2 Marks Questions

Question 12 Marks

A chord of a circle of radius $10 \ cm$ subtends a right angle at the centre. Find the area of the corresponding:
$i$. minor segment
$ii.$ major sector.

Answer

$i. r =10 \ cm, \theta=90^{\circ}$
Area of minor sector $=\frac{\theta}{360} \times \pi r^2$
$=\frac{90}{360} \times 3.14 \times 10 \times 10$
$=78.5 \ cm^2$
Area of $\triangle O A B=\frac{O A \times O B}{2}$
$=\frac{10 \times 10}{2}=50 \ cm^2$
$\therefore$ Area of the minor segment
$=$ Area of minor sector $-$ Area of $\triangle O A B$
$=78.5 \ cm^2-50 \ cm^2=28.5 \ cm^2$
$ii$. Area of major sector $=\pi r^2-$ area of minor sector
$=3.14 \times 10 \times 10-78.5$
$=314-28.5=285.5 \ cm^2$

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Question 22 Marks

Prove the trigonometric identity: $\tan ^2 A \sec ^2 B-\sec ^2 A \tan ^2 B=\tan ^2 A-\tan ^2 B$

Answer

$\text { LHS }=\tan ^2 A \sec ^2 B-\sec ^2 A \tan ^2 B$
$=\tan ^2 A\left(1+\tan ^2 B\right)-\left(1+\tan ^2 A\right) \tan ^2 B\left [\because 1+\tan ^2 \theta=\sec ^2 \theta\right]$
$=\tan ^2 A+\tan ^2 A \tan ^2 B-\tan ^2 B-\tan ^2 A \tan ^2 B$
$=\tan ^2 A-\tan ^2 B$
$=\text { RHS }$
Hence proved.

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Question 32 Marks

Find the difference of the areas of a sector of angle $120^{\circ}$ and its corresponding major sector of a circle of radius $21 \ cm$ .

Answer

Let $A_1$ and $A_2$ be the areas of the given sector and the corresponding major sector respectively.
Given, $\theta=120^{\circ}$ and it's radius is $21 \ cm$ .
So, $r =21 \ cm$.
$\therefore A_1=\frac{\theta}{360} \times \pi r^2$
$=\frac{120}{360} \times \pi \times(21)^2$
$=147 \pi \ cm^2$
and, $A _2=$ Area of the circle $- A _1$
$\Rightarrow A_2=\left\{\pi \times(21)^2-147 \pi\right\} \ cm^2$
$=\pi(441-147) \ cm^2$
$=294 \pi \ cm^2$
Required differences $= A _2- A _1$
$=(294 \pi-147 \pi) \ cm^2=147 \pi \ cm^2$
$=\left(147 \times \frac{22}{7}\right) \ cm^2$
$=462 \ cm^2$

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Question 42 Marks

If $A=B=60^{\circ}$, verify that $\sin (A-B)=\sin A \cos B-\cos A \sin B$

Answer

$\text { L.H.S. }=\sin (A-B)=\sin \left(60^{\circ}-60^{\circ}\right)$
$=\sin 0^{\circ}=0$
$\text { R. H.S. }=\sin A \cos B -\cos A \sin B$
$=\sin 60^{\circ} \cos 60^{\circ}-\cos 60^{\circ} \sin 60^{\circ}$
$=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}$
$=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}=0$
$\therefore \text { L.H.S. }$
$=\text { R.H.S }$

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Question 52 Marks

A circle is inscribed in a $\triangle A B C$, touching $BC , CA$ and $AB$ at $P , Q$ and $R$ respectively, as shown in the given figure. If $AB =10 \ cm, AQ =7 \ cm$ and $CQ =5 \ cm$ then find the length of $BC$ .

Answer

Here it is given that, $AB =10 \ cm, AQ =7 \ cm$ and $CQ =5 \ cm$.
Now we know that the lengths of tangents drawn from an external point to a circle are equal.
$\therefore A R=A Q=7 \ cm$
$B R=(A B-A R)$
$=(10-7) \ cm=3 \ cm.$
$\therefore B P=B R=3 \ cm,$
$C P=C Q=5 \ cm.$
$B C=(B P+C P)$
$=(3+5) \ cm=8 \ cm$
Hence the length of $BC =8 \ cm$.

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Question 62 Marks

In Fig. $D E \| B C$ and $C D \| E F$. Prove that $AD ^2= AB \times AF$.

Answer

In $\triangle ABC$, it is given that
$D E \| B C$
$\Rightarrow \frac{A B}{A D}=\frac{A C}{A E} .........(i)$
In $\triangle ADC$, it is given that
$F E \| D C$
$\Rightarrow \frac{A D}{A F}=\frac{A C}{A E} .........(ii)$
From $(i)$ and $(ii),$ we get
$\frac{A B}{A D}=\frac{A D}{A F}$
$\Rightarrow AD^2=AB \times AF$

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Question 72 Marks

Find the $LCM$ and $HCF$ of the pairs of integers $336$ and $54$ and verify that $LCM \times HCF =$ product of the two numbers.

Answer

$336$ and $54$
$336=2 \times 2 \times 2 \times 2 \times 3 \times 7=2^4 \times 3 \times 7$
$54=2 \times 3 \times 3 \times 3=2 \times 3^3$
$H C F=2 \times 3=6$
$L C M=2^4 \times 3^3 \times 7=3024$
Product of two numbers 336 and $54=336 \times 54=18144$
$HCF \times LCM=6 \times 3024=18144$
Hence, product of two numbers $= HCF \times LCM ^`$

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