2 Marks Questions

Question 12 Marks

In the given figure, $\triangle AHK \sim \Delta ABC$. If $AK =8 \ cm, BC =3.2 \ cm$ and $HK =6.4 \ cm$, then find the length of $AC .$

Answer

$\because \triangle AHK \sim \triangle ABC \text { (given) }$
$\therefore \frac{H K}{B C}=\frac{A K}{A C}$
$\Rightarrow \frac{6.4}{3.2}=\frac{8.0}{A C}$
$\Rightarrow AC =4 \ cm$

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Question 22 Marks

In the given figure, $O$ is the centre of the circle. If $\angle AOB =145^{\circ}$, then find the value of $x$.

Answer

Take a point $P$ on circumference and join $AP \ BP$
$\angle APB=\frac{1}{2} \times 145^{\circ}=72.5^{\circ}$
$\angle APB+\angle ACB=180^{\circ}$
$\Rightarrow \angle ACB=107.5^{\circ} \text { or } x=107.5^{\circ}$

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Question 32 Marks

If $\sin ( A - B )=\frac{1}{2}, \cos (A+ B )=\frac{1}{2} ; 0< A + B \leq 90^{\circ}, A > B$; find $\angle A$ and $\angle B$.

Answer

$
\begin{array}{c}
\sin (A-B)=\sin 30^{\circ} \\
A-B=30^{\circ} \quad \quad \ldots \ldots(i)\\
\cos (A+B)=\cos 60^{\circ} \\
A+B=60^{\circ}\quad \quad \ldots \ldots(ii)
\end{array}
$
Solving (i) and (ii)
$
A=45^{\circ}, B=15^{\circ}
$

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Question 42 Marks

Evaluate : $\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\sin ^2 60^{\circ}}$

Answer

$\frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}$
$=\frac{67}{12}$

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Question 52 Marks

Find the length of the arc of a circle which subtends an angle of $60^{\circ}$ at the centre of the circle of radius 42 cm .

Answer

$\begin{aligned} \text { Length of arc } & =2 \times \frac{22}{7} \times 42 \times \frac{60}{360} \\ & =44 cm\end{aligned}$

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Question 62 Marks

The minute hand of a clock is 14 cm long. Find the area on the face of the clock described by the minute hand in 5 minutes.

Answer

Angle subtended in $5 min .=30^{\circ}$
Area described by minute hand $=\frac{30}{360} \times \frac{22}{7} \times 14 \times 14$ $=\frac{154}{3} cm^2$ or $51.33 cm^2$ approx.

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Question 72 Marks

Three bells toll at intervals of $9,12$ and $15$ minutes respectively. If they start tolling together, after what time will they next toll together?

Answer

$9=3^2$
$12=2^2 \times 3$
$15=3 \times 5$
$\text { L.C.M }=2^2 \times 3^2 \times 5=180$
Three bells will toll together after $180$ min .

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Maths 2 Marks Questions

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