Pythagoras Theorem — Exercise 10

In △ABC, ∠C = 90°.

If BC = a, AC = b and AB = c, find :

(i) c when a = 8 cm and b = 6 cm

(ii) a when c = 25 cm and b = 7 cm

(iii) b when c = 13 cm and a = 5 cm

Answer

By Pythagoras theorem,

⇒ Hypotenuse² = Perpendicular² + Base²

In △ ABC,

⇒ AB² = AC² + BC²

Given, BC = a, AC = b and AB = c

⇒ c² = b² + a²

(i) Given,

a = 8 cm and b = 6 cm

⇒ c² = 6² + 8²

⇒ c² = 36 + 64

⇒ c² = 100

⇒ c = $\sqrt{100}$

⇒ c = 10 cm

Hence, c = 10 cm.

(ii) Given,

c = 25 cm and b = 7 cm

⇒ 25² = 7² + a²

⇒ 625 = 49 + a²

⇒ a² = 625 - 49

⇒ a² = 576

⇒ a = $\sqrt{576}$

⇒ a = 24 cm

Hence, a = 24 cm.

(iii) Given,

c = 13 cm and a = 5 cm

⇒ 13² = b² + 5²

⇒ 169 = b² + 25

⇒ b² = 169 - 25

⇒ b² = 144

⇒ b = $\sqrt{144}$

⇒ b = 12 cm

Hence, b = 12 cm.

Length of the sides of triangles are given. Determine which of them is a right angled triangle. In case of right angled triangle determine the hypotences.

(i) 5 cm, 4 cm, 3 cm

(ii) 10 cm, 15 cm, 13 cm

(iii) 60 cm, 80 cm, 100 cm

Answer

Choose the greatest length. Check whether the square of greatest length is equal to the sum of squares of other two lengths.

(i) Here greatest length is 5 cm and other lengths are 3 cm, 4 cm.

Note that 5² = 25 and 3² + 4² = 9 + 16 = 25.

Thus, 5² = 3² + 4²

Hence, the triangle with given lengths of sides is a right triangle.

(ii) Here greatest length is 15 cm and other lengths are 10 cm, 13 cm.

Note that 15² = 225 and 10² + 13² = 100 + 169 = 269.

Thus, 225 ≠ 269

Hence, the triangle with given lengths of sides is not a right triangle.

(iii) Here greatest length is 100 cm and other lengths are 80 cm, 60 cm.

Note that 100² = 10000 and 80² + 60² = 6400 + 3600 = 10000

Thus, 10000 = 10000

Hence, the triangle with given lengths of sides is a right triangle.

A rectangular field is 40 m long and 30 m broad. Find the length of its diagonal.

Answer

Let, diagonal of the rectangle be BD.

BC = 30 m and CD = 40 m

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In right-angled triangle BCD,

⇒ BD² = BC² + CD²

⇒ BD² = 30² + 40²

⇒ BD² = 900 + 1600

⇒ BD² = 2500

⇒ BD = $\sqrt{2500}$

⇒ BD = 50 m

Hence, the length of its diagonal is 50 m.

A man goes 15 m due west and then 8 m due north. How far is he from the starting point?

Answer

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AC² = BC² + AB²

⇒ AC² = 8² + 15²

⇒ AC² = 64 + 225

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 m.

Hence, the man is 17 m far from the starting point.

A ladder 17 m long reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.

Answer

Let AB be the ladder, BC be the building and B be the window.

Then, AB = 17 m, BC = 15 m

AC be the distance of the foot of the ladder from the building.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = BC² + AC²

⇒ 17² = 15² + AC²

⇒ 289 = 225 + AC²

⇒ AC² = 289 - 225

⇒ AC² = 64

⇒ AC = $\sqrt{64}$

⇒ AC = 8 m.

Hence, the distance of the foot of the ladder from the building is 8 m.

A ladder 13 m long rests against a vertical wall. If the foot of the ladder is 5 m from the foot of the wall, find the distance of the other end of the ladder from the ground.

Answer

Let AB be the ladder, BC be the vertical wall and AC be the distance of the foot of the ladder from the wall.

Then, AB = 13 m, AC = 5 m

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ABC,

⇒ AB² = BC² + AC²

⇒ 13² = BC² + 5²

⇒ 169 = BC² + 25

⇒ BC² = 169 - 25

⇒ BC² = 144

⇒ BC = $\sqrt{144}$

⇒ BC = 12 m.

Hence, the distance of the other end of the ladder from the ground is 12 m.

A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

Answer

From figure,

Let width of the street be AB = AC + BC

Let CD and CE be the ladder at different positions.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ADC,

⇒ CD² = AD² + AC²

⇒ 15² = 9² + AC²

⇒ 225 = 81 + AC²

⇒ AC² = 225 - 81

⇒ AC² = 144

⇒ AC = $\sqrt{144}$

⇒ AC = 12 m

In triangle BCE,

⇒ CE² = BE² + BC²

⇒ 15² = 12² + BC²

⇒ 225 = 144 + BC²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 m

AB = AC + BC = 12 + 9 = 21 m.

Hence, the width of the street is 21 m.

In the given figure, ABCD is a quadrilateral in which BC = 3 cm, AD = 13 cm, DC = 12 cm and ∠ABD = ∠BCD = 90°. Calculate the length of AB.

Answer

In right angled triangle BCD,

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle BCD,

⇒ BD² = CD² + BC²

⇒ BD² = 12² + 3²

⇒ BD² = 144 + 9

⇒ BD² = 153

⇒ BD = $\sqrt{153}$ cm.

In triangle ABD,

⇒ AD² = AB² + BD²

⇒ 13² = AB² + ($\sqrt{153}$)²

⇒ 169 = AB² + 153

⇒ AB² = 169 - 153

⇒ AB² = 16

⇒ AB = $\sqrt{16}$

⇒ AB = 4 cm.

Hence, length of AB = 4 cm.

In the given figure, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm, calculate the length of PR.

Answer

In right angled triangle PQS,

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = PS² + QS²

⇒ 10² = PS² + 6²

⇒ 100 = PS² + 36

⇒ PS² = 100 - 36

⇒ PS² = 64

⇒ PS = $\sqrt{64}$

⇒ PS = 8 cm.

From figure,

RS = RQ + QS = 9 + 6 = 15 cm

In right angled △ PRS,

Hypotenuse² = Perpendicular² + Base²

⇒ PR² = RS² + PS²

⇒ PR² = 15² + 8²

⇒ PR² = 225 + 64

⇒ PR² = 289

⇒ PR = $\sqrt{289}$

⇒ PR = 17 cm.

Hence, the length of PR = 17 cm.

In a rhombus PQRS, side PQ = 17 cm and diagonal PR = 16 cm. Calculate the area of the rhombus.

Answer

We know that,

The diagonals of a rhombus bisect each other at right angles.

From figure,

OP = OR = 8 cm

PQ = 17 cm

In right angled △ POQ,

Hypotenuse² = Perpendicular² + Base²

⇒ PQ² = OP² + OQ²

⇒ 17² = 8² + OQ²

⇒ 289 = 64 + OQ²

⇒ OQ² = 289 - 64

⇒ OQ² = 225

⇒ OQ = $\sqrt{225}$

⇒ OQ = 15 cm.

Since, OQ = OS = 15 cm (the diagonals of a rhombus bisect each other)

QS = OQ + OS = 15 + 15 = 30 cm

∴ Diagonals are PR = 16 cm and QS = 30 cm.

As we know,

Area of rhombus = $\dfrac{1}{2}$ × Product of diagonals

= $\dfrac{1}{2} \times (30 × 16) = \dfrac{1}{2} \times 480$ = 240 cm².

Hence, the area of rhombus = 240 cm².

From the given figure, find the area of trapezium ABCD.

Answer

From figure,

AB || DC

Distance between parallel sides, AE = BC = 4 cm

In right angled △ AED,

Hypotenuse² = Perpendicular² + Base²

⇒ AD² = AE² + ED²

⇒ 5² = 4² + ED²

⇒ 25 = 16 + ED²

⇒ ED² = 25 - 16

⇒ ED² = 9

⇒ ED = $\sqrt{9}$

⇒ ED = 3 cm.

∴ DC = EC - ED = 5 - 3 = 2 cm

As we know,

Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides} \times \text{distance between them})$

$= \dfrac{1}{2} \times (\text{AB + DC}) \times \text{BC} \\[1em] = \dfrac{1}{2} \times (5 + 2) \times 4 \\[1em] = \dfrac{1}{2} \times 7 \times 4 \\[1em] = 7 \times 2 \\[1em] = 14 \text{ cm}^2.$

Hence, the area of trapezium ABCD is 14 cm².

The sides of a right triangle containing the right angle are 5x cm, (3x - 1) cm. If the area of the triangle be $60 \text{ cm}^2$, calculate the length of the sides of the triangle.

Answer

Consider △ABC as a right angled triangle.

AB = 5x cm and BC = (3x - 1) cm

We know that,

Area of △ABC = ${1}{2}$ × base × height = $\dfrac{1}{2}$ × BC × AB

Substituting the values we get,

⇒ 60 = $\dfrac{1}{2}$ × (3x - 1) × 5x

⇒ 120 = 5x(3x - 1)

⇒ 120 = 15x² - 5x

⇒ 15x² - 5x - 120 = 0

⇒ 5(3x² - x - 24) = 0

⇒ 3x² - x - 24 = 0

⇒ 3x² - 9x + 8x - 24 = 0

⇒ 3x(x - 3) + 8(x - 3) = 0

⇒ (3x + 8)(x - 3) = 0

⇒ 3x + 8 = 0 or x - 3 = 0

⇒ 3x = -8 or x = 3

⇒ x = $-\dfrac{8}{3}$ or x = 3.

Since, x cannot be negative as length of a side cannot be negative. So, x = 3.

AB = 5 × 3 = 15 cm

BC = (3 × 3 - 1) = 9 - 1 = 8 cm

In right angled △ABC,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AC² = AB² + BC²

Substituting the values we get,

⇒ AC² = 15² + 8²

⇒ AC² = 225 + 64

⇒ AC² = 289

⇒ AC = $\sqrt{289}$

⇒ AC = 17 cm.

Hence, the length of the sides of the triangle is 17 cm, 15 cm and 8 cm.

Find the altitude of an equilateral triangle of side 5 $\sqrt{3}$ cm.

Answer

In △ ABC,

Draw altitude AD perpendicular to BC.

In an equilateral triangle, the altitude also acts as the median (bisecting the base).

∴ BD = $\dfrac{1}{2}$ × BC = $\dfrac{1}{2} \times 5\sqrt{3} = \dfrac{5\sqrt{3}}{2}$

In right angled △ABD,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AD² + BD²

$\Rightarrow (5\sqrt{3})^2 = \text{AD}^2 + \Big(\dfrac{5\sqrt{3}}{2}\Big)^2 \\[1em] \Rightarrow 25 \times 3 = \text{AD}^2 + \dfrac{25 \times 3}{4} \\[1em] \Rightarrow 75 = \text{AD}^2 + \dfrac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = 75 - \dfrac{75}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{300 - 75}{4} \\[1em] \Rightarrow \text{AD}^2 = \dfrac{225}{4} \\[1em] \Rightarrow \text{AD} = \sqrt{\dfrac{225}{4}} \\[1em] \Rightarrow \text{AD} = \dfrac{15}{2} \\[1em] \Rightarrow \text{AD} = 7.5 \text{ cm}.$

Hence, the altitude is 7.5 cm.

In a rhombus ABCD, prove that AC² + BD² = 4AB².

Answer

In rhombus, AB = BC = CD = AD (All sides of a rhombus are equal)

Diagonal AC and BD intersect at point O.

Diagonals bisect each other at right angles.

So, AO = OC = $\dfrac{1}{2}$ × AC and BO = OD = $\dfrac{1}{2}$ × BD

In right angled △AOB,

Using Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

⇒ AB² = AO² + BO²

$\Rightarrow \text{AB}^2 = \Big(\dfrac{\text{AC}}{2}\Big)^2 + \Big(\dfrac{\text{BD}}{2}\Big)^2 \\[1em] \Rightarrow \text{AB}^2 = \dfrac{\text{AC}^2}{4} + \dfrac{\text{BD}^2}{4} \\[1em] \Rightarrow \text{AB}^2 = \dfrac{\text{AC}^2 + \text{BD}^2}{4}$

∴ 4AB² = AC² + BD²

Hence, proved that AC² + BD² = 4AB².

In △ABC, ∠B = 90° and D is the mid-point of BC. Prove that

(i) AC² = AD² + 3 CD²

(ii) BC² = 4 (AD² - AB²)

Answer

(i) Given,

ABC is a triangle such that ∠ABC = 90°. D is the mid-point of BC.

In △ ABD, using Pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AB² + BD² ......(1)

Similarly, in △ ABC,

⇒ AC² = AB² + BC²

⇒ AC² = AB² + (BD + DC)²

⇒ AC² = AB² + BD² + DC² + 2 × BD × DC

As D is the midpoint of BC, BD = DC.

⇒ AC² = AB² + BD² + CD² + 2 × CD × CD

⇒ AC² = AB² + BD² + CD² + 2CD²

⇒ AC² = AB² + BD² + 3CD²

Using equation (1), we get

⇒ AC² = AD² + 3CD²

Hence, proved that AC² = AD² + 3 CD².

(ii) In △ ABD,

⇒ AD² = AB² + BD²

BD = $\dfrac{\text{BC}}{2}$

$\Rightarrow \text{AD}^2 = \text{AB}^2 + \Big(\dfrac{\text{BC}}{2}\Big)^2 \\[1em] \Rightarrow \text{AD}^2 = \text{AB}^2 + \dfrac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{AD}^2 - \text{AB}^2 = \dfrac{\text{BC}^2}{4} \\[1em] \Rightarrow \text{BC}^2 = 4 (\text{AD}^2 - \text{AB}^2)$

Hence, proved that BC² = 4 (AD² - AB²).

Two poles of height 9 m and 14 m stand vertically on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.

Answer

Let AB be the smaller pole and CD the bigger pole.

From figure,

⇒ CE = AB = 9 m and AE = BC = 12 m

⇒ CD = CE + ED

⇒ 14 = 9 + ED

⇒ ED = 14 - 9

⇒ ED = 5 m

From figure,

△ADE is right angled triangle.

By pythagoras theorem,

Hypotenuse² = Base² + Height²

⇒ AD² = AE² + ED²

⇒ AD² = (12)² + (5)²

⇒ AD² = 144 + 25

⇒ AD² = 169

⇒ AD = $\sqrt{169}$

⇒ AD = 13 m.

Hence, the distance between their tops is 13 m.

In △ABC, if AB = AC and D is a point on BC. Prove that AB² - AD² = BD × CD.

Answer

ABC is a triangle in which AB = AC and D is any point on BC.

In △ ABE and △ ACE,

⇒ AB = AC (Given)

⇒ AE = AE (Common)

⇒ ∠AEB = ∠AEC (both are 90°)

Using RHS congruency criterion,

△ ABE ≅ △ ACE

⇒ BE = CE (by C.P.C.T.)

In △ ABE, using Pythagorean theorem,

Hypotenuse² = Base² + Height²

⇒ AB² = AE² + BE² ......(1)

In △ ADE, using Pythagorean theorem,

⇒ AD² = AE² + DE² ......(2)

Subtracting equation (2) from (1), we get:

⇒ AB² - AD² = (AE² + BE²) - (AE² + DE²)

⇒ AB² - AD² = AE² + BE² - AE² - DE²

⇒ AB² - AD² = BE² - DE²

⇒ AB² - AD² = (BE - DE)(BE + DE)

⇒ AB² - AD² = (BE - DE)(CE + DE) [∴ BE = CE]

⇒ AB² - AD² = BD × CD

Hence, proved that AB² - AD² = BD × CD.