3 Marks Question

Question 13 Marks

If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.

Answer

Let $\text{ABCD}$ be a parallelogram such that its sides touch a circle with centre $O$.
We know that the tangents to a circle from an exterior point are equal in length.
Therefore $,AP = AS \ [$From $A] ...(i)$
$BP = BQ \ [$ From $B ] \ldots (ii)$
$CR = CQ\ [$ From $C ] \ldots...iii)$
and $, DR = DS\ [$From $D] ...(iv)$
Adding $(i), (ii), (iii)$ and $(iv),$ we get,
$AP+BP+CR+DR$
$=AS+BQ+CQ+DS$
$\Rightarrow(AP+BP)+(CR+DR)$
$=(AS+DS)+(BQ+CQ)$
$\Rightarrow AB+CD=AD+BC$
$\Rightarrow 2 AB=2 BC$
$\Rightarrow AB=BC$
Therefore, $AB = BC = CD = AD$
Thus, $\text{ABCD}$ is a rhombus.

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Question 23 Marks

If two pipes function simultaneously, a reservoir will be filled in $12$ hours. One pipe fills the reservoir $10$ hours faster than the other. How many hours will the second pipe take to fill the reservoir?

Answer

Let, the slower pipe takes $x$ hours to fill the reservoir.
Hence, the faster pipe will take $(x-10)$ hours to fill the reservoir.
Since, the slower pipe takes $x$ hours to fill the reservoir.
$\therefore$ Portion of the reservoir filled by the slower pipe in $1$ hour $=\frac{1}{x}$
$\therefore$ Portion of the reservoir filled by the slower pipe in $12$ hours $=\frac{1}{x} \times 12=\frac{12}{x}$.
Now, portion of the reservoir filled by the faster pipe in $1 hr =\frac{1}{x-10}$
$\therefore$ Portion of the reservoir filled by faster pipe in $12$ hours $=\frac{1}{x-10} \times 12=\frac{12}{x-10}$
It is given that the reservoir is completely filled in $12$ hours by simultaneously operating both pipes.
$\therefore \frac{12}{x}+\frac{12}{x-10}=1$
$\Rightarrow \frac{12(x-10)+12 x}{x(x-10)}=1$
$\Rightarrow \frac{12 x-120+12 x}{x^2-10 x}=1$
$\Rightarrow x^2-10 x=24 x-120$
$\Rightarrow x^2-10 x-24 x+120=0$
$\Rightarrow x^2-34 x+120=0$
$\Rightarrow x^2-30 x-4 x+120=0$
$\Rightarrow x(x-30)-4(x-30)=0$
$\Rightarrow(x-30)(x-4)=0$
$\Rightarrow x-30=0 .$
$[\therefore x-4 \neq 0, \text { otherwise }(x-10)$ i.e time taken by faster pipe will become $ -6 hr]$
i.e. negative, which is not possible
$\Rightarrow x=30$
Hence, the second pipe take $30$ hours to fill the reservoir.

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Question 33 Marks

The angle of elevation of the top of a tower from a point $A$ on the ground is $30^{\circ}$. On moving a distance of $20$ metre towards the foot of the tower to a point $B$ the angle of elevation increases to $60^{\circ}$. Find the height of the tower and the distance of the tower from the point $A .$

Answer

Let height of tower be $h m$ and distance $BC$ be $x m$
In $\triangle \text{DBC,} \frac{h}{x}=\tan 60^{\circ}$
$\Rightarrow h=\sqrt{3 x} \ldots . .(i)$
$\frac{h}{x+20}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\Rightarrow \sqrt{3 h}=x+20 \ldots . \text { (ii) }$
Substituting the value of $h$ from eq. $(i)$ in eq. $(ii),$ we get
$3 x=x+20$
$3 x-x=20$
Or $2 x=20$
$\Rightarrow x=10 m \ldots \text { (iii) }$
Again $h=\sqrt{3 x}$
or, $h=\sqrt{3} \times 10=10 \sqrt{3}$
$=10 \times 1 \cdot 732$
$=17.32 m$
$[$from $(i)$ and $($in$)]$
Hence, height of tower is $17.32 m$ and distance of tower from point $A$ is $30 m$

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Question 43 Marks

Prove that $(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2=7+\tan ^2 A+\cot ^2 A$.

Answer

$\text { LHS }=(\sin A+\operatorname{cosec} A)^2+(\cos A+\sec A)^2$
$=\sin ^2 A+\operatorname{cosec}^2 A+2 \sin A \operatorname{cosec} A+\cos ^2 A+\sec ^2 A+2 \cos A \sec A$
$=\sin ^2 A+\cos ^2 A+\operatorname{cosec}^2 A+\sec ^2 A+2+2$
$=1+\operatorname{cosec}^2 A+\sec ^2 A+4$
$=\left(1+\cot ^2 A\right)+\left(1+\tan ^2 A\right)+5$
$=7+\tan ^2 A+\cot ^2 A=\text { RHS }$

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Question 53 Marks

In Figure, $X Y$ and $X^{\prime} Y^{\prime}$ are two parallel tangents to a circle with centre $O$ and another tangent $A B$ with point of contact $C$ intersects $XY$ at $A$ and $X ^{\prime} Y ^{\prime}$ at $B$ . Prove that $\angle A O B=90^{\circ}$.

Answer

According to the question $, XY$ and $X 'Y\ '$ are x two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersects $XY$ at $A$ and $X 'Y\ '$ at $B$.

In quad. $\text{APQB} ,$ we have
$\angle A P O=90^{\circ}$
and $\angle B Q O=90^{\circ}\ [\because$ tangent at any point is perpendicular to the radius through the point of contact $]$
Now, $\angle A P O+\angle B Q O+\angle Q B C+\angle P A C=360^{\circ}$
$\Rightarrow \angle P A C+\angle Q B C=360^{\circ}-(\angle A P O+\angle B Q O)=180^{\circ}$
We have,
$\angle C A O=\frac{1}{2} \angle P A C$
and $\angle C B O=\frac{1}{2} \angle Q B C\ [\because$ tangents from an external point are equally inclined to the line segment joining the centre to that point$]$
$\therefore \angle CAO+\angle CBO$
$=\frac{1}{2}(\angle P A C+\angle Q B C)$
$=\frac{1}{2} \times
180^{\circ}=90^{\circ}$
In $\triangle A O B$, we have
$\angle C A O+\angle A O B+\angle C B O=180^{\circ}$
$\Rightarrow 90^{\circ}+\angle A O B=180^{\circ}$
$\Rightarrow \angle A O B=90^{\circ}$

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Question 63 Marks

Had Aarush scored $8$ more marks in a Mathematics test, out of $35$ marks, $7$ times these marks would have been $4$ less than square of his actual marks. How many marks did he get in the test?

Answer

Let the actual marks be $x$
Accroding to the question,
$7(x+8)=x^2-4$
$\Rightarrow 7 x+56=x^2-4$
$\Rightarrow x^2-7 x-60=0$
$\Rightarrow x^2-12 x+5 x-60=0$
$\Rightarrow x(x-12)+5(x-12)=0$
$\Rightarrow(x-12)(x+5)=0$
$\Rightarrow x-12=0 \text { or } x+5=0$
$\Rightarrow x=12 \text { or } x=-5$
$\Rightarrow x=12 \ldots[\because \text { Marks can't be negative }]$
Hence, Aarush scored $12$ marks in Mathematics test.

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Question 73 Marks

The mid $-$ point $P$ of the line segment joining the points $A (-10,4)$ and $B (- 2,0)$ lies on the line segment joining the points $C (- 9, - 4)$ and $D (- 4, y)$. Find the ratio in which $P$ divides $CD$. Also, find the value of $y$

Answer

Coordinates of the midpoint $P$ of $A$ and $B$ are
$\left(\frac{-10+(-2)}{2}, \frac{4+0}{2}\right)=(-6,2)$
$P$ lies on the line joining $C$ and $D$.
Let $P(-6,2)$ divide $C(-9,-4)$ and $D(-4, y)$ in the ratio of $r: 1$
Using the section formula for the $x -$ coordinate we get
$-6=\frac{-4 r-9}{r+1} $
$\Rightarrow-6 r-6$
$=-4 r-9$
$\Rightarrow 2 r=3 $
$\Rightarrow r=\frac{3}{2}$
Hence, $P (-6,2)$ divides $C (-9,-4)$ and $D (-4, y )$ in the ratio of $3: 2$
Using the section formula for $y -$ coordinate we get
$-6=\frac{-4 r-9}{r+1} $
$\Rightarrow-6 r-6$
$=-4 r-9$
$\Rightarrow 2 r=3 $
$\Rightarrow r=\frac{3}{2}$
Hence, $P (-6,2)$ divides $C (-9,-4)$ and $D (-4, y )$ in the ratio of $3:2$
Using the section formula for $y-$ coordinate we get
$2=\frac{3 y-8}{3+2} $
$\Rightarrow 10=3 y-8 $
$\Rightarrow 3 y=18$
$\Rightarrow y=6$

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Question 83 Marks

Find the $\text{LCM}$ and $\text{HCF}$ of $404$ and $96$ and verify that $\text{LCM} \times \text{HCF} =$ product of the two numbers

Answer

Prime factorisation of $404$ and $96$ is:
$404=2 \times 2 \times 101$
or $404=2^2 \times 101$
$96=2 \times 2 \times 2 \times 2 \times 2 \times 3$
or $96=2^5 \times 3$
$\therefore \operatorname{HCF}\ (404,96)=2^2=4$
$\operatorname{LCM}\ (404,96)=101 \times 2^5 \times 3$
$\operatorname{LCM}\ (404,96)=9696$
Now we have to verify that,
$\operatorname{HCF}\ (404,96) \times \operatorname{LCM}\ (404,96)=404 \times 96$
Hence $, \text{LHS} = \text{HCF} \times \text{LCM} =4 \times 9696=38784$
$\text { RHS }=$ Product of numbers $=404 \times 96=38784$
Since, $\text{LHS} = \text{RHS}$
$\therefore \text{HCF} \times \text{LCM} =$ Product of $404$ and $96$.
Hence verified.

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