Mathematics

Class 92025–26 SNC23 Questions SolvedEnglish Medium

Exercise 6.5 — Class 9 Mathematics Solutions

All 23 questions below are worked step by step — every step clearly worked out with mathematical reasons — for the Punjab Board 9th class Mathematics textbook.

Exercise 6.5

9 Questions (14 Sub-parts)

Quick Solution Step by Step

Every step explained, with hints and the method used.

Question 1

EasyUnder Review

Find the values of $x$ , $y$ and $z$ from the following right-angled triangles.

Hint

Identify the hypotenuse, perpendicular and base relative to the given angle, then use the trigonometric ratio that links the known side to each unknown side.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q1Part (i)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = 30^{\circ}$ . The base $A B = y = 4 cm$ , the side $B C = x$ is opposite $∠ A$ , and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Hint

Use $\tan 30^{\circ}$ for the opposite side and $\cos 30^{\circ}$ for the hypotenuse.

Solution

Find $x$ using tangent.

Relative to $∠ A$ , $B C = x$ is the perpendicular and $A B = 4$ cm is the base, so $\tan θ = \frac{Perpendicular}{Base}$ .
$\begin{matrix} \tan 30^{\circ} & = \frac{x}{4} \\ \frac{1}{\sqrt{3}} & = \frac{x}{4} \\ x & = \frac{4}{\sqrt{3}} \\ = \frac{4 \sqrt{3}}{3} \\ \approx 2.31 cm \end{matrix}$
Find $z$ using cosine.

The base $A B$ and hypotenuse $A C$ are linked by $\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 30^{\circ} & = \frac{4}{z} \\ \frac{\sqrt{3}}{2} & = \frac{4}{z} \\ \sqrt{3} z & = 8 \\ z & = \frac{8}{\sqrt{3}} \\ = \frac{8 \sqrt{3}}{3} \\ \approx 4.62 cm \end{matrix}$

Answer

$x = \frac{4 \sqrt{3}}{3} \approx 2.31 cm, z = \frac{8 \sqrt{3}}{3} \approx 4.62 cm$

Quick Solution

Q1Part (ii)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = \frac{π}{4}$ . The side $B C = y = \sqrt{3} cm$ is opposite $∠ A$ , the base is $A B = x$ and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Hint

Change $\frac{π}{4}$ to $45^{\circ}$ , then use $\tan 45^{\circ}$ and $\sin 45^{\circ}$ .

Solution

Convert the angle to degrees.

Since $π$ radians $= 180^{\circ}$ :
$\begin{matrix} \frac{π}{4} & = \frac{180^{\circ}}{4} \\ = 45^{\circ} \end{matrix}$
Find $x$ using tangent.

Relative to $∠ A$ , $B C$ is the perpendicular and $A B = x$ is the base, so use $\tan θ = \frac{Perpendicular}{Base}$ .
$\begin{matrix} \tan 45^{\circ} & = \frac{\sqrt{3}}{x} \\ 1 & = \frac{\sqrt{3}}{x} \\ x & = \sqrt{3} cm \end{matrix}$
Find $z$ using sine.

The perpendicular and hypotenuse are linked by $\sin θ = \frac{Perpendicular}{Hypotenuse}$ .
$\begin{matrix} \sin 45^{\circ} & = \frac{\sqrt{3}}{z} \\ \frac{1}{\sqrt{2}} & = \frac{\sqrt{3}}{z} \\ z & = \sqrt{2} \times \sqrt{3} \\ = \sqrt{6} cm \end{matrix}$

Answer

$x = \sqrt{3} cm, z = \sqrt{6} cm \approx 2.45 cm$

Quick Solution

Q1Part (iii)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ C = 60^{\circ}$ . The hypotenuse is $C A = z = 2 cm$ , the base next to $∠ C$ is $C B = x$ , and the side $A B = y$ is opposite $∠ C$ . Find $x$ and $y$ .

Hint

With the hypotenuse known, use $\cos 60^{\circ}$ for the adjacent side and $\sin 60^{\circ}$ for the opposite side.

Solution

Find $x$ using cosine.

Relative to $∠ C$ , $C B = x$ is the base and $C A$ is the hypotenuse, so $\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 60^{\circ} & = \frac{x}{2} \\ \frac{1}{2} & = \frac{x}{2} \\ x & = 1 cm \end{matrix}$
Find $y$ using sine.

$A B = y$ is the perpendicular, so $\sin θ = \frac{Perpendicular}{Hypotenuse}$ .
$\begin{matrix} \sin 60^{\circ} & = \frac{y}{2} \\ \frac{\sqrt{3}}{2} & = \frac{y}{2} \\ y & = \sqrt{3} cm \end{matrix}$

Answer

$x = 1 cm, y = \sqrt{3} cm \approx 1.73 cm$

Quick Solution

Q1Part (iv)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ and $m ∠ A = 45^{\circ}$ . The base is $A B = y = 4 cm$ , the side $B C = x$ is opposite $∠ A$ , and the hypotenuse is $A C = z$ . Find $x$ and $z$ .

Hint

Use $\tan 45^{\circ} = 1$ for $x$ and $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$ for $z$ .

Solution

Find $x$ using tangent.

Relative to $∠ A$ , $\tan θ = \frac{Perpendicular}{Base}$ .
$\begin{matrix} \tan 45^{\circ} & = \frac{x}{4} \\ 1 & = \frac{x}{4} \\ x & = 4 cm \end{matrix}$
Find $z$ using cosine.

The base and hypotenuse are linked by $\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 45^{\circ} & = \frac{4}{z} \\ \frac{1}{\sqrt{2}} & = \frac{4}{z} \\ z & = 4 \sqrt{2} cm \\ \approx 5.66 cm \end{matrix}$

Answer

$x = 4 cm, z = 4 \sqrt{2} cm \approx 5.66 cm$

Quick Solution

Question 2

MediumUnder Review

Find the unknown side and angles of the following triangles.

Hint

Use Pythagoras' theorem for the missing side, an inverse trigonometric ratio for one acute angle, and $m ∠ A + m ∠ C = 90^{\circ}$ for the other.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q2Part (i)

MediumUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $a = B C = \sqrt{3} cm$ and $c = A B = \sqrt{13} cm$ . Find the hypotenuse $b = A C$ , $m ∠ A$ and $m ∠ C$ . (Give angles to one decimal place.)

Hint

Find $b$ first, then $\tan A = \frac{a}{c}$ .

Solution

Find $b$ by Pythagoras' theorem.

The hypotenuse is opposite the right angle: $b^{2} = a^{2} + c^{2}$ .
$\begin{matrix} b^{2} & = (\sqrt{3})^{2} + (\sqrt{13})^{2} \\ = 3 + 13 \\ = 16 \\ b & = 4 cm \end{matrix}$
Find $m ∠ A$ using tangent.

Relative to $∠ A$ , $a$ is the perpendicular and $c$ is the base.
$\begin{matrix} \tan A & = \frac{a}{c} \\ = \frac{\sqrt{3}}{\sqrt{13}} \\ \approx 0.4804 \\ m ∠ A & = \tan^{- 1} (0.4804) \\ \approx {25.7}^{\circ} \end{matrix}$
Find $m ∠ C$ .

The two acute angles of a right triangle add up to $90^{\circ}$ .
$\begin{matrix} m ∠ C & = 90^{\circ} - m ∠ A \\ = 90^{\circ} - {25.66}^{\circ} \\ \approx {64.3}^{\circ} \end{matrix}$

Answer

$b = 4 cm, m ∠ A \approx {25.7}^{\circ}, m ∠ C \approx {64.3}^{\circ}$

Quick Solution

Q2Part (ii)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $a = B C = 4 cm$ and $c = A B = 4 cm$ . Find the hypotenuse $b = A C$ , $m ∠ A$ and $m ∠ C$ .

Hint

Equal legs mean $\tan A = 1$ .

Solution

Find $b$ by Pythagoras' theorem.

$b^{2} = a^{2} + c^{2}$ .
$\begin{matrix} b^{2} & = 4^{2} + 4^{2} \\ = 32 \\ b & = \sqrt{16 \times 2} \\ = 4 \sqrt{2} cm \end{matrix}$
Find $m ∠ A$ using tangent.

Relative to $∠ A$ , $a$ is the perpendicular and $c$ is the base.
$\begin{matrix} \tan A & = \frac{a}{c} \\ = \frac{4}{4} \\ = 1 \\ m ∠ A & = 45^{\circ} \end{matrix}$
Find $m ∠ C$ .

The acute angles of a right triangle are complementary.
$\begin{matrix} m ∠ C & = 90^{\circ} - 45^{\circ} \\ = 45^{\circ} \end{matrix}$

Answer

$b = 4 \sqrt{2} cm \approx 5.66 cm, m ∠ A = m ∠ C = 45^{\circ}$

Quick Solution

Question 3

EasyUnder Review

Each side of a square field is $60 m$ long. Find the lengths of the diagonals of the field.

Hint

A diagonal is the hypotenuse of a right triangle whose legs are two sides of the square.

Solution

Set up a right triangle.

Let $A B C D$ be the square field. Diagonal $A C$ cuts it into right-angled $△ A B C$ with $m ∠ B = 90^{\circ}$ , $A B = B C = 60 m$ , and $A C$ as hypotenuse. $A B = B C = 60 m, A C = ?$
Apply Pythagoras' theorem.

$(Hypotenuse)^{2} = (Base)^{2} + (Perpendicular)^{2}$ .
$\begin{matrix} A C^{2} & = 60^{2} + 60^{2} \\ = 3600 + 3600 \\ = 7200 \\ A C & = \sqrt{3600 \times 2} \\ = 60 \sqrt{2} \\ \approx 84.85 m \end{matrix}$
Both diagonals.

The diagonals of a square are equal, so $B D = A C$ . $B D = 60 \sqrt{2} m$

Answer

$Each diagonal = 60 \sqrt{2} m \approx 84.85 m$

Quick Solution

Question 4

MediumUnder Review

Solve the following triangles when $m ∠ B = 90^{\circ}$ . (In $△ A B C$ , $a = B C$ , $b = C A$ and $c = A B$ ; side $b$ is the hypotenuse.)

Hint

Solving a triangle means finding all its unknown sides and angles: use the angle sum for the third angle, then trigonometric ratios (or Pythagoras' theorem) for the sides.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q4Part (i)

EasyUnder Review

$m ∠ C = 60^{\circ}$ , $c = 3 \sqrt{3} cm$

Hint

$c$ is opposite $∠ C$ , so use $\tan 60^{\circ}$ and $\sin 60^{\circ}$ .

Solution

Find $m ∠ A$ .

The angles of a triangle add up to $180^{\circ}$ .
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Find $a$ using tangent of $∠ C$ .

Relative to $∠ C$ , $c = A B$ is the perpendicular and $a = B C$ is the base.
$\begin{matrix} \tan 60^{\circ} & = \frac{c}{a} \\ \sqrt{3} & = \frac{3 \sqrt{3}}{a} \\ a & = 3 cm \end{matrix}$
Find $b$ using sine of $∠ C$ .

$\sin θ = \frac{Perpendicular}{Hypotenuse}$ .
$\begin{matrix} \sin 60^{\circ} & = \frac{c}{b} \\ \frac{\sqrt{3}}{2} & = \frac{3 \sqrt{3}}{b} \\ \sqrt{3} b & = 6 \sqrt{3} \\ b & = 6 cm \end{matrix}$

Answer

$m ∠ A = 30^{\circ}, a = 3 cm, b = 6 cm$

Quick Solution

Q4Part (ii)

EasyUnder Review

$m ∠ C = 45^{\circ}$ , $a = 8 cm$

Hint

$a$ is adjacent to $∠ C$ : use $\tan 45^{\circ}$ for $c$ and $\cos 45^{\circ}$ for $b$ .

Solution

Find $m ∠ A$ .

Angle-sum property of a triangle.
$\begin{matrix} m ∠ A & = 180^{\circ} - 90^{\circ} - 45^{\circ} \\ = 45^{\circ} \end{matrix}$
Find $c$ using tangent of $∠ C$ .

Relative to $∠ C$ , $c$ is the perpendicular and $a$ is the base.
$\begin{matrix} \tan 45^{\circ} & = \frac{c}{8} \\ 1 & = \frac{c}{8} \\ c & = 8 cm \end{matrix}$
Find $b$ using cosine of $∠ C$ .

$\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 45^{\circ} & = \frac{8}{b} \\ \frac{1}{\sqrt{2}} & = \frac{8}{b} \\ b & = 8 \sqrt{2} cm \\ \approx 11.31 cm \end{matrix}$

Answer

$m ∠ A = 45^{\circ}, c = 8 cm, b = 8 \sqrt{2} cm \approx 11.31 cm$

Quick Solution

Q4Part (iii)

MediumUnder Review

$a = 12 cm$ , $c = 6 cm$

Hint

Two legs are given: Pythagoras for $b$ , then $\tan A = \frac{a}{c}$ .

Solution

Find $b$ by Pythagoras' theorem.

$b$ is the hypotenuse: $b^{2} = a^{2} + c^{2}$ .
$\begin{matrix} b^{2} & = 12^{2} + 6^{2} \\ = 144 + 36 \\ = 180 \\ b & = \sqrt{36 \times 5} \\ = 6 \sqrt{5} \\ \approx 13.42 cm \end{matrix}$
Find $m ∠ A$ using tangent.

Relative to $∠ A$ , $a$ is the perpendicular and $c$ is the base.
$\begin{matrix} \tan A & = \frac{12}{6} \\ = 2 \\ m ∠ A & = \tan^{- 1} (2) \\ \approx {63.4}^{\circ} \end{matrix}$
Find $m ∠ C$ .

Acute angles of a right triangle are complementary.
$\begin{matrix} m ∠ C & = 90^{\circ} - {63.43}^{\circ} \\ \approx {26.6}^{\circ} \end{matrix}$

Answer

$b = 6 \sqrt{5} cm \approx 13.42 cm, m ∠ A \approx {63.4}^{\circ}, m ∠ C \approx {26.6}^{\circ}$

Quick Solution

Q4Part (iv)

EasyUnder Review

$m ∠ A = 60^{\circ}$ , $c = 4 cm$

Hint

$c$ is adjacent to $∠ A$ : use $\tan 60^{\circ}$ and $\cos 60^{\circ}$ .

Solution

Find $m ∠ C$ .

Angle-sum property of a triangle.
$\begin{matrix} m ∠ C & = 180^{\circ} - 90^{\circ} - 60^{\circ} \\ = 30^{\circ} \end{matrix}$
Find $a$ using tangent of $∠ A$ .

Relative to $∠ A$ , $a$ is the perpendicular and $c$ is the base.
$\begin{matrix} \tan 60^{\circ} & = \frac{a}{4} \\ \sqrt{3} & = \frac{a}{4} \\ a & = 4 \sqrt{3} \\ \approx 6.93 cm \end{matrix}$
Find $b$ using cosine of $∠ A$ .

$\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 60^{\circ} & = \frac{4}{b} \\ \frac{1}{2} & = \frac{4}{b} \\ b & = 8 cm \end{matrix}$

Answer

$m ∠ C = 30^{\circ}, a = 4 \sqrt{3} cm \approx 6.93 cm, b = 8 cm$

Quick Solution

Q4Part (v)

EasyUnder Review

$m ∠ A = 30^{\circ}$ , $c = 4 cm$

Hint

Use $\tan 30^{\circ}$ for $a$ and $\cos 30^{\circ}$ for $b$ , then rationalise.

Solution

Find $m ∠ C$ .

Angle-sum property of a triangle.
$\begin{matrix} m ∠ C & = 180^{\circ} - 90^{\circ} - 30^{\circ} \\ = 60^{\circ} \end{matrix}$
Find $a$ using tangent of $∠ A$ .

Relative to $∠ A$ , $a$ is the perpendicular and $c$ is the base.
$\begin{matrix} \tan 30^{\circ} & = \frac{a}{4} \\ \frac{1}{\sqrt{3}} & = \frac{a}{4} \\ a & = \frac{4}{\sqrt{3}} \\ = \frac{4 \sqrt{3}}{3} \\ \approx 2.31 cm \end{matrix}$
Find $b$ using cosine of $∠ A$ .

$\cos θ = \frac{Base}{Hypotenuse}$ .
$\begin{matrix} \cos 30^{\circ} & = \frac{4}{b} \\ \frac{\sqrt{3}}{2} & = \frac{4}{b} \\ b & = \frac{8}{\sqrt{3}} \\ = \frac{8 \sqrt{3}}{3} \\ \approx 4.62 cm \end{matrix}$

Answer

$m ∠ C = 60^{\circ}, a = \frac{4 \sqrt{3}}{3} cm \approx 2.31 cm, b = \frac{8 \sqrt{3}}{3} cm \approx 4.62 cm$

Quick Solution

Q4Part (vi)

MediumUnder Review

$b = 10 cm$ , $a = 6 cm$

Hint

Hypotenuse and one leg are given: Pythagoras for $c$ , then $\sin A = \frac{a}{b}$ .

Solution

Find $c$ by Pythagoras' theorem.

$b$ is the hypotenuse, so $c^{2} = b^{2} - a^{2}$ .
$\begin{matrix} c^{2} & = 10^{2} - 6^{2} \\ = 100 - 36 \\ = 64 \\ c & = 8 cm \end{matrix}$
Find $m ∠ A$ using sine.

Relative to $∠ A$ , $a$ is the perpendicular and $b$ is the hypotenuse.
$\begin{matrix} \sin A & = \frac{6}{10} \\ = 0.6 \\ m ∠ A & = \sin^{- 1} (0.6) \\ \approx {36.9}^{\circ} \end{matrix}$
Find $m ∠ C$ .

Acute angles of a right triangle are complementary.
$\begin{matrix} m ∠ C & = 90^{\circ} - {36.87}^{\circ} \\ \approx {53.1}^{\circ} \end{matrix}$

Answer

$c = 8 cm, m ∠ A \approx {36.9}^{\circ}, m ∠ C \approx {53.1}^{\circ}$

Quick Solution

Question 5

MediumUnder Review

Let $Q$ and $R$ be two points on the same bank of a canal. The point $P$ is placed on the other bank, straight across from $R$ , so that $P R$ is perpendicular to the bank and $m ∠ P R Q = 90^{\circ}$ . If $Q R = 5 m$ and $P Q = 13 m$ , find the width of the canal and $m ∠ P Q R$ in radians.

Hint

The width is the leg $P R$ of a right triangle with hypotenuse $13$ m; then use $\cos Q = \frac{5}{13}$ .

Solution

Find the width $P R$ by Pythagoras' theorem.

In right-angled $△ P R Q$ , $P Q$ is the hypotenuse, so $P R^{2} = P Q^{2} - Q R^{2}$ .
$\begin{matrix} P R^{2} & = 13^{2} - 5^{2} \\ = 169 - 25 \\ = 144 \\ P R & = 12 m \end{matrix}$
Find $m ∠ P Q R$ using cosine.

Relative to $∠ Q$ , $Q R$ is the base and $P Q$ is the hypotenuse.
$\begin{matrix} \cos θ & = \frac{Q R}{P Q} \\ = \frac{5}{13} \\ θ & = \cos^{- 1} (\frac{5}{13}) \\ \approx {67.38}^{\circ} \end{matrix}$
Convert to radians.

Multiply degrees by $\frac{π}{180}$ .
$\begin{matrix} θ & = 67.38 \times \frac{π}{180} \\ \approx 1.18 rad \end{matrix}$

Answer

$Width = 12 m, m ∠ P Q R \approx 1.18 rad ({67.38}^{\circ})$

Quick Solution

Question 6

MediumUnder Review

Calculate the length $x$ in the figure. The figure is a quadrilateral $A B C D$ with diagonal $B D$ . In $△ A B D$ , $m ∠ A B D = 90^{\circ}$ , $A B = 10 cm$ and $A D = 17 cm$ . In $△ B C D$ , $m ∠ B C D = 90^{\circ}$ , $C D = 8 cm$ and $B C = x$ .

Hint

Use Pythagoras' theorem twice: $B D$ is a leg of one triangle and the hypotenuse of the other.

Solution

Find $B D^{2}$ in $△ A B D$ .

$A D$ is the hypotenuse (opposite the right angle at $B$ ), so $B D^{2} = A D^{2} - A B^{2}$ .
$\begin{matrix} B D^{2} & = 17^{2} - 10^{2} \\ = 289 - 100 \\ = 189 \end{matrix}$
Find $x$ in $△ B C D$ .

Now $B D$ is the hypotenuse (opposite the right angle at $C$ ), so $x^{2} = B D^{2} - C D^{2}$ . There is no need to square-root $189$ first.
$\begin{matrix} x^{2} & = 189 - 8^{2} \\ = 189 - 64 \\ = 125 \\ x & = \sqrt{25 \times 5} \\ = 5 \sqrt{5} \\ \approx 11.18 cm \end{matrix}$

Answer

$x = 5 \sqrt{5} cm \approx 11.18 cm$

Quick Solution

Question 7

EasyUnder Review

A ladder is placed against a wall such that the foot of the ladder is $2 m$ away from the wall. If the length of the ladder is $8 m$ , find the height of the wall (the height at which the top of the ladder touches the wall).

Hint

The ladder is the hypotenuse; the wall and the ground are the two legs.

Solution

Set up the right triangle.

Let $A$ be the foot of the ladder, $B$ the foot of the wall and $C$ the top of the ladder. The wall is vertical, so $m ∠ B = 90^{\circ}$ , with hypotenuse $A C = 8$ m and base $A B = 2$ m. $A C = 8 m, A B = 2 m, B C = ?$
Apply Pythagoras' theorem.

$B C^{2} = A C^{2} - A B^{2}$ .
$\begin{matrix} B C^{2} & = 8^{2} - 2^{2} \\ = 64 - 4 \\ = 60 \\ B C & = \sqrt{60} \\ = 2 \sqrt{15} \\ \approx 7.75 m \end{matrix}$

Answer

$Height = 2 \sqrt{15} m \approx 7.75 m$

Quick Solution

Question 8

MediumUnder Review

The diagonal of a rectangular field $A B C D$ is $(x + 9) m$ and the sides are $(x + 7) m$ and $x m$ . Find the value of $x$ .

Hint

Apply Pythagoras' theorem to the sides and diagonal, then solve the resulting quadratic by factorisation.

Solution

Set up Pythagoras' theorem.

Every angle of a rectangle is $90^{\circ}$ , so in $△ A B C$ with $A B = x + 7$ , $B C = x$ and diagonal $A C = x + 9$ : $(x + 7)^{2} + x^{2} = (x + 9)^{2}$
Expand and simplify.

Use $(p + q)^{2} = p^{2} + 2 p q + q^{2}$ on both sides.
$\begin{matrix} x^{2} + 14 x + 49 + x^{2} & = x^{2} + 18 x + 81 \\ 2 x^{2} + 14 x + 49 - x^{2} - 18 x - 81 & = 0 \\ x^{2} - 4 x - 32 & = 0 \end{matrix}$
Solve by factorisation.

Find two numbers with product $- 32$ and sum $- 4$ : $- 8$ and $4$ .
$\begin{matrix} x^{2} - 8 x + 4 x - 32 & = 0 \\ x (x - 8) + 4 (x - 8) & = 0 \\ (x - 8) (x + 4) & = 0 \\ x = 8 & or x = - 4 \end{matrix}$
Reject the negative value.

A length cannot be negative, so $x = - 4$ is rejected. Check: sides $15$ m and $8$ m, diagonal $17$ m, and $15^{2} + 8^{2} = 289$ and $17^{2} = 289$ . $x = 8$

Answer

$x = 8 (sides 15 m and 8 m, diagonal 17 m)$

Quick Solution

Question 9

EasyUnder Review

Calculate the value of $x$ in each case.

Hint

Identify the hypotenuse in each right triangle and apply Pythagoras' theorem.

Solution

(This question consists of multiple subparts. See solutions below.)

Quick Solution

Q9Part (i)

EasyUnder Review

In right-angled $△ A B C$ , $m ∠ B = 90^{\circ}$ , $A B = x$ , $B C = 12 cm$ and the hypotenuse $A C = 20 cm$ .

Hint

Subtract the square of the known leg from the square of the hypotenuse.

Solution

Apply Pythagoras' theorem.

$A C$ is opposite the right angle, so $A B^{2} = A C^{2} - B C^{2}$ .

\begin{matrix} x^{2} & = 20^{2} - 12^{2} \\ = 400 - 144 \\ = 256 \\ x & = 16 cm \end{matrix}

Answer

$x = 16 cm$

Quick Solution

Q9Part (ii)

EasyUnder Review

In $△ A D C$ , $D B$ is perpendicular to $A C$ at the point $B$ , and $B$ is the midpoint of $A C$ , so $A B = B C$ . Given $B C = 4 cm$ , $D C = 5 cm$ and $A D = x$ .

Hint

Find the common side $D B$ from $△ D B C$ first, then use it in $△ D B A$ .

Solution

Find $D B$ in $△ D B C$ .

$m ∠ D B C = 90^{\circ}$ and $D C$ is the hypotenuse, so $D B^{2} = D C^{2} - B C^{2}$ .
$\begin{matrix} D B^{2} & = 5^{2} - 4^{2} \\ = 25 - 16 \\ = 9 \\ D B & = 3 cm \end{matrix}$
Find $x$ in $△ D B A$ .

$A B = B C = 4$ cm and $m ∠ D B A = 90^{\circ}$ , with $A D = x$ the hypotenuse.
$\begin{matrix} x^{2} & = A B^{2} + D B^{2} \\ = 4^{2} + 3^{2} \\ = 25 \\ x & = 5 cm \end{matrix}$

Answer

$x = 5 cm$

Quick Solution

The exercises and question numbering reproduced on these pages are from Mathematics for Class 9 (National Curriculum of Pakistan 2023), published by the Punjab Curriculum and Textbook Board (PCTB), Lahore, authored by Muhammad Akhtar Shirani, Madiha Mahmood, and Ghulam Murtaza. PCTB holds the copyright in the original textbook. PrepSure is not affiliated with, endorsed by, or sponsored by PCTB. The worked solutions, explanations, hints and method notes are PrepSure's own original work, written and reviewed by our team. If you hold rights in this material and believe anything here exceeds fair use, write to us and we will take it down.

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