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) m2 – n2, 2mn, m2 + n2
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 PA2 + PC2 = BA2 + AD2. 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) 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.
- 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.
a2 + c2 = b2
52 + 122 = 132
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,
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˚
- 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,
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
- 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˚
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,
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˚
- ABCD 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˚
