2 Marks Questions

Question 12 Marks

In the given figure, $\text{ABCD}$ is a square of side $14 \ cm .$ Semi$-$circles are drawn with each side of square as diameter. Find the area of the shaded region.
$\left(\text { Use } \pi=\frac{22}{7}\right)$

Answer

We have marked the four shaded region as $\text{I, II, III}$ and $\text{IV}$ and the centres of the semicircles as $\text{P, Q, R}$ and $\text{S}$ in the figure above. It is given that the side of the square is $14 \ cm .$
Now,
Area of the region $I\ +$ Area of the region $III =$ Area of the square $-$ Area of the semicircles with centres $S$ and $Q$ and radius $=7\ cm$
$=14 \times 14-2 \times \frac{1}{2} \times \pi \times 7^2$
$=196-49 \times \frac{22}{7}$
$=196-154$
$=42 \ cm^2$
Similarly, Area of the region $II\ +$ Area of the region $IV\ =$ Area of the square $-$ Area of the semicircles with centres $P$ and $R$ and radius $= 7 \ cm$
$=14 \times 14-2 \times \frac{1}{2} \times \pi \times 7^2$
$=196-49 \times \frac{22}{7}$
$=196-154$
$=42 \ cm^2$
So, the area of the shaded region $=$ Area of region $I\ +$ Area of region $III\ +$ Area of region $II\ +$ Area of region $IV$
$=42 \ cm^2+42 \ cm^2$
$=84 \ cm^2$
Hence, the area of the shaded region is $84 \ cm^2$

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Question 22 Marks

In the given figure, a square $0\ \text{ ABC}$ is inscribed in a quadrant $O P B Q$ of a circle. If $O A=20 \ cm, $ find the area of the shaded region. $($Use $\pi=3.14$ )

Answer

In $\triangle OAB$

$O B^2=O A^2+A B^2$
$=(20)^2+(20)^2=2 \times(20)^2$
$O B=20 \sqrt{2}$
Radius of the circle,
$r=20 \sqrt{2}$
Area of quadrant $\text{OPBQ}$
$=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(20 \sqrt{2})^2$
$=0.25 \times 3.14 \times 800 \ cm^2$
$=628 \ cm^2$
Area of square
$\text{OABC}=(\text { Side })^2=(20)^2 \ cm^2=400 \ cm^2$
Area of the shaded region $=$ Area of quadrant $\text{OPBQ} -$ Area of square $\text{OABC}$
$=628-400$
$=228 \ cm^2$
Hence, the area of shaded area is $228 \ cm^2$

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Question 32 Marks

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions $14 cm \times 7 cm$. Find the area of the remaining card board.$\left[\right.$ Use $\left.\pi=\frac{22}{7}\right]$

Answer

Two circular pieces of equal radii which are cut out from rectangular sheet have the maximum area.
Hence.
The diameter of each circle is, $\left( d =\frac{14}{2}=7 cm\right.$
Therefore, radius of each circular sheet,
Thas the correct answer is (C). $r=\frac{d}{2}=\frac{7}{2} cm$

Area of a circular piece
$
=\pi \times\left(\frac{7}{2}\right)^2=\frac{22}{7} \times \frac{49}{4}=\frac{77}{2} cm^2
$ $
\left(A_2\right)=2 \times \frac{77}{2}=77 cm^2
$
Now,
Area of the remaining card board $= A _1- A _2$
$
=98-77=21 cm^2
$
Thus the area of the remaining card board is
$
21 cm^2
$

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Question 42 Marks

In the given figure $,\text{ ABCD}$ is a square of side $4 \ cm$ . A quadrant of a circle of radius $1 \ cm$ is drawn at each vertex of the square and a circle of diameter $2 \ cm$ is also drawn. Find the area of the shaded region. $($Use $\pi=3.14 )$

Answer

Let $A$ be the area of each quadrant of the circle of radius $1 \ cm$ .
$\therefore $ Area of a quadrant $=\frac{1}{4} \pi r^2$
$\Rightarrow A=\frac{1}{4} \times 3.14 \times 1 \times 1$
$\Rightarrow A=0.785 \ cm^2$
Therefore, area of the $4$ quadrants
$=4 A=4 \times 0.785=3.14 \ cm^2$
Area of the circle inside the
square $=\pi r^2=3.14 \times 1 \times 1=3.14 \ cm^2$
Now, Area of thesquare $=(\text { side })^2=(4)^2=16 \ cm^2$
So, Area of the shaded part of the square
$=\text { Area of the square }-\binom{\text { Area of } 4 \text { quadrants }+}{\text { Area of the circle }}$
$=16-(3.14+3.14)$
$=16-6.28$
$=9.72 \ cm^2$
the area of the shaded region is $9.72 \ cm^2$

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Question 52 Marks

Find the area of a quadrant of a circle, where the circumference of circle is $44 \ cm . [$Use $\pi=\frac{22}{7} ]$

Answer

Let $r$ be the radius of the circle.
Given that circumference of circle is $44 \ cm .$
$\Rightarrow 2 \pi r=44$
$\Rightarrow r=\frac{44}{2 \pi}$
$\Rightarrow r=\frac{44 \times 7}{2 \times 22}$
$\Rightarrow r=7 \ cm$
Now, the area of the quadrant of the circle
$=\frac{\pi r^2}{4}$
$\Rightarrow \frac{22 \times(7)^2}{7 \times 4}=\frac{77}{2}=38.5$
Therefore, the area of the quadrant of the circle is $38.5 \ cm^2$

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Question 62 Marks

In the given figure, $APB$ and $CQD$ are semi circles of diameter $7 \ cm$ each, while $ARC$ and $BSD$ are semi circles of diameter $14 \ cm$ each. Find the perimeter of the shaded region.$[$Use $\pi=\frac{22}{7} ]$

Answer

Let $r$ and $R$ be the radii of the semicircle $APB$ and $ARC$ respectively.
$r=\frac{7}{2} \ cm$ and $ R=\frac{14}{2}=7 \ cm$
The given shape is symmetric.
Required perimeter of the shaded region
$=2(\pi R-\pi r) ($Hint: Perimeter of the semicircle $=\pi r )$
$\Rightarrow 2 \pi(R-r)$
$\Rightarrow 2 \times \pi\left[7-\left(\frac{7}{2}\right)\right]$
$\Rightarrow 2 \times \pi\left(\frac{7}{2}\right)$
$\Rightarrow 2 \times \frac{22}{7} \times\left(\frac{7}{2}\right)$
$\Rightarrow 22 \ cm$
Therefore, the perimeter of the shaded region is $22 \ cm$

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