Case study (4 Marks)

Question 14 Marks

A backyard is in the shape of a triangle $\text{ABC}$ with right angle at $B$. $AB =7 m$ and $BC =15 m$. 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 $AP =x m$.

Based on the above information, answer the following questions:
$(i)$ Find the length of $AR$ in terms of $x$.
$(ii)$ Write the type of quadrilateral $\text{BQOR} $.
$(iii) \ (a)$ 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

$(i) AR =x m$
$(ii)$ Quad. $\text{ORBQ}$ is a square.
$(iii) \ (a) \ PC =8+x$
$AC^2=(8+2 x)^2=49+225=274$
$\Rightarrow 8+2 x=\sqrt{274}$
$\Rightarrow x=\frac{-8+\sqrt{274}}{2}$ or $ 4.28$ approx.
OR
$\text { (iii) (b) } AC^2=(8+2 x)^2$
$=49+225=274$
$\Rightarrow 8+2 x=\sqrt{274}$
$\Rightarrow x=\frac{-8+\sqrt{274}}{2} $ or $ 4.28$ approx.
Hence, radius $r=7- x =7-\left(-4+\frac{\sqrt{274}}{2}\right)$
$=\left(11-\frac{\sqrt{274}}{2}\right)$ or $ 2.72$ approx.
Therefore, radius of the circle is $\left(11-\frac{\sqrt{274}}{2}\right) m$ or $2.72 m$ approx.

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Question 24 Marks

BINGO is game of chance. The host has 75 balls numbered 1 through 75. Each player has a BINGO card with some numbers written on it. The participant cancels the number on the card when called out a number written on the ball selected at random. Whosoever cancels all the numbers on his/her card, says BINGO and wins the game.

The table given below, shows the data of one such game where 48 balls were used before Tara said 'BINGO'.

Numbers announced	Number of times
0-15	8
15-30	9
30-45	10
45-60	12
60-75	9

Based on the above information, answer the following:
(i) Write the median class.
(ii) When first ball was picked up, what was the probability of calling out an even number?
(iii) (a) Find median of the given data.
OR
(b) Find mode of the given data.

Answer

Number announced	0-15	15-30	30-45	45-60	60-75
Number of times (1)	8	9	10	12	9
cf	8	17	27	39	48=N

(i) $\frac{N}{2}=24$
$\therefore$ median class is $30-45$
(ii) $\quad P$ (picking up an even number) $=\frac{37}{75}$
(iii) (a) Median $=30+\frac{\left(\frac{48}{2}-17\right)}{10} \times 15$ $=40.5$
OR
(iii) (b) Modal class is $45-60$
$
\begin{aligned}
\text { Mode } & =45+\frac{12-10}{2 \times 12-10-9} \times 15 \\
& =51
\end{aligned}
$

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Question 34 Marks

A rectangular floor area can be completely tiled with $200$ square tiles. If the side length of each tile is increased by $1$ unit, it would take only $128$ tiles to cover the floor.

$(i)$ Assuming the original length of each side of a tile be $x$ units, make a quadratic equation from the above information.
$(ii)$ Write the corresponding quadratic equation in standard form.
$(iii) (a)$ Find the value of $x$, the length of side of a tile by factorisation.
OR
$(b)$ Solve the quadratic equation for $x$, using quadratic formula.

Answer

$(i)\ 200 x^2=128(x+1)^2$
$(ii)\ 25 x^2=16 x^2+32 x+16$
$\Rightarrow 9 x^2-32 x-16=0$
$(iii)\ (a)\ 9 x^2-32 x-16=0$
$\Rightarrow(9 x+4)(x-4)=0$
$x \neq \frac{-4}{9}$
so, $x=4$
OR
$(iii)\ (b)\ x =\frac{32 \pm \sqrt{1024+576}}{18}=\frac{32 \pm 40}{18}$
$x \neq \frac{-4}{9}$
so, $x=4$

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Maths Case study (4 Marks)

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