Case study (4 Marks)

Question 14 Marks

$A$ backyard is in the shape of a triangle $A B C$ with right angle at $B, A B=7 m$ and $B C=15 m$. A circular pit was dug inside it such that it touches the walls $A C, B C$ and $A B$ at $P, Q$ and $R$ respectively such that $A P=x m$.
Based on the above information, answer the following questions:
(i) Find the length of $A R$ in terms of $x$.
(ii) Write the type of quadrilateral $B Q O R$.
(iii) (a) Find the length PC in terms of $x$ and hence find the value of $x$.
OR
(b) Find $x$ and hence find the radius $r$ of circle.

Answer

(i) Lengths of tangents drawn from a point to a circle are equal.
$\therefore \quad A P=A R \Rightarrow A R=x \qquad[\because A P=x]$
(ii) In quadrilateral $B Q O R$, we find that
$B Q=Q R$$\qquad$ [Lengths of tangents drawn from $B$ ]
Also, $B Q=O R=r$ and $B R=O Q=r$
$
\therefore \quad B Q=O Q=B R=O R=r
$
Also, $\angle Q B R=90^{\circ}$. Hence, Quad rilateral $B Q O R$ is a square.
(iii) (a) $A P=A R=x$
$
\begin{aligned}
\therefore & & B Q & =B R=7-x \\
& & P C & =Q C \Rightarrow P C=15-(7-x) \Rightarrow P C=8+x
\end{aligned}
$
In right triangle $A B C$, we have
$
A B=7, B C=15 \text { and } A C=A P+P C=x+8+x=8+2 x
$
$
\therefore \quad A C^2=A B^2+B C^2\qquad[Using Pythagoras Theorem]
$
$
\begin{array}{ll}
\Rightarrow & (8+2 x)^2=7^2+15^2 \\
\Rightarrow & 64+32 x+4 x^2=49+225 \Rightarrow 4 x^2+32 x-210=0 \Rightarrow 2 x^2+16 x-105=0 \\
\therefore & x=\frac{-16 \pm \sqrt{16^2-4 \times 2 \times-105}}{4} \Rightarrow x=\frac{-16 \pm \sqrt{1096}}{4}=\frac{-16 \pm 33.10}{2}=4.27
\end{array}
$
(b) We obtain: $x=4.27$
$
\therefore \quad r=B Q=7-x \Rightarrow r=7-4.27=2.73
$

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Question 24 Marks

The discus throw is an event in which an athlete attempts to throw a discus. The athlete spins anti-clockwise around one and a half times through a circle, then releases the throw. When released the discus travels along tangent to the circular spin orbit.

In the given figure, $A B$ is one such tangent to a circle of radius 75 cm . Point $O$ is centre of the circle and $\angle A B O=30^{\circ}$. $P Q$ is parallel to $O A$.
Based on above information find the lengths of:
(i) $A B$ $\qquad$ (ii) $O B$ $\qquad$ (iii) $A P$ $\qquad$ (iv) $P Q$

Answer

$A B$ is tangent to the circle at $A$ and $O A$ is radius. Therefore, $\angle O A B=90^{\circ}$
(i) In right triangle $O A B$, we obtain
$
\tan 30^{\circ}=\frac{O A}{A B} \Rightarrow \frac{1}{\sqrt{3}}=\frac{75}{A B} \Rightarrow A B=75 \sqrt{3} cm
$
(ii) Applying Pythagoras theorem in $\triangle O A B$, we obtain
$
\begin{array}{ll}
& O B^2=O A^2+A B^2 \\
\Rightarrow \quad & O B=\sqrt{75^2+(75 \sqrt{3})^2}=\sqrt{75^2(1+3)}=75 \times 2=150 cm \\
\therefore \quad & B Q=O B-O Q=(150-75)=75 cm
\end{array}
$
(iii) $\triangle B A O \sim \triangle B P Q$ by AAA criterion of similarity.
$
\begin{array}{ll}
\therefore & \frac{B A}{B P}=\frac{B O}{B Q}=\frac{O A}{P Q} \\
\Rightarrow & \frac{75 \sqrt{3}}{B P}=\frac{150}{75}=\frac{75}{P Q} \Rightarrow \frac{75 \sqrt{3}}{B P}=2 \text { and } 2=\frac{75}{P Q} \Rightarrow B P=\frac{75}{2} \sqrt{3} \text { and } P Q=37.5 cm \\
\therefore & A P=A B-B P=75 \sqrt{3}-\frac{75}{2} \sqrt{3}=\frac{75}{2} \sqrt{3} cm
\end{array}
$
(iv) $P Q=37.5 cm$.

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Question 34 Marks

Raghav has been selected by his school to design logo for sports day T-shirts for students and staff. The logo design is as given in the figure and he is working on the fronts and different colours according to the theme. In the given figure, a circle with centre $O$ is inscribed in a $\triangle A B C$, such that it touches the sides $A B, B C$ and $C A$ at points $D, E$ and $F$ respectively. The length of side $A B, B C$ and $C A$ are $12 cm, 8 cm$ and 10 cm respectively.

(i) The length of $A D$ is
(a) 7 cm $\qquad$ (b) 8 cm $\qquad$ (c) 5 cm $\qquad$ (d) 9 cm
(ii) The length of $B E$ is
(a) 8 cm $\qquad$ (b) 5 cm $\qquad$ (c) 2 cm $\qquad$ (d) 9 cm
(iii) CF is of length
(a) 9 cm $\qquad$ (b) 5 cm $\qquad$ (c) 2 cm $\qquad$ d) 3 cm
(iv) If radius of the circle is 4 cm , then area of $\triangle O A B$ is
(a) 20 cm $\qquad$ (b) 36 cm $\qquad$ (c) 24 cm $\qquad$ (d) 48 cm

Answer

Let $s$ be the semi-perimeter of $\triangle A B C$ and $a=B C=8 cm, b=C A=10 cm$ and $c=A B=12 cm$. From example 11 (page 8.13 of main book), we obtain
$
A D=A F=s-a, B E=B D=s-b \text { and } C F=C E=s-c
$
(i) (a): $A D=(s-a)=(15-8) cm =7 cm$
(ii) (b): $B E=(s-b)=(15-10) cm =5 cm$
(iii) (d): $C F=(s-c)=(15-12) cm =3 cm$
(iv) (c): Area of $\triangle O A B=\frac{1}{2}(A B \times O D)=\frac{1}{2}(12 \times 4)=24 cm^2$

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Question 44 Marks

London eye is an amusement ride consisting of a rotating upright big wheel with multiple passenger-carrying components (commonly referred to as passenger cars, cabins, tubs, capsules, gondolas or pods) attached to the rim in such a way that as the wheel turns, they are kept upright usually by gravity. After taking a ride in London eye, Anu came out from the crowd and was observing her friends who were enjoying the ride. She was curious about the different angles and measures that the wheel will form. She makes a figure as given below.

(i) In Fig. the measure of $\angle R O Q$ is
(a) $60^{\circ}$ $\qquad$ (b) $100^{\circ}$ $\qquad$ (c) $150^{\circ}$ $\qquad$ (d) $90^{\circ}$
(ii) In Fig. 8.33, The measure of $\angle R Q P$ is
(a) $75^{\circ}$ $\qquad$ (b) $60^{\circ}$ $\qquad$ (c) $30^{\circ}$ $\qquad$ (d) $90^{\circ}$
(iii) In Fig. 8.33, the measure of $\angle R S Q$ is
(a) $60^{\circ}$ $\qquad$ (b) $75^{\circ}$ $\qquad$ (c) $100^{\circ}$ $\qquad$ (d) $30^{\circ}$
(iv) In Fig. 8.33, the measure of $\angle O R P$ is
(a) $90^{\circ}$ $\qquad$ (b) $70^{\circ}$ $\qquad$ (c) $100^{\circ}$ $\qquad$ (d) $60^{\circ}$

Answer

(i) (c): Clearly, $\angle R P Q+\angle R O Q=180^{\circ}$
$
\Rightarrow \quad 30^{\circ}+\angle R O Q=180^{\circ} \Rightarrow \angle R O Q=150^{\circ}
$
(ii) (a): In $\triangle P Q R$, we have $\angle P R Q=\angle P Q R$ and $\angle Q P R=30^{\circ}$
$
\therefore \quad \angle R Q P+\angle P R Q+\angle Q P R=180^{\circ} \Rightarrow 2 \angle R Q P=180^{\circ}-30^{\circ} \Rightarrow \angle R Q P=75^{\circ}
$
(iii) (b): In $\triangle O Q R$, we find that $\angle O Q R=\angle O R Q=90^{\circ}-75^{\circ}=15^{\circ}$
$
\therefore \quad \angle Q O R=180^{\circ}-15^{\circ}-15^{\circ}=150^{\circ} \Rightarrow \angle R S Q=\frac{1}{2} \angle Q O R=75^{\circ}
$
(iv) (a): The angle between the tangent at a point and the radius is $90^{\circ}$. Therefore, $\angle O R P=90^{\circ}$

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Question 54 Marks

The chain and gears of bicycles or motorcycles or belt around pulleys are some real-life illustrations of tangents to circles.

(i) $P I$ and $P A$ are tangents to the circle from point $P$. If arc $I Z A$ subtends an angle $240^{\circ}$ at the centre of the circle, then $\angle I P A=$
(a) $120^{\circ}$ $\qquad$ (b) $90^{\circ}$ $\qquad$ (c) $60^{\circ}$ $\qquad$ (d) $30^{\circ}$
(ii) If $I P=15 cm$, then $A I=$
(a) 7.5 cm $\qquad$ (b) 15 cm $\qquad$ (c) 30 cm $\qquad$ (d) 18 cm
(iii) If $I P=21 cm$ and measure of $A P$ is $x^2+5$, then $x=$
(a) 4 $\qquad$ (b) 16 $\qquad$ (c) $\sqrt{26}$ $\qquad$ (d) $\sqrt{30}$
(iv) $\angle O I P+\angle A P O=$
(a) $90^{\circ}$ $\qquad$ (b) $60^{\circ}$ $\qquad$ (c) $120^{\circ}$ $\qquad$ (d) $150^{\circ}$

Answer

(i) (c) (ii) (b) (iii) (a) (iv) (c)

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Question 64 Marks

A solar eclipse occurs when the Moon passed between the Earth and the Sun. Following diagram represents the total and partial eclipse.

(i) The tangents to the Moon surface from the point $A$ are
(a) $A B$ and $A C$ $\qquad$ (b) $A P$ and $A Q$
(c) $A P$ and $A B$ $\qquad$ (d) $A Q$ and $A C$
(ii) If $\angle P A Q=40^{\circ}$, then the measure of $\angle P O Q$ is
(a) $70^{\circ}$ $\qquad$ (b) $40^{\circ}$ $\qquad$ (c) $50^{\circ}$ $\qquad$ (d) $140^{\circ}$
(iii) If $\angle P O Q=110^{\circ}$, then $\angle Q A O=$
(a) $55^{\circ}$ $\qquad$ (b) $35^{\circ}$ $\qquad$ (c) $70^{\circ}$ $\qquad$ (d) $110^{\circ}$
(iv) If $\angle P O Q=130^{\circ}$, then $\angle O P Q=$
(a) $130^{\circ}$ $\qquad$ (b) $50^{\circ}$ $\qquad$ (c) $65^{\circ}$ $\qquad$ (d) $25^{\circ}$

Answer

(i) (a) (ii) (d) (iii) (b) (iv) (d)

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Maths Case study (4 Marks)

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