Class 8

Students can Download Maths Chapter 16 Mensuration Ex 16.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.1

Question 1.
Find the total surface area of a cuboid with l= 4m, b = 3m and h = 1.5 m
Answer:
T.S.A = 2 (lb + bh + lh)
= 2(4 × 3 + 3 × 1.5 + 1.5 × 4)
= 2(12 + 4.5 + 6)
= 2(22.5)
T.S.A = 45 m²

Question 2.
Find the area of four walls of a room whose length is 3.5m, breath 2.5m and height is 3m.
Answer:
L = 3.5m, b = 2.5m, h = 3m
Area of the 4 walls = L.S.A of cuboid
= 2h(l + b)
= 2 × 3(3.5 + 2.5)
= 6(6) = 36m²

Question 3.
The dimensions of a room are l = 8m, b = 5m and h = 4m. Find the cost of distempering its four walls at the rate of 40/m2.
Answer:
L.S.A of cuboid = 2h (l + b)
= 2 × 4 (8 + 5) = 8 (13)
L.S.A = 104 m²
The cost of distempering 1 m² = Rs 40
The cost of distempering 104m² = 104 × 40 = Rs. 4160.

Question 4.
A room is 4.8m long, 3.6m broad and 2m high. Find the cost of laying tiles on its floor and its four walls at the rate of 100/m²
Answer:
L = 4.8m, b = 3.6m , h = 2m
Required area = Area of the floor + L.S.A of cuboid
= lb + 2h(l + b)
= 4.8 × 3.6 + 2 × 2(4.8 + 3.6)
= 17.28 + 33.6 = 50.88 m²
The cost of laying tiles for 1m² = Rs 100
The cost of laying tiles for 50.88 m² = 50.88 × 100 = Rs. 5088

Question 5.
A closed box is 40 cm long, 50 cm wide and 60 deep. Find the area of the foil needed for covering it.
Answer:
L = 40 cm, b = 50cm, h = 60 cm
Area of the foil = T.S. A of cuboid
= 2 (lb + bh + lh)
= 2(40 × 50 + 50 × 60 + 60 × 40)
= 2(7400)= 14,800 cm²
Area of the foil required 14,800 cm²

Question 6.
The total surface area of a cube is 384 cm2 calculate the side of the cube.
Answer:
T.S.A of cube = 384 cm²
61² = 384
l² = \(x=\frac{384}{6}\)
l² = 64
l = √64
l = 8cm

Question 7.
The L.S.A of a cube is 64m², calculate the side of the cube.
Answer:
L.S.A of cube = 41²
64 = 41²
\(x=\frac{64}{4}\) = l²
∴ l = √16 = 4m
The side of the cube is 4m.

Question 8.
Find the cost of whitewashing the four walls of a cubical room of side 4m at the rate of Rs.20/m2.
Answer:
l = 4m
Area to be whitewashed
= L.S.A of the cube
= 4l² = 4 × 4² = 4 × 16 = 64 m²
cost of whitewashing 1m² = 20
∴ The cost of whitewashing 64m² = 64 × 20 = Rs. 1280.

Question 9.
A cubical box has edge 10cm and another cuboidal box is 12.5 cm long, 10cm wide and 8cm high
i. Which box has a smaller total surface area?
ii. If each edge of the cube is doubled, how many times will its T.S.A increase?
Answer:
T.S.A of cube 61²= 6 × 10² = 6 ×100 = 600 cm²
T.S.A of a cuboid 61² = 6 × 102 = 6 x 100 = 600 cm²
T.S.A of a cuboid = 2(lb + bh + Ih)
= 2[12.5 x 10 +10 x 8 + 8 x 12.5]
= 2[125 + 80 + 100]
= 2[305]
t = 610 cm²
1. The cubical box has smaller total surface area.
2. If the side of the cube is doubled its side will be 20cm.
T.S.A 6l² = 6 × 202 = 6 × 400 = 2400 cm²
original area = 600cm²
\(\frac { 2400 }{ 600 }\) = 4
∴ Its area will increase by 4 times

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.2

Question 1.
In a triangle ABC, if ∠A= 55° and∠B.= 40° find ∠C
Answer:
∠A +∠B + ∠C = 180°
[Sum of the angles of a triangle 180°]
55 + 40 + ∠C = 180°
95 + ∠C = 180“
∠C = 180 – 95
∠C = 85°

Question 2.
In a right-angled triangle, if one of the other two angles is 35°, find the remaining angle.
Answer:
Let the angles be ∠A, ∠B and ∠C
Then ∠A = 90°,∠B = 35° and ∠C = ?
∠A + ∠B + ∠C = 180°
90 + 35 + ∠C = 180°
125 + ∠C = 180
∠C = 180 – 125 = 55°
∠C = 55°

Question 3.
If the vertex angle of an isosceles triangle is 50° find the other two angles.
Answer:
In an isosceles triangle, the base angles are equal, Let the each base angle be ‘x’
∠A + ∠B + ∠C = l80°
50 + x + x = 180°
[Sum of the angles of a traingle 180°]
50 + 2x = 180
2x = 180 – 50
2x = 130
x = \(\frac{130}{2}\)
x = 65°
∴ The other two angles are equal to 65° & 65°

Question 4.
The angles of a triangle are in the ratio 1: 2 : 3. Determine the three angles.
Answer:
Let the common ratio be ‘x’
The three angles are x, 2x and 3x
x + 2x + 3x = 180°
[Sum of the angles of a triangle 180°]
6x = 180°
x = \(\frac{180}{6}\) = 30
x = 30°
2x = 2 × 30 = 60°
3x = 3 × 30 = 90°
∴ The angles are 30°, 60° and 90°

Question 5.
In the adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle.
Answer:
∠A + ∠B + ∠C = 180°
[Sum of the angles of a triangle 180°]
x + 15 + x – 15 + x + 30 = 180°
3x + 30 = 180°
3x = 180° – 30
3x = 150
x = \(\frac{150}{3}\) = 50°
∠A = x +15 = 50 + 15 = 65°
∠B = x – 15 = 50 – 15 = 35°
∠C = x + 30 = 50 + 30 = 80°
∴ The angles are 65°, 3 5° and 80°

Question 6.
The angles of a triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10° find the three angles.
Answer:
Let the first angle be x, second angle is x + 10° and third angle is x + 20
x + x + 10 + x + 20 = 180°
[Sum of the angles of a triangle 180°]
3x + 30 = 180°
3x = 180 – 30
3x = 150°
x = \(\frac{150}{3}\) = 50°
x = 50°
∴ First angle x = 50°
Second angle = x + 10 = 50 + 10 = 60°
Third angle = x + 20 = 50 + 20 = 70°
Three angles are 50°, 60° & 70°

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.1

Question 1.
Match the following

Answer:
1 – c
2 – d
3 – a
4 – b.

Question 2.
Based on the sides, classify the following triangles (figures not drawn of the scales)

Answer:
Equilateral triangle (viii)
Isosceles triangle (iv), (ix), (x)
Scalene triangle (i), (ii), (iii), (v), (vi), (vii).

Students can Download Maths Chapter 13 Statistics Ex 13.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.2

Question 1.
Draw a histogram to represent the following frequency distribution.

Class interval	Frequency
20-25	5
25-30	10
30-35	18
35-40	14
40-45	12

Answer:

Question 2.
Draw a histogram to represent the following frequency distribution.

Class interval	Frequency
10- 19	7
20-29	10
30-39	20
40-49	5
50-59	15

Answer:
The given distribution is in the inclusive form.
d = lower limit of a class – upper limit of a class before it = 20 – 19
d = 1 ,  =  = 0.5

Stated class interval	Actual class	Frequency
10- 19	9.5 – 19.5	7
20-29	19.5-29.5	10
30-39	29.5 – 3u9.5	20
40-49	39.5 -49.5	5       .
50-59	49.5 – 59.5	15

Students can Download Maths Chapter 13 Statistics Ex 13.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 13 Statistics Ex 13.1

Question 1.
The marks scored by 40 candidates in an examination (out of 100) is given below.
75, 65, 57, 50, 32, 54, 75, 67, 75, 88, 80, 42, 40, 41, 34, 78, 43, 61, 42, 46, 68, 52, 43, 49, 59, 49, 67, 34, 33, 87, 97, 47, 46, 54, 48, 45, 51, 47, 41, 43.
Prepare a frequency distribution table with the class size 10. Take the class intervals as 30 – 39, 40 – 49 and answer the following questions.
(i) Which class intervals have highest and lowest frequency.
(ii) Write the upper and lower limits of the class interval 30 – 39.
(iii) What is the range of the given distribution?
Answer:

(i) The class interval 40-49 has highest frequency and 90-99 has the lowest frequency.
(ii) The lower limit of the class interval 30 – 39 is 29.5 and upper limit is 39.5
(iii) Range = Highest score – lowest score = 97- 32 = 65
Range = 65.

Question 2.
Prepare the frequency distribution table for the given set of scores. 39, 16, 30, 37, 53, 15, 16, 60, 58, 26, 28, 19, 20, 12, 14, 24, 59, 21, 57, 38, 25, 36,34, 15, 25, 41, 52, 45, 60, 63, 18, 26, 43, 18, 27, 59, 63, 46, 48, 25, 33, 46, 27, 46, 42, 48, 35, 64, 24 Take class intervals as (10 – 20), (20, 30)…. and answer the following.
(i) What does the frequency corresponding to the third class interval mean?
(ii) What is the size of each class interval? Find the midpoint of the class interval 30 – 40.
(iii) What is the range of the given set of scores?
Answer:

i) The frequency of third class interval 30 – 40 is 10
(ii) Size of each class interval is 10 and mid point of the class interval 30 – 40 is
 =  = 35
(iii) Range = 64 – 12 = 52.

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.3

Question 1.
Find all the angles in the following

Answer:
∠CMP + ∠PMD = 180° [Linear pair]
∠CMP+ 135° = 180°
∠CMP = 180° – 135
∠CMP = 45°
∠LMD = ∠CMP = 45° [Vertical opposite angles]
∠LMC =∠PMD = 135° [Vertical opposite angles]
∠ALM = ∠ LMC = 135° [Alternate angles]
∠BLM = ∠ LMC = 135° [Alternate angle]
∠QLB = ∠LMD = 45° [Corresponding angles]
∠QLA = ∠ LMC = 135° [Corresponding angles

Question 2.
Find the value of x in the diagram below.

Answer:
∠PQS = ∠EPT = 130° [Corresponding angle]
∠PQS + ∠SQR = 180° [Linear pair]
SQR = 180 – 130
∠SQR = 50°
∠QRS + ∠FRS = 180° [Linear pair]
∠QRS + 90° =180°
∠QRS = 180 – 90
∠QRS = 90°
∠SQR+ ∠QRS +∠QSR = 180°
[Sum of the angles of triangle is 180° ]
50 + 90 + ∠QSR = 180°
140 + ∠QSR = 180°
∠QSR = 180 – 140
∠QSR =40°
∠TSD = ∠QSR [Vertically opposite angles]
x = 40°.

Question 3.
Show that if a straight line is perpendicular to one of the two or more parallel lines, then it is also perpendicular to the remaining lines.
Answer:

\(\overrightarrow{\mathrm{AB}}\|\overrightarrow{\mathrm{CD}}\| \overrightarrow{\mathrm{EF}} \| \overrightarrow{\mathrm{GH}} \cdot \overline{\mathrm{XY}}\) interrectsthese lines at P, Q, R and S and \(\overrightarrow{\mathrm{XY}} \perp \overline{\mathrm{AB}}\)
Toprove: \(\overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{CD}}, \overline{\mathrm{XY}} \perp \overrightarrow{\mathrm{GH}}\)
Proof :∠XPB = 90° (data)
∠XPB = ∠ PQD = ∠QRF = ∠RSH = 90°

Question 4.
Let \(\overrightarrow{\mathrm{A}} \mathrm{B} \text { and } \overrightarrow{\mathrm{Cb}}\) be two parallel lines and \(\overrightarrow{\mathrm{PQ}}\) be a transversal. Show that the angle bisectors of a pair of two internal angles on the same side of the transversal are perpendicular to each other.
Answer:

Let the bisectors of ∠BRS and∠RSD intersect at T
To prove: RT⊥ST i.e., ∠RTS = 90°
Proof: ∠BRS +∠RSD = 180°
\(\frac { 1 }{ 2 }\) ∠BRS + \(\frac { 1 }{ 2 }\) ∠RSD = \(\frac { 1 }{ 2 }\) × 180°
\(\frac { 1 }{ 2 }\) ∠BRS = ∠TRS =∠TRB
\(\frac { 1 }{ 2 }\) ∠RSQ = ∠TSR =∠TSD
In triangle TRS,
∠TRS +∠TSR + ∠RTS = 180°
90 + ∠RTS = 180°
∠RTS = 90°
∴ RT ⊥ ST

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.2

I. Draw diagrams illustrating each of the following situations

Question a.
Three straight lines which do not pass through a fixed point.
Answer:

Question b.
A point and rays emanating from that point such that the angle between any two adjacent rays is an acute angle.
Answer:

Question c.
Two angles which are not adjacent angles, but still supplementary.
Answer:

Question d.
Three points in the plane which are equidistant from each other.
Answer:

Question 2.
Recognise the type of angles in the following figures.
Answer:
i. Adjacent angles

AOX is an acute angle and BOX is an obtuse angle
ii. Exterior angles

PRY and BOX are supplementary angles
iii. Reflex angle

Question 3.
Find the value of ‘x’ in each of the following diagrams.
(i) x + 2x = 180°

Answer:
3x = 180°
x = \(\frac{180^{\circ}}{3}\) = 60°
∴ x= 60°
(ii) ∠COB = 90°

∠COD +∠DOB = 90°
4x + x=90°
5x = 90
x = \(\frac { 90 }{ 5 }\)
x = 18°
∠AOC + ∠COB + ∠BOD + ∠DOA = 360° Complete angle

∠ AOC + x + x +∠ DOA = 360°
90 + 2x = 360° [∠ AOC +∠DOA = 90° ]
2x = 360 – 90
2x = 270
x = \(\frac { 270 }{ 2 }\)
x = 135°
(iv) ∠AOB = 180°

∠AOC + ∠BOC = 180°
x – y + x + y = 180°
2x = 180°
x = \(\frac { 180 }{ 2 }\)
x = 90°
(v) ∠COF= ∠EOD

∠COF = 3x
∴∠EOD = 3x
∠AOB = 180°
∠AOE + ∠EOD + ∠DOB = 180°
x + 3x + x + 30 = 180°
5x + 30 = 180°
5x = 180 – 30
5x = 150
x = \(\frac { 150 }{ 5 }\)
x = 30°
(vi)

EF is a straight line
∴ x + y = 180° ………..(i)
DC is a straight line
∴ y+ 65° = 180° …(ii)
From(i)and(ii)
x + y = y + 65°
x + y – y = 65°
x = 65°

Question 4.
which pair of angles are supplementary in the following diagram ? Are they supplementary rays ?

Answer:
∠AOB and ∠COD are supplementary angles.
∠COB and ∠AOD are supplementary angles.
There are no supplementary rays in the diagram.

Question 5.
Suppose two adjacent angles are supplementary. Show that if one of them is an obtuse angle, then the other angle must be acute.

Answer:
In the figure
∠APQ + ∠BPQ = 180° and ∠APQ is an obtuse angle.
To prove ∠BPQ < 90°
Construction: Draw a perpendicular SP at P
Proof: In the fig
∠ APS +∠SPQ + ∠QPS = 180°
90 + ∠SPQ + ∠QPB = 180°
∠SPQ + ∠QPB = 180-90
∠SPQ +∠QPB = 90
∴∠SPQ < 90° and ∠QPB < 90°
∴∠QPB is an acute angle
∴∠APQ is an obtuse angle

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.5

Question 1.
Construct rectangle ABCD with the following data.
(a) AB = 4 cm, BC = 6 cm.
Answer:
Opposite sides of a rectangle are equal as each angle is a right angle.
∴ AB = CD = 4 cm.
Steps:

1. Draw a line segment AB = 6 cm.
2. Construct a perpendicular at B.
3. With a 7 cm radius at A as center draw an arc on the perpendicular. Let the point be C.
4. With BC as radius draw an arc from A.
5. With AB as radius cut the arc from A.
6. Join AD and CD.

(b) AB = 6 cm AC = 7.2 cm
Steps:

1. Draw a line segment AB = 6 cm.
2. Draw perpendicular at B.
3. With a 4 cm radius cut the line at A.
4. With AD = 6 cm draw an arc.
5. With CD – 4 cm cut the arc.
6. Join AD and CD.

Question 2.
Construct square ABCD
(a) Which has side length 2 cm.
(b) Which has diagonal 6 cm.
Answer:
(a). Steps

1. Draw a line segment AB = 2 cm.
2. At A, construct a perpendicular
3. With A as center cut the perpendicular
4. With a 2 cm radius from B and D draw the arc.
5. Join DC and BC.

(b) Steps:

1. Draw a line segment AC = 6 cm.
2. Construct a perpendicular bisector to AC intersecting at O.
3. With O as the center and 3cm radius draw the line on both sides cutting at B and D.
4. Join AB, AD, DC & BC.

Students can Download Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 3 Axioms, Postulates and Theorems Ex 3.1

Question 1.
What are undefined objects in Euclid’s geometry?
Answer:
Certain objects in mathematics which cannot be defined using the terms an already know. Lines, points, plane, space are undefined objects in Euclidean geometry.

Question 2.
What is the difference between an axiom and a postulate?
Answer:
Axiom :

1. Axioms are elementary statements, which are self-evident and which are accepted without questions.
2. Axioms are applicable to all branches of mathematics and science.

Postulate:

1. The statements which are particular to geometry and accepted without question are called geometrical postulates.
2. Postulates are applicable to geometry only.

Question 3.
Give an example of the following axioms from your experience.
a. If equals are added to equals, the wholes are equal.
b. The whole is greater than the part.
Answer:
a. In the fig. AB = CD = 4cm BE = DF = 2 cm

AB + BE = 4 + 2 = 6cm ……….(i)
CD + DF = 4 + 2 = 6cm ……….(ii)

From (i) and (ii) AE = CF
(b) In the adjoining figure
AB = AC + CB

Compare AB,AC and CB you find AB > AC and AB > CB.

Question 4.
What is the need of introducing axioms?
Answer:
The Greek mathematicians faced great difficulty while developing geometry as a pure deductive science. They had to depend on certain primitive notions like points, line and planes and space. But this was not enough to deduce everything. They had to set up certain statements whose validity was accepted unquestionably. Thus there was a need of introducing axioms.

Question 5.
You have seen earlier that the set of all-natural numbers is closed under addition (closure property). Is this an axiom or something you can prove?
Answer:
Yes, this can be considered as an axiom

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4

Question 1.
The sides of a rectangle are in the ratio 2: 1. The perimeter is 30 cm. Calculate the measure of all the sides.
Answer:
The ratio of the sides is 2 : 1
Let the sides be 2x and x
Perimeter = 30
2x + x + 2x + x = 30
6x = 30
x = \(\frac{3}{6}\) = 5 cm
2x = 2 x 5 = 10cm
∴ The sides aer 10 cm, 5 cm, 10 cm, 5 cm.

Question 2.
In the adjacent rectangle ABCD ∠OCD = 30° . Calculate ∠BOC. What type of triangle is BOC.

Answer:
∠BCD = 90° [Angle of a rectangle]
∠OCD + ∠OCB = 90°
30 + ∠QCB = 90°
∠OCB = 90 – 30
∠OCB = 60°
∠OCB = ∠OBC = 60°
OC = OB [Diagonals of a rectangle bisect each other]
∴ ∠BOC = 180 – (60 + 60) = 180 – 120
∠BOC = 60°
In Δ BOC, ∠BOC = ∠OBC = ∠OCB = 60° & equiangular triangle
∴ BOC is an equiangular triangle.

Question 3.
All the rectangles are parallelograms but all the parallelograms are not rectangles. Justify this statement.
Answer:
A rectangle has all the properties of a parallelogram, therefore, all rectangles are parallelograms. No angle of a Parallelogram is a right angle but all angles of a rectangle are right angles.
∴ All parallelograms are not rectangular.

Question 4.
The side of a rectangular park are in the ratio 4 : 3. If the area is 1728 m find the cost of fencing it at the rate of Rs . 2.50/m
Answer:
The ratio of the sides = 4 : 3
Let the sides be 4x and 3x
Area = 1728
l × b = 1728
4x × 3x = 1728
12x² = 1728
x² = \(\frac{12}{1728}\)
x² = 144
x = √144 = 12m
4x = 4 × 12 = 48m
3x = 3x × 12 = 36m
Perimeter = 4x + 3x + 4x + 3x
= 48 + 36 + 48 + 36
= 168m
The cost of fencing 1m is Rs 2.50
The cost of fencing 168m = 168 × 250 = Rs. 420.

Question 5.
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remaining portion is a yard of a trapezoidal shape whose parallel sides have length 15m and 25m. What fraction of the yard is occupied by the flower bed?

Answer:
In the fig AB = 25 m and FE = 15 m
∴ DF + EC = 25 – 15 = 10m
But DF = EC
∴DF = EC = 5m also AD = BC = 5m
Area of rectangle ∆BCD = AB x BC = 1 x b = 25 x 5 = 125 m²
Area of ∆ADF = \(\frac{1}{2}\) base × height
= \(\frac{1}{2}\) 5 × 5 = \(\frac{25}{2}\)m²
∆ADF ≅ ABCE [data]
Area of ∆BCE = \(\frac{25}{2}\) m²
Sum of Area of the two triangles
∴ \(\frac{1}{5}\) of the field is occupied by the flower bed.

Question 6.
In a rhombus ABCD  ∠C = 70°. Find the other angles of the rhombus.
Answer:
∠A = ∠C = 70° [Opposite angles]
∠A +∠B = 180° [Adjacent angles of a rhombus]
70 + ∠B = 180°
∠B = 180 – 70
∠B = 110°
∠B = ∠D = 110° [Opposite angles are equal].
Angles are 70°, 110°, 70° & 110°

Question 7.
In a rhombus PQRS, if PQ = 3x – 7 and QR = x + 3 find PS.

Answer:
PQ = QR [Sides of a rhombus]
3x – 7 = x + 3
3x – x = 3 + 7
2x = 10
\(\frac{10}{2}\)
x = 5
PQ = 3x – 7 = 3 × 5 – 7 = 15 – 7
PQ = 8 cm
∴ SP = 8

Question 8.
Rhombus is a parallelogram. Justify.
Answer:
In a rhombus opposite sides are equal and parallel to each other therefore it is a parallelogram.

Question 9.
In a given square ABCD, if the area of triangle ABD is 36cm2 Find
1. The area of the triangle BCD and
2. Area of the square ABCD
Answer:
The diagonal of a square divides the square into two congruent triangles.

∴ Area of ∆ABD = ∆BCD = 36cm²
Area of square ABCD = 36 + 36 = 72 cm².

Question 10.
The side of square ABCD is 5cm and another square PQRS has a perimeter equal to 40cm. Find the ratio of perimeter ABCD to the perimeter of PQRS. Find the ratio of the area of ABCD to the area of PQRS.
Answer:
Side of ABCD = 5 cm
Perimeter of ABCD = 5 × 4 = 20 cm
Perimeter of PQRS = 40 cm 4 × side = 40
side = \(\frac { 40 }{ 4 }\)
side = 10cm
Perimeter of ABCD : Perimeter PQRS
= 20 : 40 = 1 : 2
Area of ABCD:Area of PQRS
= 52 : 102 = 25 : 100 = 1 : 4

Question 11.
A square field has a side of 20m. Find the length of the wire required to fence it four times.
Answer:
Perimeter = 4a = 4 × 20 = 80m
Length of the wire required to fence 4 times. = 4 × 80 = 320 m.

Question 12.
List out the difference between square and rhombus
Answer:

Square	Rhombus
i. All the angles are equal.	i. Opposite angles are equal
ii. All the angles are right angles	ii. No angle is a right angle
iii. Diagonals are equal	iii. Diagonals are not equal.