## Pythagoras Theorem – Exercise 12.2 – Questions:

- 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) m^{2} – n^{2}, 2mn, m^{2} + n^{2}

- In 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. The 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˚

5. 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˚

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

- ABCD is a rectangle. P is any point outside it such that PA
^{2}+ PC^{2}= BA^{2}+ AD^{2}. Prove that ∠APC = 90˚

## Pythagoras Theorem – Exercise 12.2 – Solutions:

**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) m ^{2} – n^{2}, 2mn, m^{2} + n^{2}**

Solution:

(i)1, 2, √3

Sides are: 1, 2, √3

Squares of the sides are: 1^{2}, 2^{2}, (√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}, 12^{2}, 6^{2}

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) m^{2} – n^{2}, 2mn, m^{2} + n^{2}

Sides are: m^{2} – n^{2}, 2mn, m^{2} + n^{2}

Squares of the sides are: (m^{2} – n^{2})^{2}, (2mn)^{2}, (m^{2} + n^{2})^{2}

i.e., m^{4} – 2m^{2}n^{2} + n^{2}, 4m^{2}n^{2}, m^{4} + 2m^{2}n^{2} + n^{2}

Sum of areas of squares on the two smaller sides: m^{4} – 2m^{2}n^{2} + n^{2} + m^{4} + 2m^{2}n^{2} + n^{2} = 4m^{2}n^{2}

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

Hence the sides m^{2} – n^{2}, 2mn and m^{2} + n^{2} form the sides of a right angled triangle with hypotenuse 2mn units , m^{2} – n^{2} and m^{2} + n^{2} units as the sides containing the right angle.

**In 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.

a^{2} + c^{2} = b^{2}

5^{2} + 12^{2} = 13^{2}

25 + 144 = 169

169 = 169

**In ∆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,

CA^{2} = CD^{2} + DA^{2}

(2m)^{2} = (CD)^{2} + m^{2}

4m^{2} = CD^{2} + m^{2}

4m^{2} – m^{2} = CD^{2}

3m^{2} = CD^{2}

CD = √3m

In ∆CBD, by Pythagoras Theorem,

BC^{2} = BD^{2} + CD^{2}

BC^{2} = (3m)^{2} + (√3m)^{2}

BC^{2} = 9m^{2} + 3m^{2}

BC^{2} = 12m^{2}

BC = √12.m

In ∆BCA, by Pythagoras Theorem,

BA^{2} = BC^{2} + AC^{2}

(4m)^{2} = (√12m)^{2} + (2m)^{2}

16m^{2} = 12m^{2} + 4m^{2}

16m^{2} = 16m^{2}

LHS = RHS

Thus, ∠BCA = 90˚

**The 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,

QR^{2} = QA^{2} + AR^{2}

= 15^{2} + 20^{2}

= 225 + 400

= 625

Therefore, QR = 25.

In triangle QAR, by Pythagoras theorem,

QR^{2} = QA^{2} + RA^{2}

25^{2} = 15^{2} + 20^{2}

625 = 225 + 400

625 = 625

**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,

AE^{2} = AD^{2} + DE^{2}

20^{2} = 12^{2} + DE^{2}

DE^{2} = 20^{2} – 12^{2}

DE^{2} = 400 – 144

DE^{2} = 256

DE = 16 cm

In triangle ADB, by Pythagoras Theorem,

AB^{2} = BD^{2} + DA^{2}

AB^{2} = 9^{2} + 12^{2}

AB^{2 }= 81 + 144

AB^{2 }= 225

AB = 15 cm

In triangle ABE, by Pythagoras theorem,

BE^{2} = AB^{2} + AE^{2}

25^{2} = 15^{2} + 20^{2}

625 = 225 + 400

625 = 625

Therefore, ∠BAE = 90˚

**6.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,

AC^{2} = AD^{2} + DC^{2}

AC^{2} = 6^{2} + 3^{2}

= 36 + 9

= 45

AC = √45

In triangle ABC ,by Pythagoras theorem,

AB^{2} = BC^{2} + AC^{2}

9^{2} = 45 + 36

81 = 81

Therefore, ∠ACB = 90˚

**ABCD is a rectangle. P is any point outside it such that PA**^{2}+ PC^{2}= BA^{2}+ AD^{2}. Prove that ∠APC = 90˚

Solution:

Given, ABCD is a rectangle. P is any point outside it such that PA^{2} + PC^{2} = BA^{2} + AD^{2}

Construction: Join AC

Proof:

In triangle ABC

AC^{2} = AB^{2} + BC^{2}

AC^{2} = AB^{2} + AD^{2} (BC = AD) …….(1)

Given:

(PA)^{2}+ PC^{2} = BA^{2} + AD^{2} ………….(2)

From (1) and (2),

AC^{2} = PA^{2} + PC^{2}

Therefore, ∠APC = 90˚