KSEEB Solutions

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1.
Solve the following pairs of equations by reducing them to a pair of linear equations :
(i) \(\frac{1}{2 x}+\frac{1}{3 y}=2\)
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\)
\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\)
(iii) \(\frac{4}{x}+3 y=14\)
\(\frac{3}{x}-4 y=23\)
(iv) \(\frac{5}{x-1}+\frac{1}{y-2}=2\)
\(\frac{6}{x-1}-\frac{3}{y-2}=1\)
(v) \(\frac{7 x-2 y}{x y}=5\)
\(\frac{8 x+7 y}{x y}=15\)
(vi) 6x + 3y = 6xy
2x + 4y = 5xy
(vii) \(\frac{10}{x+y}+\frac{2}{x-y}=4\)
\(\frac{15}{x+y}+\frac{5}{x-y}=-2\)
(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)
Solution:
(i) \(\frac{1}{2 x}+\frac{1}{3 y}=2\) ……………. (i)
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\) ……………… (ii)
From eqn. (i),
\(\frac{1}{2 x}+\frac{1}{3 y}=2\)
\(\frac{3 y+2 x}{6 x y}=2\)
3y + 2x = 12xy
2x + 3y – 12xy = 0
2x + 3y = 12xy …………….. (i)
From eqn. (ii),
\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
\(\frac{2 y+3 x}{6 x y}=\frac{13}{6}\)
6(2y + 3x) = 13 × 6xy
12y + 18x = 78xy
18x + 12y – 78xy = 0
18x + 12y = 78xy ……………… (ii)
Multiplying eqn. (i) by 4,
4(2x + 3y) = 4 × 12xy
8x + 12y = 78xy …………….. (iii)
From eqn. (ii) – eqn. (iii),

\(x=\frac{30 x y}{10}=3 x y\)
\(\frac{x}{x}=3 y\)
1 = 3y
\(y=\frac{1}{3}\)
2x + 3y = 12xy
\(2 x+3\left(\frac{1}{3}\right)=12 \times x \times \frac{1}{3}\)
2x + 1 = 4x
2x – 4x = 1
-2x = -1
2x = 1
\(x=\frac{1}{2}\)
\(x=\frac{1}{2}, \quad y=\frac{1}{3}\)

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\) ………….. (i)
\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\) ………….. (ii)
From eqn (i)
\(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\)
\(\frac{2 \sqrt{y}+3 \sqrt{x}}{\sqrt{x y}}=2\)
\(3 \sqrt{x}+2 \sqrt{x y}=2 \sqrt{x y}\) …………… (i)
From eqn (ii),
\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\)
\(\frac{4 \sqrt{y}-9 \sqrt{x}}{\sqrt{x y}}=-1\)
\(-9 \sqrt{x}+4 \sqrt{y}=-\sqrt{x y}\)
\(9 \sqrt{x}-3 \sqrt{y}=\sqrt{x y}\) ……………….. (ii)
Multiplying eqn (i) with 2
\(2(3 \sqrt{x}+2 \sqrt{y})=2 \times 2 \sqrt{x y}\)
\(6 \sqrt{x}+4 \sqrt{y}=4 \sqrt{x y}\) ……………….. (iii)
By adding eqn (ii) + eqn (iii)

\(\sqrt{x}=\frac{5 \sqrt{x y}}{15}\)
\(\frac{\sqrt{x}}{\sqrt{x}}=\frac{5 \sqrt{y}}{15}\)
\(1=\frac{\sqrt{y}}{3}\)
\(3=\sqrt{y} \quad \sqrt{y}=3\)
∴ y = 9
Substituting the value of ‘y’ in eqn (i)

∴ x = 4, y = 9

(iii) \(\frac{4}{x}+3 y=14\)
\(\frac{3}{x}-4 y=23\)
by putting the value of By multiplying eqn. (i) by 4, multiplying eqn. (ii) by 3, and then adding them,

∴ p = 5
\(\frac{1}{x}=p=5 \quad \quad x=\frac{1}{5}\)
Substituting the value of ‘p’ in eqn. (i)
4p + 3y = 14
4 × 5 + 3y = 14
20 + 3y = 14
3y = 14 – 20
3y = -6
∴ y = -2
∴ \(\quad x \cdot=\frac{1}{5}, y=-2\)

(iv) \(\frac{5}{x-1}+\frac{1}{y-2}=2\)
\(\frac{6}{x-1}-\frac{3}{y-2}=1\)
Let \(\frac{1}{x-1}=p \text { and } \frac{1}{y-2}=q\)
\(5\left(\frac{1}{x-1}\right)+\frac{1}{y-2}=2\) ……………. (i)
\(6\left(\frac{1}{x-1}\right)-3\left(\frac{1}{y-2}\right)=1\) …………… (ii)
5p + 1 = 2 …………… (iii)
6p – 3q = 1 …………… (iv)
By solving equation \(\mathrm{p}=\frac{1}{3}, \quad \mathrm{q}=\frac{1}{3}\)
Substituting \(\frac{1}{x-1}\) for p,
\(p=\frac{1}{x-1}=\frac{1}{3}\)
x – 1 = 3
∴ x = 4
Substituting \(\frac{1}{y-2}\) for q,
\(q=\frac{1}{y-2}=\frac{1}{3}\)
y – 2 = 3
∴ y = 5
∴ x = 4, y = 5

(v) \(\frac{7 x-2 y}{x y}=5\) ……………. (i)
\(\frac{8 x+7 y}{x y}=15\) ………………… (ii)
From eqn. (i)
\(\frac{7 x-2 y}{x y}=5\)
7x – 2y = 5xy ………….. (i)
From eqn. (ii)
\(\frac{8 x+7 y}{x y}=15\)
8x + 7y = 15xy ……………… (ii)
Eqm (i) × 7 and Eqn. (ii) × 2, we have
49x – 14y = 35xy ………….. (iii)
16x + 14y = 30xy ……………. (iv)
From eqn. (iii) + eqn. (iv)

∴ \(\frac{65 \mathrm{x}}{65 \mathrm{x}}=\mathrm{y}\)
1 = y
∴ y = 1
Substituting the value of ‘y’ in eqn. (i)
7x – 2y = 5xy
7x – 2 (1) = 5 × x × 1
7x – 2 = 5x
7x – 5x = 2
2x = 2
∴ x = 1
∴ x = 1, y = 1

(vi) 6x + 3y = 6xy ………………. (i)
2x + 4y = 5xy ………………… (ii)
Multiplying eqn. (ii) by 3
3(2x + 4y = 5xy)
6x + 12y = 18xy ………………. (iii)
From eqn. (i) – eqn. (ii)

9y = 9xy
\(\frac{9 y}{9 y}=x\)
1 = x
∴ x = 1
Substituting the value of ‘x’ in eqn (i),
6x + 3y = 6xy
6 × 1 + 3y = 6 × 1 × y
6 + 3y = 6y
6y – 3y = 6
3y = 6
∴ \(\quad y=\frac{6}{3}\)
∴ y = 2
∴ x = 1, y = 2

(viii) \(\frac{10}{x+y}+\frac{2}{x-y}=4\)
\(\frac{15}{x+y}+\frac{5}{x-y}=-2\)
Let \(\mathrm{u}=\frac{1}{\mathrm{x}+\mathrm{y}}, \mathrm{v}=\frac{1}{\mathrm{x}-\mathrm{y}}\) then,
10u + 2v = 4 …………. (i)
15u – 5v = -2 …………. (ii)
Multiplying eqn. (i) by 5 and multiplying eqn. (ii) by 2 and then by adding

∴ \(\quad \mathbf{u}=\frac{1}{5}\)
Substituting the value of ‘u’ in eqn. (i)
\(10 \times \frac{1}{5}+2 v=4\)
2 + 2v = 4
2v = 4 – 2
2v = 2
∴ v = 1
∴ u = 1/5 , v =1
\(\frac{1}{x+y}=\frac{1}{5} \quad \frac{1}{x-y}=1\)
∴ x + y = 5 x – y = 1
x + y = 5 …………….. (iii)
x – y = 1 …………….. (iv)
From eqn. (iii) + eqn. (iv)

x = 3
∴ Substituting the value of ‘x’ in (iii)
x + y = 5
3 + y = 5
y = 5 – 3
y = 2
∴ x = 3, y = 2

(viii) \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ………….. (i)
\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\) ……………. (ii)

Substituting the value of ‘p’ in eqn. (iii)
\(p+q=\frac{3}{4}\)
\(\frac{1}{4}+q=\frac{3}{4}\)
∴ \(\quad q=\frac{3}{4}-\frac{1}{4}\)
∴ \(\quad q=\frac{1}{2}\)
∴ \(\quad p=\frac{1}{3 x+y}=\frac{1}{4}\)
3x + y = 4 ……………. (v)
\(q=\frac{1}{3 x-y}=\frac{1}{2}\)
3x – y = 2 ……………… (vi)
From eqn. (v) + eqn. (vi)
3x + y = 4
+3x – y = 2
6x = 6
∴ \(\quad x=\frac{6}{6}=1\)
Substituting the value of ‘x’ in eqn. (v)
3x + y = 4
3 × 1 + y = 4
3 + y = 4
∴ y = 4 – 3
∴ y = 1
∴ x = 1, y = 1

Question 2.
Formulate the following problems as a pair of equations, and hence find their solutions :
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Solution:
Let speed of Ritu
in still water be x km/hr.
in current water be y km/hr
Speed of water while
upstream is (x + y) km/hr,
downstream is (x – y) km/hr,
2(x + y) = 20
x + y = 10 ………… (i)
2(x – y) = 4
x – y = 2 …………… (ii)
By Adding eqn. (i) and eqn. (ii),

∴ \(\quad x=\frac{12}{2}=6\)
Substituting the value, x = 6 in eqn. (i),
x + y = 10
6 + y = 10
y = 10 – 6
∴ y = 4
∴ Speed of Ritu
in still water, x = 6 km/hr.
in current water, y = 4 km/hr.

(ii) 2 women and 5 men can together finish an embroidery work in 4 days,, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone. ‘
Solution:
Number of days required
for women be ’x’
for men be ‘y’
Work done by woman in 1 day \(=\frac{1}{x}\)
Work done by man in 1 day \(=\frac{1}{y}\)
\(4\left(\frac{2}{x}+\frac{5}{4}\right)=1\)
\(\frac{2}{x}+\frac{5}{4}=\frac{1}{4}\) ………….. (i)
\(3\left(\frac{3}{x}+\frac{6}{y}\right)=1\)
\(\frac{3}{x}+\frac{6}{y}=\frac{1}{3}\) ……………… (ii)
Let \(\frac{1}{{x}}={p} \text { and } \frac{1}{y}={q}\)
∴ \(2 p+5 q=\frac{1}{4}\)
8p + 20q = 1 ……….. (iii)
\(3 p+6 q=\frac{1}{3}\)
9p + 18q = 1 ……………. (iv)
By Cross-multiplication
\(\frac{\mathrm{p}}{-20-(-18)}=\frac{\mathrm{q}}{-9-(-8)}=\frac{1}{144-180}\)
\(\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}\)
If \(\frac{p}{-2}=\frac{1}{-36}\) then \(p=\frac{-2}{-36}=\frac{1}{18}\)
If \(\frac{q}{-1}=\frac{1}{-36}\) then \(q=\frac{-1}{-36}=\frac{1}{36}\)
\(p=\frac{1}{x}=\frac{1}{18} \quad ∴ \quad x=18\)
\(q=\frac{1}{y}=\frac{1}{36} \quad ∴ \quad y=36\)
∴ Number of days taken by one woman to complete work, x = 18
Number of days taken by one man to complete the work, y = 36.

(iii) Roohl travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
Let speed of the train be x’ km/hr. and Speed of the bus is ‘y’ km/hr.
\(\frac{60}{\mathrm{x}}+\frac{240}{\mathrm{y}}=4\) …………… (i)
\(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\) …………… (ii)
If \(\frac{1}{\mathbf{x}}=\mathbf{p} \quad \frac{1}{\mathbf{y}}=\mathbf{q}\) then
60p + 240q = 4 ……………… (iii)
\(100 p+200 q=\frac{25}{6}\) ……………….. (iv)
Multiplying eqn. (iii) by 10, Mulitplying eqn. (iv) by 6 and then eqn. (iii) – eqn. (iv), we have

∴ \(\quad q=\frac{15}{1200}\)
∴ \(\quad q=\frac{1}{80}\)
Substituting the value of \(q=\frac{1}{80}\) in eqn. (iii)
\(60 p+240 \times \frac{1}{80}=4\)
60p + 3 = 4
60p =4 – 3
∴ \(\mathrm{p}=\frac{1}{60}\)
∴ \(\quad \mathrm{p}=\frac{1}{60}=\frac{1}{\mathrm{x}}\)
∴ x = 60 km/hr.
∴ \(\quad q=\frac{1}{80}=\frac{1}{y}\)
∴ y = 80 km/hr.
∴ Speed of train, x = 60 km/hr.
Speed of bus, y = 80 km/hr.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5

Question 1.
Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
i) 7 cm, 24 cm, 25 cm,
ii) 3 cm, 8 cm, 6 cm.
iii) 50 cm, 80 cm, 100 cm.
iv) 13 cm, 12 cm, 5 cm.
Solution:
In ⊥∆ABC, ∠B = 90°.
Let AB = a, BC = b, Hypotenuse AC = c then
AC² = AB² + BC²
c² = a² + b²
∴ Here diagonal is the greatest side.
i) a, b,c
7 cm, 24 cm, 25 cm,
c² = a² + b²
25² = (7)² + (24)²
625 = 49 + 576
625 = 625
625 = 49 + 576 625 = 625
∴ This is right angled triangle.
Measurement of Hypotenuse, c = 25 cm.

ii) a c b
3 cm, 8 cm, 6 cm.
c² = a² + b²
8² = (3)² + (6)²
64 = 9 + 36
64 ≠ 45
∴ These are not sides of right angled triangle.

iii) a b c
50 cm, 80 cm, 100 cm.
c² = a² + b²
100²= (50)² + (80)²
10000 = 2500 + 6400
10000 ≠ 8900
∴ These are not sides of right angled triangle.

iv) a b c
12 cm, 5 cm, 13 cm,
c² = a² + b²
13² = (12)² + (5)²
169 = 144 + 25
169 = 169
∴ These are sides of right angled triangle.
Measurement of Hypotenuse =13 cm.

Question 2.
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.
Solution:

Data: PQR is a triangle 9 right angled at P and M is a point on QR such that PM ⊥ QR.
To Prove: PM² = QM.MR
Proof: In ⊥D QPR, ∠P = 90°,
QR² = QP² + PR² ……………. (i)
In ⊥D PMQ, ∠M = 90°
QP² = PM² + QM² ………. (ii)
In ⊥D PMR, ∠M = 90°
PR² = PM² + MR² ……………. (iii)
By Adding eqn. (ii) and (iii)
QP² + PR² = PM² + QM² + PM² + MR²
QR² = 2PM² + QM² + MR² (. Eqn. (i))
(QM + MR)² = 2PM² + QM² + MR²
QM² + MR² + 2QM.MR = 2PM² + QM² + MR²
QM² – QM² + MR² – MR² + 2QM.MR = 2PM²
2QM.MR = 2PM²
2PM² = 2QM.MR
∴ PM² = QM.MR

Question 3.
In the following figure, ABD is a triangle right angled at A and AC ⊥BD. Show that
i) AB²= BC.BD
ii) AC² = BC.DC
iii) AD² = BD.CD

Solution:
Data: In ∆ABD, ∠A = 90°,
AC ⊥ BD.
To Proved: AB² = BC.BD
ii) AC² = BC.DC
iii) AD² = BD.CD

i) AB² = BC.BD
∆ACB ~ ∆BAD (. Theorem7)
\(\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{\mathrm{BC}}{\mathrm{AB}}\)
∴ AB²= BC × BD.

ii) AC² = BC.DC
∆BCA ~ ∆ACD
\(\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{\mathrm{BC}}{\mathrm{AC}}\)
∴ AC × AC = BC × CD
∴ AC² = BC × CD

iii) AD² = BD.CD
∆ACD ~ ∆BAD
\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{CD}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}\)
∴ AD × AD = BD × DC
∴ AD² = BD × DC

Question 4.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
Solution:
Data: ABC is an isosceles triangle right angled at C.

To Prove: AB² = 2AC².
In ⊥∆ACB, ∠C = 90°
∴ AB²= AC² + BC²
(∵ Pythagoras Theorem)
AB² = AC² + AC² (∵ AC = BC)
AB² = 2 AC².

Question 5.
ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right-angled triangle.
Solution:
Data: ABC is an isosceles triangle with AC = BC.

AB² = 2 AC².
To Prove: ∆ABC is a right angled triangle
AB² = 2AC² (Data)
AB² = AC² + AC²
AB² = AC² + BC² (∵ AC = BC)
Now, in ∆ABC, square of one side is equal to squares of other two sides.
∆ABC is a right angled triangle, Opposite angle to AB, i.e., ∠C is 90°.

Question 6.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:
Data: ABC is an equilateral triangle of side 2a.

To Prove: Altitude of ∆ABC,
AD =?
In equilateral triangle
D bisect base.
∴ AB = BC = CA = 2a. 3 D a
If BC = 2a,
\(\frac{1}{2} \mathrm{BC}=\mathrm{a}\) unit
∴ BD = DC = a.
Now, in ⊥∆ADB, ∠D = 90°
∴ AD² + BD² = AB²
AD² + a² = (2a)²
AD² + a² = 4a²
∴ AD² = 4a² – a²
AD² = 3a².
\(\sqrt{A D^{2}}=\sqrt{3 a^{2}}\)
∴Altitude,\(\mathrm{AD}=\sqrt{3} \mathrm{a}\) unit.

Question 7.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution:
Data: ABCD is a rhombus.
Here, AB = BC = CD = DA.
Diagonals AC and BD intersects at ’O’.
To Prove: AB² + BC² + CD² + DA² = AC² + BD².
In rhombus diagonals bisects perpendicularly.
∴ ∠AOB = ∠AOD = 90°.
In ⊥∆AOB,
AB² = AO² + BO² ……..(i)
In ⊥∆BOC,
AC² = BO² + CO² ……..(ii)
In ⊥∆COD,
CD² = OC² + OD² ……….(iii)
In ⊥∆AOD,
AD² = AO² + OD²…………(iv)
By Adding equations (i) + (ii) + (iii) + (iv)
AB² + BC² + CD² + DA² =
= AO² + BO² + BO² + CO² + CO² + DO² + AO² + OD²
= 2AO² + 2BO² + 2CO² + 2DO²
= 2AO² + 2CO² + 2BO² + 2DO²
Now, RHS = 2AO² + 2CO² + 2BO² + 2DO²
\(=2 \times\left(\frac{1}{2} \mathrm{AC}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{AC}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{BD}\right)^{2}+2 \times\left(\frac{1}{2} \mathrm{BD}\right)^{2}\)
\(=2 \times \frac{1}{4} \mathrm{AC}^{2}+2 \times \frac{1}{4} \mathrm{AC}^{2}+2 \times \frac{1}{4} \mathrm{BD}^{2}+2 \times \frac{1}{4} \mathrm{BD}^{2}\)
\(=\frac{1}{2} A C^{2}+\frac{1}{2} A C^{2}+\frac{1}{2} B D^{2}+\frac{1}{2} B D^{2}\)
RHS = AC² + BD²
∴ LHS = RHS
∴ AB² + BC² + CD² + DA² = AC² + BD².

Question 8.
In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, OF ⊥ AB. Show that
i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
ii) AF² + BD² + CE² = AE² + CD² + BF².

Solution:
Data: O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC, OF⊥ AB.
To Proved: i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²
ii) AF² + BD² + CE² = AE² + CD² + BF²

(i) OA² + OB² + OC² – OD² – OE² – OF² = AF² + BD² + CE²

OA² = AF² + OF² → (i)
OB² = BD² + OD² → (ii)
OC² = OE² + EC² → (iii)
By Adding equations (i) + (ii) + (iii),
OA² + OB²+ OC² = AF² + OF² + BD² + OD² + OE² +EC²
∴ OA² + OB² + OC² – OE² – OF² – OD² = AF² + BD² + CE²

(ii) AF² + BD² + CE² = AE² + CD² + BF².
OA² = AF² + OF²
∴ AF² = OA² – OF² → (i)
OB² = BD² + OD²
∴BD² = OB² – OD² → (ii)
OC² = OE² + EC²
∴ CE² = OC² – OE² → (iii)
From adding equations (i) + (ii) + (iii),
AF² + BD² + CE² = OA² – OF² + OB² – OD² + OC² – OE²
AF² + BD² + CE² = OA² – OE² + OB² – OF² + OC² – OE²
∴ AF² + BD² + CE² = AE² + FB² + CD² .

Question 9.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Solution:
In ⊥∆ACB, ∠C = 90°, BC = ?
AC² + CB² = AB²
(8)² + CB² = (10)²
64 + CB² = 100
CB² = 100 – 64
CB² = 36
∴ CB = 6
∴ Ladder is at a distance of 6m from the base of the wall.

Question 10.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution:
In ⊥∆PQR, ∠Q = 90°, QR = ?
∴ PQ² + QR² = PR²
(18)² + QR² = (24)²
324 + QR² = 576
QR² = 576 – 324
QR² = 252
∴ QR = 15.8 m.

Question 11.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?

Solution:
Distance travelled by aeroplace trowards North is \(1 \frac{1}{2}\) hours :
\(=1000 \times 1 \frac{1}{2}\)
\(=1000 \times \frac{3}{2}\)
= 1500 km.
Diatance travelled by aeroplane towards West in \(1 \frac{1}{2}\) Hours :
\(=1200 \times 1 \frac{1}{2}\)
\(=1200 \times \frac{3}{2}\)
= 1800 km.
In ⊥∆AOB,
AB² = OA² + OB²
= (1500)² + (1800)²
= 2250000 + 3240000 = 5490000
\(\mathrm{AB}=\sqrt{5490000}\)
\(A B=\sqrt{90000 \times 61}\)
\(A B=300 \sqrt{61} \mathrm{km}\) km
∴ Two planes are 300V6T km. apart after 14 hours

Question 12.
Two poles of heights 6 m and 11m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution:
Pole AB = 6 m.
Pole CD = 11 m.
Distance between poles BD = 12 m.
Distance between feet of the poles, AC = ?
ABDM is a rectangle, AB = MD = 6 m.
BD = AM = 12 m.
In ⊥∆AMC, ∠AMC = 90°
AM = 12 m, CM = 5 m. AC = ?
AC² = AM² + CM²
= (12)² + (5)²
= 144 + 25
AC² = 169
∴ AC = 13 m.

Question 13.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.
Prove that AE² + BD² = AB² + DE².

Solution:
Data: In ∆ABC, ∠C = 90°, D and E are points on the sides CA and CB respectively
Top Prove: AE² + BD² = AB² + DE²
In ⊥∆ACE, ∠C = 90°
∴ AE² = AC² + CE² ………. (i)
In ⊥∆DEB, ∠C = 90°
∴ BD² = DC² + CB² ………… (ii)
From adding equations (i) + (ii)
AE² + BD² = AC² + CE² + DC² + CB²
= AC² + CB² + DC² + CE²
∴ AE² + BD² = AB² + DE² ( . Theorem 8).

Question 14.
The perpendicular from A on side BC of a ∆ABC intersects BC at D such that DB = 3CD (see the following figure) Prove that 2AB² = 2AC² + BC².

Solution:
In ∆ACD,
AC² = AD² + DC²
AD² = AC² – DC²……….. (1)
In ∆ABD,
AB² = AD² + DB²
AD² = AB² – DB²………. (2)
From equations (1) and (2),
AC² – DC² = AB² – DB² ………… (3)
3DC = DB (data given)
\(\mathrm{DC}=\frac{\mathrm{BC}}{4}, \text { and } \mathrm{DB}=\frac{3 \mathrm{BC}}{4}\) ……….. (3)
Substituting eqn. (4) in eqn. (3),
\(\mathrm{AC}^{2}-\left(\frac{\mathrm{BC}}{4}\right)^{2}=\mathrm{AB}^{2}-\left(\frac{3 \mathrm{BC}}{4}\right)^{2}\)
\(A C^{2}-\frac{B C^{2}}{16}=A B^{2}-\frac{9 B C^{2}}{16}\)
\(\frac{16 \mathrm{AC}^{2}-\mathrm{BC}^{2}}{16}=\frac{16 \mathrm{AB}^{2}-9 \mathrm{BC}^{2}}{16}\)
16AC² – BC² = 16 AB² – 9BC²
16AB² – 16AC² = 9 BC² – BC²
16AB² – 16AC² = 8BC²
8(2AB² – 2AC² = BC²)
2AB² – 2AC² = BC²
2AB² = 2AC² ≠ BC²

Question 15.
In an equilateral triangle ABC, D is a point on side BC such that \(B D=\frac{1}{3} B C\), Prove that 9AD² = 7AB².
Solution:
∆ABC is an equilateral triangle.

AB = BC = AC = a
In ∆ABC, AE is perpendicular line.
\(B E=E C=\frac{B C}{2}\)
Altitude, \(\mathrm{AE}=\frac{\mathrm{a} \sqrt{3}}{2}\)
\(\mathrm{BD}=\frac{1}{3} \mathrm{BC}(\mathrm{Data})\)
\(B D=\frac{a}{3}\)
DE = BE – AD
\(=\quad \frac{a}{2}-\frac{a}{3}=\frac{a}{6}\)
In ∆ADE,
AD² = AE² + DE²
\(=\left(\frac{\mathrm{a} \sqrt{3}}{2}\right)^{2}+\left(\frac{\mathrm{a}}{6}\right)^{6}\)
\(=\frac{3 a^{2}}{4}+\frac{a^{2}}{36}=\frac{27 a^{2}+a^{2}}{36}\)
\(=\frac{28 \mathrm{a}^{2}}{36}\)
\(=\frac{7}{9} \mathrm{a}^{2}\)
\(\mathrm{AD}^{2}=\frac{7}{9} \mathrm{AB}^{2} \quad 9 \mathrm{AD}^{2}=7 \mathrm{AB}^{2}\)

Question 16.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Solution:
Data: ABC is an equilateral triangle.
Here, AB = BC = CD.
AD ⊥ BC.
To Prove: 3AC² = 4AD²
In ⊥∆ADC, ∠ADC = 90°
∴ AC² = AD² + DC²
\(=A D^{2}+\left(\frac{1}{2} A C\right)^{2}\)
\(\left( \begin{array}{l}{\mathrm{DC}=\frac{1}{2} \mathrm{BC}} \\ {\mathrm{DC}=\frac{1}{2} \mathrm{AC}}\end{array}\right)\)
\(\mathrm{AC}^{2}=\mathrm{AD}^{2}+\frac{1}{4} \mathrm{AC}^{2}\)
\(\frac{\mathrm{AC}^{2}}{1}-\frac{1}{4} \mathrm{AC}^{2}=\mathrm{AD}^{2}\)
\(\frac{4 \mathrm{AC}^{2}-1 \mathrm{AC}^{2}}{4}=\mathrm{AD}^{2}\)
∴ 3AC² = 4AD².

Question 17.
Tick the correct answer and justify:
In ∆ABC, \(A B=6 \sqrt{3} \mathrm{cm}\). AC = 12 cm. and BC = 6 cm. The angle B is
A) 120°
b) 60°
C) 90°
D) 45°
Solution:
C) 90°

Justification:
a             b           c
\(6 \sqrt{3}\)        6             12
Here largest side is 12 cm.
If square of the hypotenuse is equal to square of other two sides, then it is a right angled triangle.
∴ c² = a² + b²
AC² = AB² + BC²
\((12)^{2}=(6 \sqrt{3})^{2}+(6)^{2}\)
144 = 36 × 3 + 36
144 = 108 + 36
144 = 144
∴ ∆ABC is a right angled triangle and angle opposite to hypotenuse, ∠B = 90°.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.5 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.5, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2

In each of the following, give also the justification of the construction :
Question 1.
Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Solution:
r = 6 cm,, d = 10 cm., t = ?

Tangent, PQ = PR = 8 cm.
Steps of Construction :

1. A circle with radius 6 cm is drawn taking O’ as centre.
2. Point P is marked at 10 cm apart from centre of circle.
3. With the half of compass mark M which is the mid-point of OP.
4. A circle is drawn at M taking radius MO or MP which intersect the given circle at Q and R.
5. Now join PQ from P, and Join PR, required tangents are obtained. If measured it is PQ = PR = 8 cm.

Question 2.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Solution:
Steps of Construction:

1. With ‘O’ as centre two circles are constructed with radii 4 cm and 6 cm.
2. Draw OM radius for small circle and drawn perpendicular at M which meets the big circle at P and Q.
3. Now PQ is the tangent drawn for a small circle.
   

Tangent PMQ = 9 cm
Verification : In ⊥∆OMP, ∠M = 90°.
∴ OM² + PM² = OP²
(4)² + PM² = (6)²
PM² = (6)² – (4)² = 36 – 16
PM² = 20
∴ PM = \(\sqrt{20}\) = 4.5 cm.
Similarly,
MQ = 4.5 cm.
∴ PQ = PM + MQ = 4.5 + 4.5
∴ PQ = 9 cm.

Question 3.
Draw a circle of radius 3 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.
Solution:
Draw Pw, Px tangents to circle with centre ‘O’ at P.
Draw Qy, Qz tangents to circle with centre ‘O” at Q.

Tantent, OP = OQ = 7 cm.
Steps of Construction :

1. Construct a circle with radius 3 cm with centre O’. Draw the diameter AB and on both sides mark P and Q such that OP = OQ = 7 cm.
2. Bisect PO, Mark point M and draw circle of radius MO and centre M. It intersects given circle at w and x.
   If joined Pw and Px, required tangents are constructed.
3. Similarly, bisect OQ, mark point N. Draw a circle of radius NO or NQ with centre N, it meets given circle at y and z points. Join Qy and Qz. These are the required tangents.

Question 4.
Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.
Solution:
r = 5 cm.
Angle between tangents is 60°.

Steps of Construction :

1. A circle is drawn with centre ‘O’ and radius 5 cm. Draw OA and OB radii and ∠AOB = 120°
2. Tangents are drawn at A and B and these intersect at P.
3. By measurement, ∠APB = 60°.
   Angle between tangents PA and PB is 60°.
   ∵ ∠AOB and ∠APB Supplementary angles.

Question 5.
Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Solution:

Steps of Construction :

1. Draw a line segment AB = 8 cm. Draw two circles with radii 4 cm and 3 cm at A and B.
2. To construct tangent from centre A to Centre B, it intersects at ‘O’ which is the midpoint of AB.
   Draw circle with radius OA, OB at Centre ‘O’, it intersects circle with centre A at C and D and with centre ‘B’ at E and F.
3. Now join E to A centre and F, required AE and AF tangents are constructed.
4. Similarly ‘B’ centre is joined to C and D, BC and BF tangents are constructed.
5. AE and AF are tangents drawn from centre ‘A’ to centre ‘B’.
   Tangents drawn from centre B to centre A are BC and BD.

Question 6.
Let ABC be a right triangle in which AB= 6 cm, BC = 8 cm and ∠B = 90°. BD is perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Solution:
AB and AR are tangents drawn for circle with centre P from external point A.
Steps of Construction :

1. ⊥∆ ABC is constructed with BC = 8 cm. (base), ∠B = 90°, AB = 6 cm.
2. Draw BD ⊥ AC. , Draw perpendicular bisectors of DC and BC, they intersect at P.
3. A circle is drawn with centre P through A, B, D, and C.
   
   For this circle AB and AR tangents are constructed from external point A.
   Tangent, AB = AR = 6 cm.

Question 7.
Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Solution:

Steps of Construction:

1. Draw a circle with the help of a circular bangle kept on paper.
2. To draw AB and CD chords for this circle which has no centre. Their perpendicular bisectors meet at ‘O’. Now ‘O’ is the centre of the circle.
3. Mark point A on the outside of the circle from centre. Join OA. R is the midpoint of OA.
4. A circle is drawn with radius RO or RA which intersects the first circle at P and Q. Now join P from A and from A to Q, required two tangents AP and AQ are constructed.
   Tangent, AP = Tangent, AQ.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1

Question 1.
The graph of y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeros of p(x) in each case.

Solution:
(i) Graph of y = p(x) will not intersects x-axis.
∴ No zero for this polynomial.

(ii) Graph of y = p(x) intersects x-axis at only one point.
∴ Only one zero is there for this polynomial.

(iii) y = p(x) intersects x-axis at 3 points.
∴ It has 3 zeroes.

(iv) y = p(x) intersects x-axis at 2 points.
∴ Two zeroes for this polynomial.

(v) y = p(x) intersects x-axis at four (4) points.
∴ Four zeroes for this polynomial.

(vi) y = p(x) intersects x-axis at three (3) points.
∴ This polynomial has 3 zeroes.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1

In each of the following, give the justification of the construction also :
Question 1.
Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.
Solution:
Divide a line segment of 7.6 cm length in the ratio 5 : 8 and measure.
m : n = 5 : 8
m + n = 5 + 8 = 13
AC : CB = 5 : 8.
by measurement: AC = 3cm, CB = 4.6cm.

Steps of Construction:

1. Draw any ray AB, such that AB = 7.6 cm.
2. Draw Ax ray at point A making acute angle.
3. Locate the points A₁, A₂, A₃, ………., A₁₃ from A.
4. Join BA₁₃. Draw BA₁₃ || A₅C.
   Now, AC : CB = 5 : 8
   If measured, AC = 3 cm, CB = 4.6 cm.

Question 2.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Solution:
Construct a triangle ABC having sides 4 cm, 5 cm and 6 cm. For this we have to construct a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Steps of Construction:

1. ∆ABC is constructed, having AB = 4 cm, BC = 5 cm, CA = 6 cm.
2. Draw Bx ray such that it forms acute angle to BC which is adjacent to vertex A.
3. Locate 3 points on Bx such that BB₁ = B₁B₂ = B₂B₃.
4. Join B₃C. Draw a parallel line to B₃C which intersect BC at C.
5. Draw a parallel line through C’ to CA and it intersects BA at A’.
   Now, A’BC’ is required triangle.

Question 3.
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides
of the first triangle.
Solution:
Construct an ∆ABC having sides 5 cm, 6 cm and 7 cm. Then construct another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Steps of Construction :

1. First construct a triangle ABC having sides AB = 5cm, BC = 6 cm, CA = 7 cm.
2. Draw a ray Bx such that it makes an acute angle at ‘B’.
3. Locate Points B₁, B2, B₃, B₄, B₅, B₆, By points on Bx.
4. Join B₅C. Draw B₅C || B₇C’, it intersects at ‘C’ which is produced BC line.
5. Draw AC || C’A’, it meets produced line BC at A’.
6. Now required ∆A’BC’ is obtained whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Question 4.
Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.
Solution:
Construct an isosceles triangle with base 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Steps of Construction:

1. Construct ∆ABC having base AB = 8 cm, altitude AD = 4 cm.
2. ∠BAx acute angle is constructed at A. From ‘A’ locate A₁, A₂, A₃.
3. Join A₂B. Draw A₂B || A₃B’. B is marked on line AB produced.
4. Draw AB || B’C’. C’ is marked on AC produced and draw AC’.
5. Now ∆AB’C’ is constructed which is similar to ∆ABC and corresponding sides,
   i.e., \(1 \frac{1}{2}=\frac{3}{2}\)
   ∆ABC ||| ∆AB’C’

Question 5.
Draw a triangle ABC with side BC = 6 cm, AB = 5 cm, and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Construct a triangle ABC with sides BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.

Steps of Construction:

1. Construct a AABC having BC = 6 cm, AB = 5 cm, and ∠ABC = 60°.
2. Draw a ray Bx to form acute angle at B and mark BB₁, B₂, B₃, at equal intervals of distance.
3. Join B₄C. Draw B₄C || B₃C’.
4. Draw AC || A’C’.
5. Now ∆ABC’ is equal to ∆ABC.
   ∴ ∆A’BC’ ||| ∆ABC
   3 : 4

Question 6.
Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC
Solution:
Draw a triangle ABC with side BC = cm, ∠B = 45°, ∠A = 105°. The construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ∆ABC.

Steps of Construction :

1. Construct a triangle ABC having BC = 7 cm, ∠A = 45° and ∠B = 105°.
2. Draw Bx ray at B to form acute angle.
3. Mark points B₁, B₂, B₃, B₄ from B. Draw B₄C’.
4. Draw B₄C || B₃C’. Draw AC || A’C’.
5. Now, ∆A’BC’ ||| ∆ABC is constructed.
   ∆A’BC’ : ∆ABC

Question 7.
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Steps of Construction:

1. Construct Right angled triangle ABC having sides BC = 4 cm, BA = 3 cm, and ∠B = 90°.
2. Draw a ray Bx to form acute angle at B.
3. Mark points B₁, B₂, B₃, B₄ from B at equal intervals.
4. Join B₄C. Draw B₄C || B₃C’.
5. Draw AC ‘|| C’A’, then ∆A’BC’ is constructed.
6. Now, ∆A’BC’ ||| ∆ABC.
   ∴ ∆A’BC’ : ∆ABC = 5 : 3.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 6 Constructions Ex 6.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 6 Constructions Exercise 6.1, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.4

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion :

Solution:
(i)

∴ This is terminating decimal expansion.

(ii)

∴ This is terminating decimal expansion.

(iii)
\(\frac{64}{455}=\frac{2^{6}}{5 \times 7 \times 13}\)
∴ This is non-terminating repeating expansion.

(iv)

∴ This is terminating decimal expansion.

(v)
\(\frac{29}{343}=\frac{29}{7^{3}}\)
∴ This is non-terminating repeating expansion.

(vi)

∴ This is terminating decimal expansion.

(vii)
\(\frac{129}{2^{2} 5^{7} 7^{5}}\)
∴ This is non-terminating repeating expansion.

(viii)

∴ This is terminating decimal expansion.

(ix)

∴ This is terminating decimal expansion.

(x)

∴ This is non-terminating repeating expansion.

Question 2.
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:

Question 3.
The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form \(\frac{p}{q}\), what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000………..
(iii) \(43 . \overline{123456789}\)
Solution:
(i) 43.123456789
\(\frac{p}{q}\) is a rational number
Prime factor of q:
q = 10⁹ = 2⁹ × 5⁹

(ii) 0.120120012000120000…………
This is an irrational number

(iii) \(43 . \overline{123456789}\)
This is a rational number.
Because, Prime factors of ‘q’ are contrinued in non-terminating decimal expansion.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.3

Question 1.
Prove that \(\sqrt{5}\) is irrational.
Solution:
Let us assume, to the contrary, that \(\sqrt{5}\) is rational.
∴ \(\sqrt{5}=\frac{a}{b}\)
∴ b × \(\sqrt{5}\) = a
By Squaring on both sides,
5b² = a² …………. (i)
∴ 5 divides a².
5 divides a.
∴ We can write a = 5c.
Substituting the value of ‘a’ in eqn. (i),
5b² = (5c)² = 25c²
b² = 5c²
It means 5 divides b².
∴ 5 divides b.
∴ ‘a’ and ‘b’ have at least 5 as a common factor.
But this contradicts the fact that a’ and ‘b’ are prime numbers.
∴ \(\sqrt{5}\) is an irrational number.

Question 2.
Prove that \(3+2 \sqrt{5}\) is irrational.
Solution:
Let us assume that \(3+2 \sqrt{5}\) is an irrational number.
Here, p, q, ∈ z, q ≠ 0

\(\sqrt{5}\) is rational number.
∵ \(\frac{p-3 q}{2 q}\) is rational number.
But \(\sqrt{5}\) is not a rational number.
This contradicts the fact that,
∴ \(3+2 \sqrt{5}\) is an irrational number.

Question 3.
Prove that the following are irrationals :
i) \(\frac{1}{\sqrt{2}}\)
ii) \(7 \sqrt{5}\)
iii) \(6+\sqrt{2}\)
Solution:
i) Let \(\frac{1}{\sqrt{2}}\) is a rational number.
\(\frac{1}{\sqrt{2}}=\frac{\mathrm{p}}{\mathrm{q}}\)
\(\sqrt{2}=\frac{q}{p}\)
By Squaring on both sides,
2 × p² = q²
2, divides q².
∴ 2, divides q
∵ q is an even number.
Similarly ‘p’ is an even number.
∴ p and q are even numbers.
∴ Common factor of p and q is 2.
This contradicts the fact that p and q also irrational.
∴ \(\sqrt{2}\) is an irrational number.
∴ \(\frac{1}{\sqrt{2}}\) is an irrational number.

ii) Let \(7 \sqrt{5}\)is a rational number.
∴ \(7 \sqrt{5}=\frac{p}{q}\)
\(\sqrt{5}=\frac{p}{7 q}\)
Here,\(\frac{p}{7 q}\) is one rational number.
It means \(\sqrt{5}\) which is equal also a rational number.
This contradicts to the fact that \(\sqrt{5}\) is an irrational number.
This contradicts to the fact that \(7 \sqrt{5}\) is rational number.
∴ \(7 \sqrt{5}\) is a rational number.

iii) Let \(6+\sqrt{2}\) is a rational number.

\(\frac{a-6 b}{b}\) is a rational number, b
∴ \(\sqrt{2}\) is also rational number.
This contradicts to the fact that \(\sqrt{2}\) is an irrational number.
This contradicts to the fact that \(6+\sqrt{2}\) is a rational number.
∴ \(6+\sqrt{2}\) is an irrational number.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5

Question 1.
Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0
3x – 9y – 2 = 0
(ii) 2x + y = 5
3x + 2y = 8
(iii) 3x – 5y = 20
6x – 10y = 40
(iv) x – 3y – 7 = 0
3x – 3y – 15 = 0
Solution:
(i) x – 3y – 3 = 0 …………. (i)
3x – 9y – 2 = 0 ……………… (ii)
Here, a₁ = 1, b₁ = -3, c₁ = -3
a₂ = 3, b₂ = -9, c₂ = -2
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{3} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-3}{-9}=\frac{1}{3}\)
\(\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}\)
Here, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{2}}{c_{2}}\)
∴ No solution for these.

(ii) 2x + y = 5
3x + 2y = 8
2x + y – 5 = 0 → (1)
3x + 2y – 8 = 0 → (2)
Here, a₁ = 2, b₁ = 1, c₁= – 5
a₂ = 3, b₂ = 2, c₂ = – 8

Here, 
∴ Many solutions are there for these

\(\frac{\mathbf{x}}{1 x-8-(2 x-5)}=\frac{y}{-5 \times 3-(-8 \times 2)}=\frac{1}{2 \times 2-3 \times 1}\)
\(\frac{x}{-8+10}=\frac{y}{-15+16}=\frac{1}{4-3} \)
\(\frac{\mathbf{x}}{+2}=\frac{\mathbf{y}}{1}=\frac{1}{1}\)
\(\frac{\mathbf{x}}{+2}=\frac{1}{1} \quad \mathbf{x}=2\)
\(\frac{y}{1}=\frac{1}{1} \quad y=1\)
∴ Unique solution: x = 2, y = 1

(iii) 3x – 5y = 20
6x – 10y = 40
3x – 5y – 20 = 0 → (1)
6x – 10y – 40 = 0 → (2)
Here, a₁ = 3 b₁ = – 5, c₁ = – 20
a₂ = 6, b₂ = – 10, c₂ = – 40


Here, 
Infinite solutions are there for these.

(iv) x – 3y – 7 = 0 …………… (i)
3x – 3y – 15 = 0 …………….. (ii)
a₁ = 1, b₁ = -3, c₁ = -7
a₂ = 3, b₂ = -3, c₂ = -15
\(\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-3}=\frac{1}{1}\)
\(\frac{c_{1}}{c_{2}}=\frac{-7}{-15}=\frac{7}{15}\)
Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\)
∴ Many solutions are there

\(\frac{x}{-3 x-15-(-3 x-7)}=\frac{y}{-7 \times 3-(-15 \times 1)}=\frac{1}{1 x-3-(3 x-3)}\)
\(\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}\)
\(\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}\)
If \(\frac{x}{24}=\frac{1}{6}\) then
6x = 24
\(x=\frac{24}{6}=4\)
If \(\frac{y}{-6}=\frac{1}{6}\) then
6y = -6
\(y=\frac{-6}{6}\)
∴ y = -1
∴ Unique solutions are there for these x = 4, y = -1.

Question 2.
(i) For which values of a and b does the following pair of linear equations have an infinite number of solutions ?
2x + 3y = 7
(a – b) x + (a + b) y = 3a + b – 2
(ii) For which value of k will the following pair of linear equations have no solution ?
3x + y = 1
(2k – 1)x + (k – 1)y = 2k + 1
Solution:
(i) 2x + 3y = 7
2x + 3y – 7 = 0 ………… (i)
(a – b)x + (a + b)y = (3a + b – 2) ………… (ii)
(a – b)x + (a + b)y – (3a + b – 2) ……………. (ii)
Here, a₁ =2, b₁ = 3, c₁ = -7
a₂ = (a – b), b₂ = (a + b), c₂ = -(3a + b – 2)
If it has infinite solutions, then we have
\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)
\(\frac{2}{(a-b)}=\frac{3}{(a+b)}=\frac{-7}{-(3 a+b-2)}\)
If \(\frac{2}{(a-b)}=\frac{-7}{(-3 a-b+2)}\) then
2(-3a – b + 2) = -7(a – b)
-6a – 2b + 4 = -7a + 7b
-6a + 7a – 2b – 7b = -4
a – 9b = 4 ………… (i)
If \(\frac{3}{(a+b)}=\frac{-7}{(-3 a-b+2)}\) then
3(-3a – b + 2) = -7(a + b)
-9a- 3b + 6 = -7a – 7b
-9a + 7a – 3b + 7b = -6
-2a + 4b = -6
-a + 2b = -3 …………. (ii)
From eqn. (i) + eqn. (ii),

7b = 7
\(b=\frac{7}{7}=1\)
Substituting the value of ‘b’ in eqn. (i),
a – 9b = -4
a – 9(1) = -4
a – 9 = -4
a = -4 + 9
a = 5
∴ a = 5, b= 1.

(ii) If the linear pair do not have solutions, then we have

3x + y – 1 = 0 → (1)
(2k – 1)x + (k – 1)y – (2k + 1) = 0 → (2)
Here a₁ = 3, b₁ = 1, c₁= 1
a₂ = (2k – 1), b₂ = (k – 1), c₂ = -(2k + 1)


3(k – 1) = 1(2k – 1)
3k – 3 = 2k – 2
3k – 2k = – 2 + 3
∴ k = 1.

Question 3.
Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Solution:
8x + 5y = 9 ……………. (i)
3x + 2y = 4 …………….. (ii)
(1) Substitution Method:
8x + 5y = 9
8x = 9 – 5y
\(x=\frac{9-5 y}{8}\)
Substituting the value of ‘x’ in eqn (i)
3x + 2y = 4
\(3\left(\frac{9-5 y}{8}\right)+2 y=4\)
\(\frac{27-15 y}{8}+\frac{2 y}{1}=4\)
\(\frac{27-15 y+164}{8}=4\)
27 + y = 8 × 4
27 + y = 32
y = 32 – 27
∴ y = 5
Substituting the value of ‘y’ in
\(x=\frac{9-5 y}{8}\)
\(=\frac{9-5 \times 5}{8}\)
\(=\frac{9-25}{8}\)
\(=\frac{-16}{8}\)
∴ x = -2
∴ x = -2, y = 5
(2) Cross-Multiplication Method:
8x + 5y – 9 = 0 …………. (i)
3x + 2y – 4 = 0 …………. (ii)
Here, a₁ = 8 b₁ = 5 c₁=-9
a₂ = 3, b₂ = 2, c₂ = -4
8x + 5y = 9 3x + 2y = 4

\(\frac{x}{5 x-4-(2 x-9)}=\frac{y}{-9 \times 3-(-4 \times 8)}=\frac{1}{8 \times 2-3 \times 5}\)
\(\frac{x}{-20+18}=\frac{y}{-27+32}=\frac{1}{16-15}\)
\(\frac{\mathbf{x}}{-2}=\frac{y}{5}=\frac{1}{1}\)
\(\frac{\mathrm{x}}{-2}=1 \quad \mathrm{x}=-2\)
\(\frac{\mathrm{y}}{5}=1 \quad \mathrm{y}=5\)

Question 4.
Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution:
Fixed hostel charges are Rs. x.
Amount to be paid to be Rs. y.
x + 20y = 1000 …………. (i)
x + 26y = 1180 …………. (ii)
By elimination method, we can find solution.
Subtracting eqn. (ii) from eqn. (i),

– 6y =- 180
\(\mathbf{y}=\frac{180}{6}\)
∴ y = Rs. 30
Substituting the value of ‘y’ in eqn. (i),
x + 20y = 1000
x + 20(30) = 1000
x + 600 = 1000
∴ x = 1000 – 600
∴ x = Rs. 400
Fixed charge, x = Rs. 400
Amount to be paid, y = Rs. 30.

(ii) A fraction becomes \(\frac{1}{3}\) when 1 is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its 4 denominator. Find the fraction.
Solution:
Let the fraction be 

3(x – 1) = y
3x – 3 = y
3x – y – 3 = 0 → (1)

4x = y + 8
4x – y – 8 = 0 → (2)
By Substitution method we can find out solution:
From eqn. (i),
3x – y – 3 = 0
y = 3x – 3
Substituting the value of y’ in eqn. (2),
4x – y – 8 = 0
4x – (3x – 3) – 8 = 0
4x – 3x + 3 – 8 = 0
x – 5 = 0
∴ x = 5.
Substituting the value of ‘x’ in
y = 3x – 3,
= 3(5) – 3
= 15 – 3
∴ y = 12
∴ x = 5, y = 12
∴ Fraction is 

(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution:
Let the number of right answers be ‘x’, and the number of wrong answers be y’.
3x – y = 40 ……….. (i)
4x – 2y = 50 ………. (ii)
By Elimination method we can find our solution :
3x – y = 40 ……… (i)
2x – y = 25 ……….. (ii)
Subtracting eqn. (ii) from eqn. (i),

Substituting the value of ‘x’ in eqn. (i),
3x – y = 40
3(15) – y = 40
45 – y = 40
-y = 40 – 45
-y = -5
∴ y = 5
Questions of right answers, x = 15
Questions of wrong answers, y = 5
∴ Total number of questions = x + y = 15 + 5 = 20

(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars ?
Solution:
Distance between places A and B is 100 km.
Speed of A car us ‘u’ km/hr.
Speed of B car is ‘v’ km/hr.
If the cars travel in the same direction at different speeds they meet in 5 hours about 20 km.
∴ u – v = 20 ………. (i)
If the cars travel in the same direction, total distance covered =100 km.
∴ u + v = 100 …….. (ii)
We can solve by Elimination method:
From eqn. (i) + eqn. (ii),

\(u=\frac{120}{2}\)
∴ u = 60 km/hr.
Substituting the value of ‘u’ in Eqn. (ii),
u + v = 100
60 + v = 100
v = 100 – 60
v = 40 km/hr.
∴ u = 60 km/hr.
v = 40 km/hr.

(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution:
Length of rectangle is ‘x’ unit.
Breadth of rectangle is ‘y’ unit.
Area of rectangle = length × breadth = xy
(x – 5) (y + 3) = xy – 9
xy + 3x – 5y – 15 = xy – 9
3x – 5y = -9 + 15
3x – 5y = 6 …………. (i)
(x + 3) (y + 2) = xy + 67
xy + 2x + 3y + 6 = xy + 67
2x + 3y = 67 – 6
2x + 3y = 61 ……….. (ii)
Solving the equation by Elimination method :
Multiplying eqn. (i) by and eqn. (ii) by 5
9x – 15y = 18 ………….. (iii)
10x + 15y = 305 ………… (iv)
From eqn. (iii) + eqn. (iv)

\(x=\frac{323}{19}\)
∴ x = 17 unit.
Substituting the value of ‘x’ in eqn. (i),
3x – 5y = 6
3(17) – 5y = 6
51 – 5y = 6
-5y = 6 – 51
– 5y = – 45
5y = 45
\(y=\frac{45}{5}\)
∴ y = 9 units.
∴ Length of that rectangle, x = 17 unit
breadth, y = 9 unit.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.5 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.5, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4

Question 1.
Let ∆ABC ~ ∆DEF and their areas be, respectively, 64 cm² and 121 cm², If EF = 15.4 cm., find BC.
Solution:

If ∆ABC ~ ∆DEF, then
\(\frac { { ar }(\Delta { ABC }) }{ { ar }(\Delta { DEF }) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \)
\(\frac{64}{121}=\frac{\mathrm{BC}^{2}}{(15.4)^{2}}\)
\(\mathrm{BC}^{2}=\frac{64}{121}=\frac{\mathrm{BC}^{2}}{(15.4)^{2}}\)
\(\mathrm{BC}^{2}=\frac{64}{121} \times \frac{237.16}{1}\)
∴ BC = 8 × 1.4
∴ BC = 11.2 cm

Question 2.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

Solution:
Data: Diagonals of a trapezium ABCD with AB||DC intersect each other at the point O. We have AB = 2 CD.
To prove: \(\frac { { Ar }(\Delta { A }{ OB }) }{ { Ar }(\Delta { COD }) } =?\)
As per similarity criterion, ∆AOB and ∆COD are,
∆AOB ~ ∆COD
\(\frac { { Ar }(\Delta { AOB }) }{ { Ar }(\Delta { COD }) } =\frac { { AB }^{ 2 } }{ { DC }^{ 2 } } \)
\(=\frac{(2 \mathrm{DC})^{2}}{\mathrm{DC}^{2}}(\cdot \mathrm{AB}=2 \mathrm{DC})\)
\(=\frac{4 \mathrm{DC}^{2}}{1 \mathrm{DC}^{2}}\)
\(\frac { { Ar }(\Delta { AOB }) }{ { Ar }(\Delta { COD }) } =\frac { 4 }{ 1 } \)
= 4 : 1

Question 3.
In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
\(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { DBC }) } =\frac { { AO } }{ { DO } } \)

Solution:
Data: ABC and DBC are two triangles on the same base BC. If AD intersects BC at O.
To Prove: \(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { DBC }) } =\frac { { AO } }{ { DO } } \)
Construction: Draw AP ⊥ BC and DM ⊥ BC.

Area of ∆ = \(\frac{1}{2}\) × base × height
\(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { DBC }) } =\frac { \frac { 1 }{ 2 } \times BC\times AP }{ \frac { 1 }{ 2 } \times BC\times DM } \)
\(=\frac{\mathrm{AP}}{\mathrm{DM}}\)
In ∆APO and ∆DMO,
∠APO = ∠DMO = 90° (construction)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ∆APO ~ ∆DMO
(∵ Similarity criterion is A.A.A.)
\(\frac{\mathrm{AP}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\)
\(\Rightarrow \quad \frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { DBC }) } =\frac { { AO } }{ { DO } } \)

Question 4.
If the areas of two similar triangles are equal, prove that they are congruent.

Solution:
In ∆ABC ~ ∆PQR
\(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { PQR }) } =\left( \frac { { AB } }{ { PQ } } \right) ^{ 2 }=\left( \frac { { BC } }{ { QR } } \right) ^{ 2 }=\left( \frac { { AC } }{ { PR } } \right) ^{ 2 }\) …………. (1)
Area of ∆ABC = Area of ∆PQR
\(\Rightarrow \quad \frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { PQR }) } =1\) …………. (2)
Substituting the eqn (2) in eqn (1)
\(1=\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)^{2}=\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)^{2}=\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)^{2}\)
⇒ AB = PQ, BC = QR, AC = PR
∴ ∆ABC ≅ ∆PQR (∵ S.S.S postulate)

Question 5.
D, E, and F are respectively mid-points of sides AB, BC and CA of ∆ABC. Find the ratio of the areas of ∆DEF and ∆ABC.

Solution:
Data: D. E and F are respectively midpoints of sides AB, BC and CA of ∆ABC.
To Prove: Ratio of the areas of ∆DEF and ∆ABC.
If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side and it is half of that side
\(\mathrm{DE} \| \mathrm{AC} \quad \mathrm{DE}=\frac{1}{2} \mathrm{AC}\)
\(\mathrm{EF} \| \mathrm{AB} \quad \mathrm{EF}=\frac{1}{2} \mathrm{AB}\)
\(\mathrm{DF} \| \mathrm{BC} \quad \mathrm{DF}=\frac{1}{2} \mathrm{BC}\)
\(\frac { { Ar }(\Delta { DEF }) }{ { Ar }(\Delta { ABC }) } =\frac { { DF }^{ 2 } }{ { BC }^{ 2 } } \quad ({ Theorem }6)\)
\(\frac { { Ar }(\Delta { DEF }) }{ { Ar }(\Delta { ABC }) } =\frac { \left( \frac { 1 }{ 2 } { BC } \right) ^{ 2 } }{ { BC }^{ 2 } } \)
\(\frac { { Ar }(\Delta { DEF }) }{ { Ar }(\Delta { ABC }) } =\frac { \frac { 1 }{ 4 } { BC }^{ 2 } }{ { BC }^{ 1 } } =\frac { 1 }{ 4 } =1:4\)

Question 6.
Prove that the ratio of the areas , of two similar triangles is equal to the square of the ratio of their corresponding medians.

Solution:
Let ∆ABC ~ ∆PQR
AD and PS are mid-point lines of ∆ABC and ∆PQR.
∆ABC ~ ∆PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\) ………………… (1)
∠A = ∠P, ∠R = ∠Q, and ∠C = ∠R (2)
AD and PS are mid-point lines
\(B D=D C=\frac{B C}{2}\)
and \(\quad \mathrm{QS}=\mathrm{SR}=\frac{\mathrm{QR}}{2}\)
\(\frac{A B}{P Q}=\frac{B D}{Q S}=\frac{A C}{P R}\) …………… (3)
In ∆ABD and ∆PQS,
∠B = ∠Q ………. Eqn (2)
and \(\frac{A B}{P Q}=\frac{B D}{Q S}\) ………. Eqn (3)
∴ ∆ABD ~ ∆PQS (SAS Postulate)
\(\frac{A B}{P Q}=\frac{B D}{Q S}=\frac{A D}{P S}\) ……………. (4)
\(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { PQR }) } =\left( \frac { { AB } }{ { PQ } } \right) ^{ 2 }=\left( \frac { { BC } }{ { QR } } \right) ^{ 2 }=\left( \frac { { AC } }{ { PR } } \right) ^{ 2 }\)
From Eqn. (1) and Eqn. (4), we have
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PS}}\)
\(\Rightarrow \quad \frac { { Ar }(\Delta { BC }) }{ { Ar }(\Delta { PQR }) } =\left( \frac { { AD } }{ { PS } } \right) ^{ 2 }\)

Question 7.
Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Solution:
Data: Equilateral triangle ABE is on AB of square ABCD. Equilateral triangle ACF is on diagonal AC.
To Prove: \((\Delta A B E)=\frac{1}{2} \text { Ar }(\Delta A C F)\)
\(\frac { { Ar }(\Delta { ABE }) }{ { Ar }(\Delta { ACF }) } =\frac { { AB }^{ 2 } }{ { AC }^{ 2 } } \quad { Theorem }6\)
\(=\frac{A B^{2}}{A B^{2}+B C^{2}}\)
∵ Pythagoras theorem
\(=\frac{\mathrm{AB}^{2}}{2 \mathrm{AB}^{2}}\)
∴ AB = AC sides of square
\(=\frac{1 \times A B^{2}}{2 \times A B^{2}}\)
\(\frac { { Ar }(\Delta { ABE }) }{ { Ar }(\Delta { ACF }) } =\frac { 1 }{ 2 } \)

Tick the correct answer and justify:
Question 8.
∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ∆ABC and ∆BDE is :
A) 2 : 1
B) 1 : 2
C) 4 : 1
D) 1 : 4
Solution:
C) 4 : 1

\(\frac { { Ar }(\Delta { ABC }) }{ { Ar }(\Delta { BDE }) } =\frac { { BC }^{ 2 } }{ { BD }^{ 2 } } =\frac { { BC }^{ 2 } }{ \left( \frac { 1 }{ 2 } { BC } \right) ^{ 2 } } \)
\(=\quad \frac{1}{1 / 4}=\frac{4}{1}\)
= 4 : 1

Question 9.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio :
A) 2 : 3
B) 4 : 9
C) 81 : 16
D) 16 : 81
Solution:
D) 16 : 81
∴ (4)² : (9)² = 16 : 81.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3

{Unless stated otherwise, use \(\pi=\frac{22}{7}\)}

Question 1.
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.
Solution:

PQ = 24 cm, PR = 7 cm,
‘O’ is the centre of circle.
Angle in semicircle,
∠RPQ = 90°
In ⊥∆RPQ, ∠P = 90°
RQ = RP² + PQ²
= (7)² + (24)²
= 49 + 576
RQ² = 625
∴ RQ = 25 cm.
Diameter, RQ = 25 cm.
∴ Radius, OR = OQ = \(\frac{25}{2}\) cm.
Area of shaded part :
= Area of semicircle – Area of ∆RPQ

= 161.3 sq.cm.

Question 2.
Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.
Solution:

Radius of small circle, r = 7 cm
Radius of big circle, R = 14 cm
∠AOC = θ = 40°
Area of shaded part, ABCD =?
Area of ABCD
= Area of OAC – Area of OBD

Question 3.
Find the area of the shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
Solution:

Each side of square ABCD = 14 cm.
APD, APC are semicircles.
Area of the shaded region = ?
i) Total area of square
= (Side)²
= (14)²
= 196 sq.cm.

ii) Each radius, r of semicircle
r = \(\frac{14}{2}\) =7cm
Area of two semicircles :

∴ Area of shaded region:
= Area of square – Area of two semicircles
= 196 – 154
= 42 sq. cm

Question 4.
Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.
Solution:

OAB is an equilateral Triangle with side 12 Circle with radius of 6
Each angle of an equilateral triangle is 60°
Angle at the centre in segment, θ = 60°
Area of Major segment + Area of an equilateral triangle OAB =

= 94.28 + 36 × 1.73
= 94.28 + 62.28
= 156.56 sq. cm.

Question 5.
From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in the given figure. Find the area of the remaining portion of the square.
Solution:
i) Each side of square ABCD = 4 cm.

∴ Area of Square:
= (Side)²
= (4)²
= 16 sq.cm.

(ii) Area of circle in the corner A is 1 cm.
Radius, r = 1 cm.
θ = 90°
∴ Area of quadrant of circle:

= 0.78 sq.cm.

iii) Radius of circle having diameter 2 cm, r = 1 cm.
∴ Area of circle:

= 3.14 sq. cm.

∴ Area of remaining part of Square :
= Area of Square – Area of 4 quadrants – Area of Circle.
= 16 – 4 × 0.78 – 3.14
= 16 – 3.12 – 3.14
= 16 – 6.26
= 9.74 sq.cm.

Question 6.
In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the given figure. Find the area of the design.

Solution:
Radius of the circle r = 32 cm.
∴ Area of the circle = πr²

ii) ∠BAC = 60° ∴ ∠BOC = 120°
(∵ Angle at the centre is double than the angle in the circumference).
Now, in ∆OBC, BC ⊥ OD,
∠BOD = ∠COD = 60°

∴ ∆ABC is an equilateral triangle.
Each side of this triangle,
a= 2 × \(16 \sqrt{3}=32 \sqrt{3} \mathrm{cm}\)
Area of equilateral triangle, ∆ABC :

= 1328.64 sq.cm,

iii) Area of shaded region :
= Area of circle – Area of an equilateral ∆^le
= 3218.28 – 1328.64
= 1889.6 sq.cm.

Question 7.
In the Figure given below, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
Solution:

i) Each side of square ABCD =14 cm.
∴ Area of square ABCD
= (Side)²
= (14)²
= 196 sq.cm.

ii) Measure of radii of 4 circles, r = \(\frac{14}{2}=\) = 7 cm.
∵ Distance between tangents which touches circles externally,
d = R + r = 7 + 7 = 14 cm.
Area of segment with centre, A = ?
r = 7 cm, θ = 90°

∴ Total area of 4 sectors
= 4 × 38.5 = 154 sq.cm.

iii) The Area of track :
= (Area of GHIJ – Area of ABCD)
= 196 – 154
= 42 sq.cm.

Question 8.
The Figure given below depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:
i) The distance around the track along its inner edge,
ii) The area of the track.
Solution:

The distance around the track along its inner edge:
= AB + Arc BEC + CD + Area of Arc DFA

The Area of Track:
= (Area of GHIJ – Area of ABCD) + (Area of semicircle HKI – Area of semicircle BEC) + (Area of semicircle GIJ – Area of semicircle AFD)

= 2120 + 2200
= 4320 sq.m.
∴ Total area of Track = 4320 sq.m.

Question 9.
In the figure given below, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Solution:

i) Radius of Big circle is the diameter of a small circle.
∴ Radius of small circle
= \(\frac{7}{2}\) = 3.5 cm
∴ Area of smalle circle = πr²
= \(\frac{22}{7}\) × (3.5)²
= \(\frac{22}{7}\) × 12.25
= 3.14 × 12.25
= 38.47 sq.cm.

ii) Area of segment OCB = ?
Radius, r = 7 cm, θ = 90°

∴ Area of sector OCB – Area of ∆OBC
= 38.47 – 24.50
= 13.97 sq.cm.
Similarly, Area of sector OAC – Area of ∆OAC
= 38.47 – 24.50
= 13.97 sq.cm.
∴ Total area of shaded region :
= Area of small circle + Area of Minor segment OBC + Area of Minor segment OAC
= 38.47 + 13.97 + 13.97
= 66.41 sq.cm.

Question 10.
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see the Figure). Find the area of the shaded region.
(Use π = 314 and \(\sqrt{3}\) = 1.73205).
Solution:

i) Area of an equilateral ∆ABC :
= 17320.5 cm²
\(\frac{\sqrt{3} a^{2}}{4}\) = 17320.5

ii) Area formed by centre A:
r = 100 cm, θ = 60°

Similarly Area of sectors formed by centres B and C is 5233 sq.cm.
∴ Total Area of THREE sectors :
= 3 × 5233 = 15699 sq.cm.
∴ Area of the shaded region:
= Area of an equilateral Triangle – Area of three sectors.
= 17320.5 – 15699
= 1621.5 sq. cm.

Question 11.
On a square handkerchief, nine circular designs each of radius 7 cm are made (see the Figure). Find the area of the remaining portion of the handkerchief.
Solution:

i) Total area of 9 circular designs each of radius 7 cm

ii) All the circles touches externally.
∴ Sum of the diameter of 3 circles in first Row =
14 + 14 + 14 = 42 cm.
∴ Length of each side of square ABCD,
a = 42 cm.
∴ Area of square ABCD = a²
= (42)²
= 1764 sq.cm.
Area of remaining part of handkerchief:
= Area of a square – Area of 9 circles.
= 1764 – 1386
= 378 sq.cm.

Question 12.
In the given figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm If OD = 2 cm, find the area of the
i) quadrant OACB,
ii) shaded region.
Solution:
Radius of Circle, OA = OB = 3.5 cm.
OD = 2 cm.
i) Area of a quadrant of a circle
ii) Area of the shaded region

i) Area of quadrant (OACB) of a Circle
r = 3.5 cm, θ = 90°

ii) In ⊥∆BOD, ∠BOD = 90° OB = 3.5 cm; OD = 2 cm

∴ Area of shaded region,
= Area of quadrant OACB – Area of ABOD
= 9.61 – 3.5
= 6.11 sq.cm.

Question 13.
In Figure (i) given below, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)
Solution:

i) Each side of square OABC = 20 cm.
∴ OA = 20 cm, AB = 20 cm.
∴ Area of square PABC = (Side)²
= (20)²
= 400 sq.cm.

ii) Diagonal OB is drawn in OABC square.
In ⊥∆OAB, OA = 20 cm.
AB = 20 cm.
OB =?
OB² = OA² + AB²
= (20)² + (20)²
= 400 + 400
= 800

∴ Area of shaded region = 628 – 400 = 228 cm².

Question 14.
AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Figure given below). If ∠AOB = 30°, find the area of the shaded region.
Solution:
i) Area of segment OAPB = ?
r = 21 cm, θ = 30°

= \(\frac{1}{12}\) × 3.14 × 49
= 12.82 sq.cm.
∴ Area of the shaded region = Area of Sector OAPB – Area of segment OCQD
= 115.4 – 12.82
= 102.58 sq.cm.

Question 15.
In the given figure (i), ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.
Solution:

i) Area of Part II :
Area of Segment ABC – Area of ∆ABC

= 56 sq.cm.

ii) Area of shaded region Part III :

Question 16.
Calculate the area of the designed region in the given figure common between the two quadrants of circles of radius 8 cm each

Solution:

i) Area of Square, ABCD = a² = (8)² = 64 cm².
ii) Sum of Areas of Part II and Part III = Area of the segment with centre D and radius of 8 cm.

iii) Area of Part I = (Sum of Part I, II and III) – (Sum of the area of Part II, III)
= Area of Square ABCD – (Sum of the area of part II and III)

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.3, drop a comment below and we will get back to you at the earliest.