Pythagoras Theorem Exercise 12.2 – Class 10

Pythagoras Theorem – Exercise 12.2 – Questions:

  1. Verify whether the following measures represent the sides of a right angled triangle.

(i)1, 2, √3

(ii) √2, √3, √5

(iii) 6√3,  12, 6

(iv) m2  – n2, 2mn, m2 + n2

  1. Pythagoras Theorem Exercise 12.2In triangle ∆ABC, a + b = 18 units, b + c = 25 units and c + a = 17 units. What type of triangle is ABC? Give reason.

 

 

3. In ∆ABC CD⊥AB, CA = 2AD and BD = 3AD. Prove that ∠BCA = 90˚

4. Pythagoras Theorem Exercise 12.2The shortest distance AP from a point A to AP is 1 cm. Q and R are respectively 15cm and  20 cm from A and on  opposite  sides  of AP. Prove that ∠QAR = 90˚

 

Pythagoras Theorem Exercise 12.25. In the isosceles ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm. Prove that ∠BAE = 90˚

 

 

 

Pythagoras Theorem Exercise 12.26.In the quadrilateral ABCD, ∠ADC  = 90˚, AB = 9 cm, BC = AD  = 6 cm and CD = 3 cm. Prove that ∠ACB = 90˚

 

 

 

  1. Pythagoras Theorem Exercise 12.2ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2. Prove that ∠APC = 90˚

 

 

 


Pythagoras Theorem – Exercise 12.2 – Solutions:

  1. Verify whether the following measures represent the sides of a right angled triangle.

(i)1, 2, √3

(ii) √2, √3, √5

(iii) 6√3,  12, 6

(iv) m2  – n2, 2mn, m2 + n2

Solution:

(i)1, 2, √3

Sides are: 1, 2, √3

Squares of the sides are: 12, 22, (√3)2

i.e., 1, 4, 3

Sum of areas of squares on the two smaller sides: 1 + 3 = 4

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides 1, 2 and √3 form the sides of a right angled triangle with hypotenuse √3 units , 1 and 3 units as the sides containing the right angle.

 

(ii) √2, √3, √5

Sides are: √2, √3, √5

Squares of the sides are: (√2)2, (√3)2, (√5)2

i.e., 2, 3, 5

Sum of areas of squares on the two smaller sides: 2 + 3 = 5

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides √2, √3 and √5 form the sides of a right angled triangle with hypotenuse √5 units , √2 and √3 units as the sides containing the right angle.

 

(iii) 6√3,  12, 6

Sides are: 6√3,  12, 6

Squares of the sides are: (6√3)2, 122, 62

i.e., 108, 144, 36

Sum of areas of squares on the two smaller sides: 108 + 36 = 144

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides 6√3, 12 and 6 form the sides of a right angled triangle with hypotenuse 12 units , 6√3 and 6 units as the sides containing the right angle.

 

(iv) m2  – n2, 2mn, m2 + n2

Sides are: m2  – n2, 2mn, m2 + n2

Squares of the sides are: (m2  – n2)2, (2mn)2, (m2 + n2)2

i.e., m4 – 2m2n2 + n2, 4m2n2, m4 + 2m2n2 + n2

Sum of areas of squares on the two smaller sides: m4 – 2m2n2 + n2 + m4 + 2m2n2 + n2 = 4m2n2

Therefore,  square on the longest side of the triangle is equal to the sum of squares on the other two sides.

By converse Pythagoras theorem, those two smaller sides must contain a right angle.

Hence the sides m2  – n2, 2mn and m2 + n2 form the sides of a right angled triangle with hypotenuse 2mn units , m2  – n2 and m2 + n2 units as the sides containing the right angle.


  1. Pythagoras Theorem Exercise 12.2In triangle ∆ABC, a + b = 18 units, b + c = 25 units and c + a = 17 units. What type of triangle is ABC? Give reason.

Solution:

a + b = 18————–(1)

b + c = 25————-(2)

c + a = 17————-(3)

Adding (1), (2) and (3), we get,

a + b + b + c + c + a = 17 + 25 + 18

2a + 2b + 2c = 60

a + b + c = 30 ————(*)

From (1), we have

18 + c = 30

c = 30 – 18 = 12

From (2) in (*) we have,

a + 25 = 30

a = 30 – 25

a = 5

Substitute the value of a and c in (*)

5 + b + 12 = 30

b = 30 – 12 – 5 = 13

Now, square on the longest side of the triangle is equal to the sum of squares on the other two sides.

a2 + c2 = b2

52 + 122 = 132

25 + 144 = 169

169 = 169


  1. Pythagoras Theorem Exercise 12.2In ∆ABC CD⊥AB, CA = 2AD and BD = 3AD. Prove that ∠BCA = 90˚

Solution:

Given, in ∆ABC CD⊥AB, CA = 2AD and BD = 3AD

We need to prove ∠BCA = 90˚

In ∆CDA, by Pythagoras theorem,

CA2 = CD2 + DA2

(2m)2 = (CD)2 + m2

4m2 = CD2 + m2

4m2 – m2 = CD2

3m2 = CD2

CD = √3m

In ∆CBD, by Pythagoras Theorem,

BC2 = BD2 + CD2

BC2 = (3m)2 + (√3m)2

BC2 = 9m2 + 3m2

BC2 = 12m2

BC = √12.m

In ∆BCA, by Pythagoras Theorem,

BA2 = BC2 + AC2

(4m)2 = (√12m)2 + (2m)2

16m2 = 12m2 + 4m2

16m2 = 16m2

LHS = RHS

Thus, ∠BCA = 90˚


  1. Pythagoras Theorem Exercise 12.2The shortest distance AP from a point A to AP is 1 cm. Q and R are respectively 15cm and  20 cm from A and on  opposite  sides  of AP. Prove that ∠QAR = 90˚

Solution:

AP = 12 cm

AQ = 15 cm

AR  = 20 cm

In ∆QAR,

QR2 = QA2  + AR2

= 152 + 202

= 225 + 400

= 625

Therefore, QR = 25.

In  triangle QAR, by Pythagoras theorem,

QR2 = QA2 + RA2

252 = 152 + 202

625 =  225 + 400

625 = 625


  1. In the isosceles ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm. Prove that ∠BAE = 90˚

Solution:

Given, ∆ABC , AB = AC, BC = 18 cm , AD⊥BC, AD = 12 cm , BC is produced to E and AE = 20cm.

To prove ∠BAE = 90˚

Proof:

In triangle ADE, by Pythagoras Theorem,

AE2 = AD2 + DE2

202 = 122 + DE2

DE2 = 202 – 122

DE2 = 400 – 144

DE2 = 256

DE = 16 cm

In triangle ADB, by Pythagoras Theorem,

AB2 = BD2 + DA2

AB2 = 92 + 122

AB2 = 81 + 144

AB2 = 225

AB = 15 cm

In triangle ABE, by Pythagoras theorem,

BE2 = AB2 + AE2

252 = 152 + 202

625 = 225 + 400

625 = 625

Therefore, ∠BAE =  90˚


Pythagoras Theorem Exercise 12.26.In the quadrilateral ABCD, ∠ADC  = 90˚, AB = 9 cm, BC = AD  = 6 cm and CD = 3 cm. Prove that ∠ACB = 90˚

Solution:

In triangle ADC, Pythagoras theorem,

AC2 = AD2 + DC2

AC2 = 62 + 32

= 36 + 9

= 45

AC = √45

In triangle ABC ,by  Pythagoras theorem,

AB2 = BC2 + AC2

92 = 45 + 36

81 = 81

Therefore, ∠ACB =  90˚


  1. Pythagoras Theorem Exercise 12.2ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2. Prove that ∠APC = 90˚

Solution:

Given, ABCD is a rectangle. P is any point outside it such that PA2 + PC2 = BA2 + AD2

Construction: Join  AC

Proof:

In triangle ABC

AC2 = AB2 + BC2

AC2 = AB2 + AD2 (BC = AD) …….(1)

Given:

(PA)2+ PC2 = BA2 + AD2 ………….(2)

From (1) and (2),

AC2 = PA2 + PC2

Therefore, ∠APC =  90˚