ Ale DEC is an equilateral
(Data and construction)
DE = DC = EC
Ale DEA is an isosceles Ale [∴ AED = 120 and ∠ ADE = 30° = ∠ DAE]
DE = EA
=> DC = EA = EC [∴ Axiom – 1]
∴ AC = 2DC
Now in Ale ADB, AB2 = AD2 + BD2 [pythagoras theorem]
AB2 = (AC2 – DC2) + (BC -CD)2
AB2 = 4DC2 – DC2 + BC2 – 2BC DC + DC2 [∴ AC = 2DC]
AB2 = 4DC2 + BC2 – 2BC . DC
AB2 = 4DC2 – 2BC . DC + BC2
AB2 = 2DC (2DC – BC) + BC2
= AC (AC – BC) + BC2 [2DC = AC]
AB2 = AC2 – AC . BC + BC2
AB2 = AC2 + BC2 – AC.BC

Question 15.
ABCD is a rhombus. Prove that AC2 + BD2 = 4AB2
Answer:

Given: In rhombus ABCD, AC and BD are diagonals, AB = BC = CD = DA
∴ AO = 1/2 AC and BO = 1/2 BD
To prove: AC2 + BD2 = 4AB2
In right angle ∆le AOB, ∠O = 90°
AB2 = OA2 + OB2
AB2 = (1/2 AC)2 + (1/2 BD)2

4AB2 = AC2 + BD2

Question 16.
AD is the altitude from A to BC in the ∆ ABC and DB:CD = 3:1 prove that BC2 = 2(AB2 – AC2)
Answer:

In ∆le ABD ∠D = 90°
AB2 = AD2 + BD2
AD2 = AB2 – BD2 – (1)
In ∆le ADC ∠D = 90°
AC2 = AD2 + CD2
AD2 = AC2 – CD2 – (2)
From (1) and (2)
AB2 – BD2 = AC2 – CD2
AB2 – AC2 = BD2 – CD2

BC2 = 2(AB2 – AC2)

Question 17.
In the given figure in ∆ ABC, AD ⊥ BC and AD2 = BD × CD. Prove that the ∆ ABC is a right angle triangle.
Answer:

∆ ABC, AD ⊥ BC and AD2 = BD × CD
To prove: ∆ ABC is a right angle triangle
Proof: In ∆le ABD ∠D = 90°
AB2 = BD2 + AD2 – (1)
In ∆le ADC ∠D = 90°
AC2 = AD2 + DC2 – (2)
Adding eqn (1) and (2)
AB2 + AC2 = BD2 + AD2 + AD2 + DC2
= 2 AD2 + BD2 + DC2
= 2(BD x CD) + BD2 + DC2
= (BD + DC)2
∴ AB2 + AC2 = BC2
∆ ABC is a right triangle, right angled at A.

Question 18.
In trapezium ABCD, AB || CD and DC = 2AB. EF || AB, where E and F lie on BC and AD respectively such $\frac{B E}{E C}=\frac{4}{3}$ . Diagonal BD intersect EF at G. Prove that 7EF = 11 AB.
Answer:
In trapezium ABCD, AB || CD and

Question 19.
A peacock on a pillar of 9 feet height on seeing a snake coming towards is hole situated just below the pillar from a distance 27 feet away from the pillar will fly to catch it. If both possess the same speed, how far from the pillar they are going to meet?
Answer:

AB is the height of pillar = 9ft
BC is the distance b/w pillar and the snake hole = 27 ft
AD = CD = x ft
BD = 27 – x
In ∆ ABD ∠D = 90°
AD2 = AB2 + BD2 [By pythagoras theorem]
(x)2 = (9)2 + (27 – x)2
x2 = 81 + 729 + x2 – 54x
54x = 810
x = $\frac{810}{54}$
x = 15 ft
BD = 27 – x = 27 – 15 = 12 ft
∴ Both of these going to meet at a distance. 12ft from the pillar.

Question 20.
In ∆ MGN, MP ⊥ GN. If MG = a units, MN = b units, GP = c units and PN = d unit. Prove that (a + b) (a – b) = (c + d) (c – d).
Answer:

In ∆le MGP ∠P = 90°
MG2 = GP2 + MP2
(a)2 = (c)2 + MP2
MP2 = a2 – c2 – (1)
In ∆le MNP ∠P = 90°
MN2 = PM2 + PN2
(b)2 = d2 + MP2
MP2 = b2 – d2 – (2)
From (1) and (2)
a2 – c2 = b2 – d2
a2 – b2 = c2 – d2
(a + b) (a – b) = (c + d) (c – d).

## KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Students can Download Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 1.
Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).
Answer:
Let P(x1, y1) be common point of both lines and divide the line segment joining A(2, – 2) and B (3, 7) in ratio K : 1

Question 2.
Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
Answer:
Given point are A(x, y), B(1, 2) & C(7, 0) These points will be collinear if the area of the triangle formed by them is zero.
Area of ∆ ADC

0 = $\frac{1}{2}$[x(2 – 0)+1(0 – y)+7(y – 2)]
0 = x × 2 + 1 × – y + 7y – 14
0 = 2x – y + 7y – 14
2x + 6y – 14 = 0 divide by 2
x + 3y – 7 = 0
It is the relation between x & y

Question 3.
Find the centre of a circle passing through the points (6, – 6), (3, – 7) and (3, 3).
Answer:
Let A → (6,- 6), B → (3,- 7) and C → (3, 3)

Squaring
(x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2
x2 – 12x + 36 +y2 +12y + 36
= x2 – 6x + 9 + y2 +14y + 49
x2 – x2 + y2 – y2 – 12x +6x + 12y – 14y + 72 – 58 = 0
– 6x – 2y = – 14 devide by – 2
3x + y = 7 → (1)
(x – 3)2 +(y + 7)2 = (x – 3)2 +(y – 3)2 (y + 7)2 = (y – 3)2
y2 +49 + 14y = y2 + 9 – 6y
y2 – y2 + 14y + 6y = 9 – 49
20y = – 40
y = – 2
Putting y = – 2 in eqn (1) 3x + y = 7
3x – 2 = 7
3x = 7 + 2
3x = 9
x = $\frac{9}{3}$ = 3
x = 3
Thus I(x, y) → (3, – 2)
Hence, the centre of a circle is (3, – 2)

Question 4.
The two opposite vertices of a square are (- 1, 2) and (3, 2). Find the coordinates of the other two vertices.
Answer:
Let ABCD is a square where two opposite vertices are A(- 1, 2) & C(3,2)

AB = BC = CD = AD [ABCD is a square]
AB = BC

Squaring both side
(x +1 )2 + (y – 2)2 = (3 – x )2 + (2 – y )2
x2 + 2x + 1 + y2 – 4y + 4
= 9 + x2 – 6x + 4 + y2 – 4y
2x + 6x – 4y + 4y = 13 – 5
8x = 8
x = $\frac{8}{8}$ = 1
x = 1
In ∆ ABC [B_ =90°

x2 + 1 + 2x + y2 + 4 – 4x + 9 + x2
– 6x + 4 + y2 – 4y = (4)2
2x2 – 4x + 2y2 – 8y + 18 = 16
2x(1)2 – 4(1) + 2y2 – 8y + 18 – 16 = 0
2 – 4 + 2y2 – 8y + 2 = 0
2y2 – 8y = 0
2y(y – 4) = 0
2y = 0 & y – 4 = 0
y = 0 & y = 4
Hence, the other two vertices are (x, y) &(x1, y1) are (1, 0) & (1, 4)

Question 5.
The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Gulmohar are planted on the, boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the Fig.7.4 The students are to sow seeds of flowering plants on the remaining area of the plot.

(i) Taking A as origin,find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆ PQR if C is the origin? Also calculate the areas of the triangles in these cases. What do you observe?
Answer:
(i) Taking A as origin then AD is x – axis and AB is y – axis.
Co-ordinates of P, Q and R are
P → (4, 6), Q → (3, 2) & R → (6, 5)
area of ∆ PQR

area of ∆ PQR
= $\frac{1}{2}$[4(2 – 5) + 3(5 – 6) + (6 – 2)]
= $\frac{1}{2}$[4 × – 3 + 3 × – 1 + 6 × 4]
= $\frac{1}{2}$[- 12 – 3 + 24]
area of ∆ PQR
= $\frac{1}{2}$[- 15 + 24] = 9/2 sq units

(ii) Taking C as origin then CB is x – axis & CD is y – axis.
P(x1, y1) = (12, 2), Q(X2, y2) = (13, 6), and R(x3, y3) = (10, 3),
x1 = 12, y1 = 2, x2 = 13, y2 = 6, x3 = 10 & y3 = 3
area of ∆ PQR
= $\frac{1}{2}$[12(6 – 3) + 13(3 – 2) + 10(2 – 6)]
= $\frac{1}{2}$[12 x 3 + 13 x 1 + 10 x – 4]
= $\frac{1}{2}$[36 + 13 – 40]
= $\frac{1}{2}$[49 – 40] = 9/2 Sq units
Hence we observed that area of ∆ remains same in both case.

Question 6.
The vertices of a ∆ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/ AC = 1/4 Calculate the area of the ∆ ADE and compare it with the area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6).
Answer:

D and E divide AB and AC respectively in the ration 1 : 3
Section Formula

area of ∆ ABC
= $\frac{1}{2}$[4(5 – 2) + 1(2 – 6) + 7(6 – 5)]
= $\frac{1}{2}$[4 × 3 + 1 × – 4 + 7 × 1]
= $\frac{1}{2}$[12 – 4 + 7]
= $\frac{1}{2}$ × [19 – 4]
= $\frac{15}{2}$sq units

area of ADE : area of ABC =1:16

Question 7.
Let A (4,2), B(6, 5) and C(1, 4) be the vertices of ∆ ABC.
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2:1.
(iv) What do yo observe?
[Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2:1.]
(v) If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of ∆ ABC, find the coordinates of the centroid of the triangle.
Answer:
(i)

median AD of the triangle ABC divide the side BC into two equal parts.
Therefore D is the mid – point of side BC Co-ordinates of mid Point

(ii) Form equation AP : PD = 2 : 1

Section Formula

(iii) BQ : QE = 2 : 1

co-ordinates of
$\frac{A D}{A B}=\frac{A E}{E C}=\frac{1}{4}$

(iv) The co-ordinates P, Q, & R are same (11/3, 11/3) All these points represents the same point which is called centriod.

(v) Point 0 is the centroid and AD is the midian. D is the mid point of BC and point 0 divide the AD into 2 : 1

Question 8.
ABCD is a rectangle formed by the points A(- 1, – 1), B(- 1, 4), C(5, 4) and D(5, – 1). P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Answer:
Given: A → (- 1, – 1), B → (- 1, 4), C → (5, 4) & D → (5, – 1) Co-ordinates of midpoint

∴ PQ = QR = RS = SP
∴ PQRS is either square (or) Rhomrus
Diagonal PR

and Diagonal QS

PR ≠ QS
∴ Diagonal are not equal
Therefore, PQRS is a rhombus.

## KSEEB Solutions for Class 10 Maths Chapter 8 Real Numbers Additional Questions

Students can Download Class 10 Maths Chapter 8 Real Numbers Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 8 Real Numbers Additional Questions

I. Multiple Choice Questions:

Question 1.
The H.C.F of 2 and 3 by Euclid’s division lemma
a. 2
b. 3
c. 5
d. 1
Answer:
d. 1

Question 2.
The H.C.F of 10 and 5 is
a. 5
b. 10
c. 15
d. 150
Answer:
a. 5

Question 3.
The relation of Euclid’s division lemma is
a. a = bq + r
b. a = b + q × r
c. a = bqr
d. a = (b – q) r
Answer:
a. a = bq + r

Question 4.
Which of the fololwing is a rational number.
a. $\sqrt{2}$
b. π
c. $\frac{1}{3}$
d. $\sqrt{3}$
Answer:
c. $\frac{1}{3}$

Question 5.
What are the quotient and remainder when 10 is divided by 3
a. 3, 1
b. 1, 3
c. 1, 1
d. 3, 3
Answer:
a. 3, 1

Question 6.
What is the H.C.F of two consecutive natural numbers?
a. 1
b. 2
c. 3
d. 4
Answer:
a. 1

Question 7.
The prime factors of 256 is
a. 26
b. 27
c. 28
d. 29
Answer:
c. 28

Question 8.
If a and b are two positive integers then H.C.F (a.b) × L.C.M (a, b) =
a. a + b
b. a × b
c. a2 + b2
d. 2ab
Answer:
b. a × b

Question 9.
$\sqrt{2}$ is
a. a rational number
b. an irrational number
c. an integer
d. a natural number
Answer:
b. an irrational number

Question 10.
5 – $\sqrt{3}$ is
a. an irrational number
b. a rational number
c. a composite number
d. a prime number
Answer:
a. an irrational number

Question 11.
Which of the following is not an irrational number?
a. $\sqrt{2}$
b. $\sqrt{3}$
c. $\sqrt{9}$
d. $\sqrt{5}$
Answer:
c. $\sqrt{9}$

Question 12.
The number of prime number between 1 and 10 is
a. 2
b. 3
c. 4
d. 12
Answer:
c. 4

Question 13.
π is
a. an irrational number
b. a rational number
c. a prime number
d. a composite number
Answer:
a. an irrational number

Question 14.
H.C.F (8, 9, 25) × L.C.M (8, 9, 25)
a. 1800
b. 2.25
c. 2.375
d. 3600
Answer:
a. 1800

Question 15.
The decimal expansion of $\frac{17}{8}$ is
a. 2.1
b. 2.25
c. 2.375
d. 2.0125
Answer:
a. 2.1

II. Short Answer Questions:

Question 1.
What is the use of Euclid’s division lemma?
Answer:
Euclid’s division lemma can be used only for calculating H.C.F.

Question 2.
Write the even prime number.
Answer:
2.

Question 3.
(7 × 11 × 13 + 13) is a composite number. Justify the statement.
Answer:
(7 × 11 × 13) + 13
= 13(77 + 1) =13 × 78.
∴ 78 is composite number.
∴ It is a composite number.

Question 4.
What is the HCF of 33× 5 and 32 × 52
Answer:
HCF of 33 × 5 and 32 × 52
= 32 × 5 = 45

Question 5.
Find the value of ‘a’ in the factor tree.

Answer:
2 × 5 = 10
10 × 2 = 20
20 × 5 = 100
∴ a= 100

Question 6.
Given that L.C.M (91, 26) = 182, then Find H.C.F (91,26)
Answer:

Question 7.
What the two numbers called if their HCF is 1.
Answer:
Co-Prime numbers

Question 8.
If P is prime number then, what is the LCM of P, P2 and P5.
Answer:
P5.

Question 9.
Define twin prime numbers.
Answer:
A pair of prime numbers is said to be twin primes it they differ by 2. Eg: (3,5) & (11,13)

Question 10.
Write the prime factors of 210.
Answer:

210 = 2 × 3 × 5 × 7

III. Long Answer Questions:

Question 1.
If n is odd positive integer. Show that (n5 – 1) divisible by 8?
Answer:
We know that an odd positive integer n is of the form (4q + 1) or (4q + 3) for some integer q.
Case – I
when n = (4q + 1)
In case n2 – 1 = (4q + 1)2 – 1
= 162 q2 + 8q = 8q(2q + 1)
which is clearly divisible by 8.

Case – II
When n = (4q + 3)
In this case, we have.
n2 – 1 = (4q + 3)2 – 1
= 16 q2 + 24q + 8(2q2 + 3q + 1)
which is clearly divisible by 8.
Hence (n2 – 1) is divisible by 8.

Question 2.
Using prime factorisation method, find the HCF and LCM of 30, 72 and 432. Also show that HCF × LCM ≠ product of three no. ?
Answer:
Given Numbers = 30, 72, 432.
30 = 2 × 3 × 5; 72 = 23 × 32 and 432 = 24 × 32
Here 21 and 31 are the smallest powers of the common factors 2 and 3 respectively.
So, HCF (30, 72, 432) = 21 × 31
= 2 × 3 = 6.
Again 24, 33 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively.
So, LCM (30, 72, 432)
= 24 × 33 × 51 = 2160
HCF × LCM = 6 × 2160 = 12960
Product of numbers = 30 × 72 × 432 = 933120
∴ HCF × LCM ≠ product of the number.

Question 3.
Find the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.
Answer:
It is given on dividing 398 by the required number, that is a remainder of 7. This means that 398 – 7 = 391 is exactly divisible by the required number. In other words, required number is a factor of 391.

Similarly, required positive integer is a factor of (436 – 11= 425) and (542 – 15 = 527). Clearly, required number is the HCF of 391, 425 and 527.

Using the factors tree, we get the prime factorisations of 391, 425 and 527 as follows. 391 = 17 × 23, 425 = 52 × 17 and 527 = 17 × 31
∴ HCF of 391, 425 and 527 is 17
Hence, required number = 17

Question 4.
There are 75 roses and 45 lily flowers. These are to be made into bouquets containing both the flowers. All the bouquets should contain the same number of flowers. Find the number of bouquets that can be formed and the number of flowers in them?
Answer:
The number of bouquets = HCF of (75, 45). Now by Euclid’s algorithm from 75 and 45

HCF of 75 and 45 is 15
Each bouquets contains = 15 flowers
∴ Number of roses in bouquets = $\frac{75}{15}$ = 5
∴ Number of lilly’s in bouquets = $\frac{45}{15}$ = 3
∴ Number of bouquets = 8

Question 5.
In a school, the strength in 8th, 9th and 10th standards are respectively 48, 42 and 60. Find the least number of books required to be distributed equally among the students of 8th, 9th and 10th standard?
Answer:
The least number of books required.
= LCM(48, 42 and 60)
Now 48 = 2 × 2 × 2 × 2 × 3 =24 × 3
42 = 2 × 3 × 7
60 = 2 × 2 × 3 × 5
Hence LCM (48, 42 and 60)
= 24 × 3 × 7 × 5 = 1680.
∴ Number of books required is 1680.

Question 6.
Show that any positive odd integers is of the form 4q + 1 or 4q + 3, where q is some integer.
Answer:
By Euclid’s division algorithm,
a + bq + r
Take b = 4 ∴ r = 0, 1, 2, 3
So, a = 4q, 4q + 1, 4q + 2, 4q + 3
Clearly a = 4q ; 4q + 2 are even, as they are divisible by 2.
But 4q + 1, 4q + 3 are odd, as they are not divisible by 2.
∴ Any positive odd integer is of the form (4q + 1) or (4q + 3)

Question 7.
x, y and z start at the same time in the same direction to run around a circular stadium x completes a round in 126 sec y in 154 sec and z in 231 sec all starting at the same point. After what time will they meet again at the starting point. How many rounds would have x, y and z completed by this time?
Answer:
To find LCM(126, 154, 231)
126 = 2 × 3 × 3 × 7
154 = 2 × 7 × 11
231 = 3 × 7 × 11
LCM of 126, 154, 231.
= 2 × 3 × 3 × 7 × 11
= 1386,
Hence, after 1386 sec they meet again the starting point.
Now x completed the rounds = $\frac{1386}{126}$
= 11 rounds
Y completed the rounds = $\frac{1386}{154}$
= 9 rounds
z completed the rounds = $\frac{1386}{231}$
= 6 rounds

Question 8.
Show that any positive even integers of the form 4q or 4q + 2, where is q is a whole number.
Ans:
(a) Let ‘a’ be an even positive integer,
a = 4q + r
a = 4q + r, where 0 ≤ r ≤ 4
a = 2 (2q) + r where r = 0, 1, 2, 3
Since a is an even positive integer, 2
divides ‘a’ ∴ r = 0
a = 2(2q) + 0
a = 4q.

(b) Let ‘a’ be an in even positive integer
Apply division algorithm with ‘a’ and ‘b’ where b = 4.
a = 4q + r where 0 ≤ r < 4
a = 4q + 2 = 2(2q) + 2
= 2(2q + 1).
Since ‘a’ is an even positive integer, 2 divides ‘a’
r = 2.
a = 2(2q + 1)
a = 4q + 2
Hence a = 4q + 2, when ‘a’ is an even positive integer.

Question 9.
The HCF of 65 and 117 is expressible in the form 65m – 117, find the value of m. Also find the LCM of 65 and 117 using prime factorisation method?
Answer:
We have 117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
Hence HCF (65, 117) = 13
∴ 65m – 117 = HCF = 13
⇒ 65m = 117 + 13 = 130.
m = $\frac{130}{65}$ = 2
Now = 65 = 13 × 5
117 = 32 × 13
LCM (65, 117) = 113 × 5 × 32
= 585.

Question 10.
The length and breadth of a rectangular field is 110m and 30m respectively. Calculate the length of the longest rod which can measure the length and breadth of field exactly.
Answer:
Length = 110 m; Breadth = 30 m.
We have to find HCF of 110 and 30 by Euclid’s Lemma.
110 = (30 × 3) + 20
30 = (20 × 1) + 10; 20 = (10 × 2) + 0
HCF of 110 and 30 is 10.
∴ Length of longest rod is 10 m.

Question 11.
Prove that $\frac{\sqrt{7}}{5}$ is an irrational number.
Answer:

But $\sqrt{7}$ is not a rational number. This leads us to a contradiction.
Our as supposition that $\frac{\sqrt{7}}{4}$ number is wrong.
$\frac{\sqrt{7}}{4}$ is a rational number.

## KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Additional Questions

Students can Download Class 10 Maths Chapter 9 Polynomials Additional Questions, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Additional Questions

I. Multiple Choice Questions:

Question 1.
The degree of polynomial is x + 2
a. 2
b. 1
c. 3
d. 4
Answer:
b. 1

Question 2.
The degree of a quadratic polynomial is
a. 0
b. 1
c. 2
d. 3
Answer:
c. 2

Question 3.
The degree polynomial of a bi-quadratic
a. 0
b. 2
c. 4
d. 1
Answer:
c. 4

Question 4.
The standard polynomial. form of a linear
a. ax + c
b. ax2 + c
c. ax2 + bx + c = 0
d. ax3 + bx2 + cx + d = 0
Answer:
a. ax + c

Question 5.
The standard form of a quadratic equation.
a. ax2 + bx + c = 0
b. ax + c = 0
c. ax2 + c = 0
d. ax3 + bx + c = 0
Answer:
a. ax2 + bx + c = 0

Question 6.
A polynomial of degree 3 is called
a. a linear polynomial
b. a quadratic polynomial
c. a cubic polynomial
d. a biquadratic polynomial
Answer:
c. a cubic polynomial

Question 7.
The value of px. = x2 – 3x – 4, x = -1
a. 1
b. – 4
c. 0
d. – 3
Answer:
c. 0

Question 8.
The degree of the polynomial x4 + x3 is
a. 2
b. 3
c. 5
d. 4
Answer:
d. 4

Question 9.
The number of zeroes of linear polynomial at most is
a. 0
b. 1
c. 2
d. 3
Answer:
b. 1

Question 10.
The degree of polynomial x + 2. x + 1. is
a. 1
b. 3
c. 4
d. 2
Answer:
d. 2

Question 11.
The degree of a zero polynomial is
a. not defined
b. 1
c. 2
d. 3
Answer:
a. not defined

Question 12.
The zeroes of the polynomial x3 -4x are
a. 0, ± 2
b. 0, ± 1
c. 0, ± 3
d. 0, 0
Answer:
a. 0, ± 2

Question 13.
The zeroes of the polynomials, t2 – 15 are.
a. ± $\sqrt{15}$
b. ± $\sqrt{5}$
c. ± $\sqrt{3}$
d. ± 3
Answer:
a. ± $\sqrt{15}$

Question 14.
Dividend is equal to.
a. divisor × quotient + remainder
b. divisor × quotient
c. divisor × quotient – remainder
d. divisor × quotient × remainder
Answer:
a. divisor × quotient + remainder

Question 15.
If the divisor is x2 and quotient is x while the number remainder is 1, then the dividend is.
a. x2
b. x
c. x3
d. x3 + 1
Answer:
d. x3 + 1

Question 16.
What is the co-efficient of the first term of the quotient when 2x2 + 2x + 1 is divided by x + 2?
a. 1
b. 2
c. 3
d. – 2
Answer:
b. 2

Question 17.
The zeroes of the polynomial x2 – 3x – 4 are
a. 4, – 1
b. 4, 1
c. – 4, 1
d.- 4, – 1
Answer:
a. 4, – 1

Question 18.
If f(x) = x2 – 1 the value of f(2) is
a. 1
b. 3
c. 4
d. 0
Answer:
b. 3

Question 19.
If f(x) = 8 f(x) is called
a. Constant polynomials
b. linear polynomials
c. quadratic polynomials
d. Cubic polynomials
Answer:
a. Constant polynomials

Question 20.
A cubic polynomial has at most
a. 1 zeroes
b. 2 zeroes
c. 3 zeroes
d. 4 zeroes.
Answer:
c. 3 zeroes

II. Short Answer Questions:

Question 1.
Find the value of P(x) = x2 + 2x – 5 at x = 1
Answer:
P(x) = x2 + 2x – 5
P (1) = (1)2 + 2(1) – 5 = 1 + 2 – 5
= 3 – 5
P(1) = – 2

Question 2.
If x = 1 is a zero of the Polynomial f(x) = x3 – 2x2 + 4x + K, write the value of K.
Answer:
f(x) = x3 – 2x2 + 4x + K x = 1
f(1) = (1)3 – 2(1)2 + 4(1) + K
f(1) = 0 zero of the polynomial
0 = 1 – 2 + 4 + K
0 = 3 + K
K = – 3

Question 3.
If α and β are zeroes of Polynomials P(x) = x2 – 5x + 6, then find the value of α + β – 3αβ.
Answer:
α + β = $\frac{-b}{a}=\frac{-(-5)}{1}$ = 5 and
αβ = $\frac{c}{a}=\frac{6}{1}$ = 6
α + β – 3αβ
= 5 – 3(6)
= 5 – 18 = -13
α + β – 3αβ = – 13

Question 4.
Find the zeroes of the Polynomial P(x) = 4x2 – 12x + 9
Answer:

P(x) = 4x2 – 12x + 9
4x2 – 6x – 6x + 9 = 0
2x(2x – 3) – 3(2x – 3) = 0
(2x – 3) (2x – 3) = 0
x = $\frac{3}{2}$, $\frac{3}{2}$

Question 5.
If 1 is a zero of the Polynomial P(x) = ax2 – 3(a – 1) x – 1, then find the value of a.
Answer:
P(x) = ax2 – 3(a – 1) x – 1
∴ x = 1
P(1) = a(1)2 – 3(a – 1) 1 – 1
P(1) = a – 3a + 3 – 1
0 = – 2a + 2
2a = 2
a = $\frac{2}{2}$ = 1
a = 1

Question 6.
Write the degree of the Polynomial 3x3 – x4 + 5x + 3
Answer:
3x3 – x4 + 5x + 3
– x4 + 3x3 + 5x + 3
∴ degree of the Polynomial is 4

Question 7.
Write the standard form of a cubic Polynomial.
Answer:
ax3 + bx3 + cx + d.

Question 8.
Find the zero of the Polynomial , P(x) = a2x, a ≠ 0
Answer:
P(x) = a2x
a2x = 0
x = $\frac{0}{a^{2}}$
x = 0

Question 9.
If α and β are the zeroes of the polynomial 4x2 + 3x + 7, then find the value of $\frac{1}{\alpha}+\frac{1}{\beta}$
Answer:

III. Long Answer Questions:

Question 1.
Find the zeroes of the following quadratic polynomials x2 + 4x + 4.
Answer:
f(x) = x2 + 4x + 4
0 = x2 + 2x + 2x + 4
0 = x(x + 2) + 2(x + 2)
(x + 2) (or) (x + 2) = 0
x + 2 = 0 (or) x + 2 = 0
x = – 2 x = – 2.
Zeroes of the Polynomials x2 + 4x + 4 are – 2 and – 2

Question 2.
If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + K, find the value of K.
Answer:
f(x) = x3 – 2x2 + 4x + K
∴ x = 1
f(x) = (1)3 – 2(1)2 + 4(1) + K
0 = 1 – 2(1) + 4 + K [f(1) = 0]
0 = 1 – 2 + 4 + K.
K + 3 = 0
K = – 3

Question 3.
For what value of K, – 4 is a zero of the polynomial x2 – x – (2K + 2)?
Answer:
f(x) = x2 – x – (2K + 2) and f(x) = 0
x = – 4
f(- 4) = (- 4)2 (- 4) – (2K + 2)
0 = 16 + 4 – 2K – 2
20 – 2 – 2K = 0 ⇒ 18 = 2K
K = $\frac{18}{2}$ ⇒ K = 9

Question 4.
Divide p(x) by g(x) in each of the following cases and verify division algorithm.
P(x) = x2 + 4x + 4, g(x) = x + 2
Answer:
x2 + 4x + 4 ÷ x + 2

q(x) = x + 2 and r(x) = 0

Question 5.
On dividing the polynomial. P(x) = x3 – 3x2 + x + 2 by a polynomial g(x) the quotient and remainder were (x – 2) and (- 2x + 4) respectively. Find g(x).
Answer:
P(x) = x3 – 3x2 + x + 2
q(x) = (x – 2); r(x) = (- 2x + 4)
g(x) = ?

g(x) = x2 – x + 1; r(x) = 0
∴ The divisor is x2 – x + 1

Question 6.
What must be subtracted from x3 + 5x2 + 5x + 8 so that the resulting polynomial is exactly divisible by x2 + 3x – 2?
Answer:
P(x) = x3 + 5x2 + 5x + 8
g(x) = x2 + 3x – 2
P(x) = g(x) q(x) + r(x)

q(x) = x + 2; r(x) = x + 12
If we subtract x + 12 from x3 + 5x2 + 5x + 8 it will be exactly divisible by x2 + 3x – 2

Question 7.
What should be added to (x4 – 1) so that it’ is exactly divisible by (x2 + 2x + 1)?
Answer:
P(x) = x4 – 1 g(x) = x2 + 2x + 1
x4 + 0x3 + 0x3 + 0x – 1

r(x) = – 4x – 4
– {r(x)} = {4x + 4}

Question 8.
If the polynomials 2x3 + ax2 + 3x – 5 and x3 + x2 – 4x – a leave the same remainder when divided by x – 1 find the value of a.
Answer:
x – 1 = 0 [∴ g(x) = (0)]
x = 1
P(x) = 2x3 + ax2 + 3x – 5
P(1) = 2(1)3 + a(1)3 + 3(1) – 5
P(1) = 2 + a + 3 – 5
P(1) = a ➝ (1)
g(x) = x3 + x2 – 4x – a
g(1) = (1)3 + (1)2 – 4(1) – a
= 1 + 1 – 4 – a
g(x) = – 2 – a ➝ (2)
g(1) = g(1) [∴ Polynomials are leave same remainder]
a = – 2 – a
a + a = – 2
2a = – 2
a = $\frac{-2}{2}$
a = – 1

Question 9.
If (x3 + ax2 – bx + 10) is divisible by x3 – 3x + 2, find the values of a and b.
Answer:
P(x) = x3 + ax2 – bx + 10
g(x) = x2 – 3x + 2
x2 – 2x – x + 2 = 0
x(x – 2) – 1(x – 2) = 0
(x – 2) (x – 1) = 0
x = 2 and x = 1
P(x) = x3 + ax2 – bx + 10
P(x) = (1)3 + a(1)2 – b(1) + 10
= 1 + a – b + 10
0 = a – b + 11
a – b = – 11 ➝ (1)
P(2) = (2)3 + a(2)2 – b(2) + 10
P(2) = 8 + 4a – 2b – 10
0 = 4a – 2b + 18
4a – 2b = – 18 ➝ (2)
Multiply equation (1) by 4 and subtract with equation (2)

b = 13
Put b = 13 in equation (1)
a – b = – 11
a – 13 = – 11
a = – 11 + 13
a = 2

Question 10.
If both x – 2 and x – $\frac{1}{2}$ are factors of ax2 + 5x + b show that a = b.
Answer:
P(x) = ax2 + 5x2 + b
x – 2 = 0
x = 2
P(x) = ax2 + 5x + b
P(2) = a(2)2 + 5(2) + b
0 = 4a + 10 + b ➝ (1)
P(x) = ax2 + 5x + b
x – $\frac{1}{2}$ = 0
x = $\frac{1}{2}$
P(x) = ax2 + 5x + b

a = – 2 ➝ (3)
Put a = – 2 in equation (1)
4a – 10 + b = 0
4(- 2) + 10 + b = 0
– 8 + 10 + b = 0
2 + b = 0
b = -2 ➝ (4)
From (3) and (4)
a = b.

Question 11.
If α and β are the zeroes of the quadratic polynomials f(x) = 2x2 – 5x + 7, find a polynomial whose zeroes are 2α + 3β and 3α + 2β.
Answer:
Since α and β are the zeroes of the quadratic polynomial f(x) = 2x2 – 5x + 7.

Let S and P denote respectively the sum and product of the zeroes of the required polynomial.
Then, S = (2α + 3β) + (3α + 2β)
= 5(α + β) = 5 × $\frac{5}{2}=\frac{25}{2}$ and
P = (2α + 3β) (3α + 2β)
⇒ P = (6α2 + 6β2 + 13αβ)
= 6α2 + 12αβ + αβ + 6β2
⇒ P = 6(α2 + β2 + 2αβ + αβ
= 6(α + β)2 + αβ
= 6(α2 + β2 + 2αβ) + αβ
= 6(α + β)2 + αβ

Hence the required polynomial g(x) is given by
g(x) = K(x2 – 5x + P)
or g(x) = K$\left(x^{2}-\frac{25}{2} x+41\right)$, where K and any non zero real number.

Question 12.
If α and β are the zeroes of the polynomial. 6y2 – 7y + 2, find a quadratic polynomial whose zeroes are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.
Answer:
Let P(y) = 6y2 – 7y + 2

Question 13.
If one zero of the polynomial 3x2 – 8x + 2k + 1 and seven times the other, find the value of k.
Answer:
Let α and β be the zeroes of the polynomial. Then as per question β = 7α.
Now sum of zeroes = α + β = α + 7α

Question 14.
If one zero of polynomial (a2 + 9)x2 + 13x + 6a and reciprocal of the other, find the value of a.
Answer:
Let one zero of the given polynomial be α
Then, the other zero and $\frac{1}{\alpha}$
∴ Product of zeroes = α × $\frac{1}{\alpha}$ = 1
But, as per the given polynomial product of Zeroes = $\frac{6 a}{a^{2}+9}$
$\frac{6 a}{a^{2}+9}$ = 1 ⇒ a2 + 9 = 6a
⇒a2 – 6a + 9 = 0 ⇒ (a – 3)2 = 0
⇒ a – 3 = 0 ⇒ a = 3
Hence a = 3.

Question 15.
Find the zeroes of the polynomial f(x) = x3 – 5x2 – 2x + 24 if it is given that the product of its two zeroes is 12.
Answer:
Let α, β and γ be the zeroes of polynomial f(x) such that αβ = 12

Putting αβ = 12 in αβγ = – 24 we get
12γ = – 24 ⇒ γ = $\frac{-24}{12}$ = – 2
Now α + β + γ = 5 ⇒ α + β – 2 = 5
⇒ α + β = 7
⇒ α = 7 – β
∴ αβ =12
⇒ (7 – β) β = 12
⇒ 7β = β2 = 12
⇒ β2 – 7β + 12 = 0
⇒ β2 – 3β – 4β – 12 = 0
⇒β(β -3) – 4(β – 3) = 0
⇒(β -4)(β – 3) = 0
⇒ β = 4 or β = 3
α = 3 or α = 4.

Question 16.
If α, β, γ be zeroes of polynomial 6x3 + 3x2 – 5x +1, then find the value of α– 1 + β– 1 + γ– 1
Answer:
P(x) = 6x3 + 3x2 – 5x + 1 so
a = 6, b = 3, c = – 5, d = 1
∴ α, β and γ are zeroes of the polynomial p(x)