## Pythagoras Theorem – Exercise 12.1 – Questions:

a. Numerical problems based on Pythagoras theorem.

- The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.
- Find the length often diagonal of a square of side 12 cm.
- The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side,
- In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.
- A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.
- A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly tp catch it. Of both poses the same speed, how far from the pillar they are going to meet?

b. Riders based on Pythagoras theorem.

- In triangle MGN, MP⊥GN. If MG = a units, MN = b units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)
- In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC = a units, BD = p units, CA = b units. Prove that

^{1}/_{a^2} + ^{1}/_{c^2} = ^{1}/_{p^2}

- Derive the formula for height and area of an equilateral triangle.

## Pythagoras Theorem – Exercise 12.1 – Solution:

**a. Numerical problems based on Pythagoras theorem.**

**The sides of a right angled triangle containing the right angle are 5 cm and 12 cm, find its hypotenuse.**

Solution:

In triangle ABC,

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

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

= 144 + 25

= 169

AC = √169 = 13 cm

**Find the length often diagonal of a square of side 12 cm.**

Solution:

By Baudhayana Theorem to squares,

square of the diagonal = 2 x square of length of its sides

AC^{2} = 2 x AB^{2}

= 2 x 12^{2}

AC = √(2×12^{2}) = 12√2 cm

**The length of the diagonal of a rectangular playground is 125m and the length of one side is 75m. find the length of the other side.**

Solution:

BD = 125m and AD = 75 m

In triangle ABD, by pythagoras theorem,

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

125^{2} = 75^{2} + AB^{2}

15625 = 5625 + AB^{2}

15625 – 5625 = AB^{2}

10000 = AB^{2}

AB = 100 m

**In triangle LAW, LAW = 90, LNA = 90 , LW = 26cm, LN = 6 cm and AN = 8 cm. Calculate the length of WA.**

Solution:

Given LW = 26 cm, NA = 8 cm and LN = 6 cm

In triangle LNA, by Pythagoras theorem,

LA^{2} = LN^{2} + NA^{2}

LA^{2} = 36 + 64

= 100

LA = 10 cm

In triangle LAW, by Pythagoras theorem

26^{2} = 10^{2} + WA^{2}

676 = 100 + WA^{2}

576 = WA^{2}

WA = √576 = 24 cm

**A door of width 6 meter has an arch above it having a height of 2m. Find the radius of the arch.**

Solution:

LetQ be the centre o the arch. join AQ and PQ.

Let PQ = x cm and AQ = QB = x + 2

In triangle QAP, by Pythagoras theorem,

AP^{2} = AQ^{2} + PQ^{2}

(x + 2)^{2} = x^{2} + 3^{2}

x^{2} + 4x + 4 = x^{2} + 9

4x + 4 = 9

4x = 9 – 4 = 5

x = ^{5}/_{4} = 1.25 cm

Therefore, radius of the arch = x + 2 = 1.25 + 2 = 3.25 cm

**A peacock on a pillar of 9 feet height on seeing a snake coming towards its hole situated just below the pillar will fly to catch it. Of both poses the same speed, how far from the pillar they are going to meet?**

Solution:

Given, height of the pillar PQ = 9 feet and distance of QS = 27 feet

Therefore, RQ = 27 – x

In triangle PQR, by Pythagoras theorem,

PR^{2} = PQ^{2} + QR^{2}

x^{2} = 9^{2} + (27 – x)^{2}

x^{2} = 81 + 729 – 54x + x^{2}

54x = 81 + 729 = 810

x = ^{810}/_{54} = 15 feet

Therefore QR = 27 – x = 27 – 15 = 12 feet.

**b.Riders based on Pythagoras theorem.**

**In triangle MGN, MP⊥GN. If MG = a units, MN = b units, GP = c units and PN = d units. Prove that (a + b)(a – b) = (c + d)(c – d)**

Solution:

Given, in triangle MGN, MP⊥GN. If MG = a units, MN = b units, GP = c units and PN = d units.

In triangle MPG, by Pythagoras theorem,

MG = MP + GP

a = MP + c ————(1)

In triangle MPN, by Pythagoras theorem,

MN = MP + PN

b = MP + d —————(2)

subtract (1) from (2),

a^{2} – b^{2} = (MP + c)^{2} – (MP + d)^{2}

a^{2} – b^{2} = MP^{2} – MP^{2} + c^{2} – d^{2}

a^{2} – b^{2} = c^{2} – d^{2}

(a + b)(a – b) = (c + d)(c – d)

**In right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC = a units, BD = p units, CA = b units. Prove that**

^{1}/_{a^2} + ^{1}/_{c^2} = ^{1}/_{p^2}

Solution:

Given, in right angled triangle ABC, ABC = 90, BD ⊥ AC. If AB = c units, BC = a units, BD = p units, CA = b units.

Proof:

BD^{2} = AD x DC

p^{2} = a x DC

BC^{2} = AC x DC

a^{2} = b x DC

AB^{2} = AC x AD

c^{2} = b x AD

Now, ^{1}/_{a^2} + ^{1}/_{c^2} = ^{1}/_{p^2}

= ^{1}/_{b} [ ^{1}/_{DC} + ^{1}/_{AD}]

= ^{1}/_{b} [ ^{AD + DC}/_{DC.AD}]

= ^{1}/_{b} [ ^{AC}/_{p^2}]

= ^{1}/_{b} [^{b}/_{p^2}]

= ^{1}/_{p^2}

**Derive the formula for height and area of an equilateral triangle.**

Solution:

AB=AC=a, BP=PC=^{a}∕_{2}, AP=h,

AP perpendicular to BC.

PROOF:

In triangle APC, by Pythagoras Theorem,

AC^{2 }= AP^{2 }+ AC^{2}

a^{2 } = h^{2} + (^{a}∕_{2} )^{ 2}

a^{2} = h^{2} + (^{a}∕_{4})^{ 2}

a ^{2}– (^{a}∕_{4} )^{2 }= h^{2}

4a^{2} – ^{a^2}∕_{4} = h^{2}

^{3a^2}∕_{4} = h

h = √(^{3}^{𝑎^2}⁄_{4}) = ^{a}/_{2}√3

Now area of triangle = ^{1}/_{2} x b x h

= ^{1}/_{2} x a x ^{a}/_{2} √3

= √3 ^{a}/_{4}