MCQ Questions for Class 10 Maths Areas Related To Cricles Quiz 4

The given figure is made up of a rectangle and

$2$ identical circles. If the length of the rectangle is

$28\ cm$ , what is the area of the region in the rectangle excluding the areas occupied by the circle?

$\left (Use\ \pi = \dfrac {22}{7}\right )$

0%
$84\ cm^{2}$
0%
$105\ cm^{2}$
0%
$152\ cm^{2}$
0%
$123\ cm^{2}$

Explanation

Radius of the circle

$=\dfrac{\text{length of rectangle}}{4}$

Thus radius of the circle and breath of the rectangle are

$7$ and

$14$ respectively.

Required Area

$=$ Area of rectangle

$- 2$

$\times$ area of circle( radius of circle

$= 28$

$\times$

$14 - 2$

$\times$

$\pi { r }^{ 2 }$

$=392-2\times \dfrac { 22 }{ 7 } \times 7\times 7$

$= 392 - 308$

$= 84$

${ cm }^{ 2 }$

Figure

$ABCD$ is a rectangle whose length is twice its width.

$\widehat {FC}$ and

$\widehat {AE}$ are arcs of circles centered at B and D respectively. If the length of

$\overline {AD}$ is x, then the area of the shade region is

0%
$\left (\dfrac {2 - \pi}{4}\right )x^{2}$
0%
$\left (\dfrac {4 - \pi}{2}\right )x^{2}$
0%
$\left (\dfrac {\pi - 2}{4}\right )x^{2}$
0%
$\left (\dfrac {2}{\pi}\right )x^{2}$
0%
$\left (\dfrac {4}{\pi}\right )x^{2}$

Explanation

Solution:

Given that:

$AD=x$ and

$AB=2AD$

To Find:

Area of the shaded region

$=?$

Solution:

$AB=2AD=2x$

Area of the shaded region

$=$ Area of the rectangle

$ABCD-$ Area of the quadrant

$AED$ - Area of the quadrant

$FCB$

$=AD\times AB-\cfrac{\pi}4r^2-\cfrac{\pi}{4}r^2$

$=x\times2x-2\times \cfrac{\pi}{4}x^2$

$=2x^2-\cfrac{\pi}2x^2$

$=\left(\cfrac{4-\pi}2\right){x^2}$

Hence, B is the correct option.

The figure shows a semicircle on top of an isosceles right triangle. If the length of semicircle

$\overline {AB}$ is

$16\pi$ , what is the approximate length of

$\overline {BC}$ ?

0%
$2.8$
0%
$4.0$
0%
$5.7$
0%
$11.3$
0%
$22.6$

Explanation

Let the length of

$AB$ be

$2r$ , where

$r$ is radius of semicircle

Given that

$\pi r=16\pi$

$\Rightarrow r=16$

The length of

$BC$ is equal to

$\dfrac{AB}{\sqrt2} = \dfrac{32}{\sqrt2}=22.6$

The correct option is

$E$

A lawn of circular shape of radius

$r$ contains a pond . The pond is a square inscribed in it. The owner of lawn wants to plant tress in the remaining area. Find the remaining area.

0%
${ r }^{ 2 }(\pi -2)$
0%
${ r }^{ 2 }(\pi -1 )$
0%
${ r }^{ 2 }(\pi -\cfrac { 1 }{ 2 } )$
0%
None of these

Explanation

Area of square

$\sqrt { 2 } r\times \sqrt { 2 } r$

$=2{ r }^{ 2 }$

Remaining area

$=\pi { r }^{ 2 }-2{ r }^{ 2 }$

$=(\pi -2){ r }^{ 2 }$

When the figure above is spun around its vertical axis, the total surface area of the solid formed will be

0%
$144\pi$
0%
$108\pi$
0%
$72\pi$
0%
$36\pi$
0%
$9\pi + 12$

Explanation

Figure formed when the figure is rotated around the vertical axis is a hemisphere.

$\therefore$ The total surface area of the figure is the area of the hemisphare.

Given that the radius

$r=6$

area of the solid (A)

$=2\pi r^2+\pi ^2$

$3\pi r^2$

$3\times \pi \times 36=108 \pi$

What is the length of an arc of a circle with a radius of

$5$ if it subtends an angle of

${60}^{o}$ at the center?

0%
$3.14$
0%
$5.24$
0%
$10.48$
0%
$2.62$

Explanation

Given:

Radius (r)

$=5$

Angle

$=60^o$

Arc length for a particular angle we can write as -

$=\dfrac{\theta}{360}\times (2\pi r)$

$=\dfrac{60}{360}\times 2\pi \times 5$

$=\dfrac{10\pi}{6}=5.24$

Option 'B'.

Consider the above figure. Find the area in the given rectangle excluding the

$6$ identical semicircles.

0%
$144\ cm^{2}$
0%
$126\ cm^{2}$
0%
$195\ cm^{2}$
0%
$243\ cm^{2}$

Explanation

Required area

$=$ Area of rectangle

$- 6$

$\times$ area of

$1$ semi-circle

$=$

$42\times 14-6\times \dfrac { 1 }{ 2 } \times \dfrac { 22 }{ 7 } \times 7\times 7$

$= 588 - 462 = 126$

${ cm }^{ 2 }$

When the figure given above is spun around its vertical axis, the volume of the solid formed will be:

0%
$9\pi$
0%
$36\pi$
0%
$72\pi$
0%
$144\pi$
0%
$288\pi$

Explanation

The solid formed after spin will be a Hemisphere whose radius is

$6$ cm.

Volume of hemisphere

$=$

$\dfrac{2}{3}\pi r^3$

$\therefore$ volume

$=\dfrac{2}{3}\pi\times 6^3$

$= \dfrac{2}{3}\pi \times 6\times 6\times 6=144\pi$

A memento is made as shown in the figure. Find the area of the region in the triangle excluding the part cut by the circle.

0%
$138.20\ cm^{2}$
0%
$161.5\ cm^{2}$
0%
$133.567\ cm^{2}$
0%
$108.97\ cm^{2}$

Explanation

Required area

$=$ Area of triangle

$-$ Area of sector

$=\dfrac { 1 }{ 2 } \times 20\times 20-\dfrac { { 90 }^{ 0 } }{ { 360 }^{ 0 } } \times \dfrac { 22 }{ 7 } \times 7\times 7$

$=\dfrac { 400 }{ 2 } -\dfrac { 77 }{ 2 } =\dfrac { 323 }{ 2 }$

$= 161.5$

${ cm }^{ 2 }$

ASB is a quarter circle. PQRS is a rectangle with sides PQ

$=8$ and PS

$=6$ . What is the length of the arc AQB?

0%
$5\pi$
0%
$10\pi$
0%
$25$
0%
$14$
0%
$28$

Explanation

Given,

$PQ=8$

$PS=6$

From figure,

$\triangle PQS$ is a right angled triangle.

So, using Pythagoras theorem,

$PS^2=PQ^2+PS^2$

$\Rightarrow PS^2=8^2+6^2=64+36=100$

$PS\sqrt{100}=10$

NOw, QS is also the radius of are AQB.

We known that formula for length of are

$=2\pi r\left(\dfrac{\theta}{360}\right)$

where

$r=radius \, of \, arc$

$\theta=$ angle subtended by the arc.

Here,

$r=10$ and

$\theta=90$

$\therefore$ Arc length

$AQB=2\times \pi\times 10\times \left(\dfrac{90}{360}\right)$

$2\times\pi \times 10\times\dfrac{1}{4}=5\pi$

Hence the length of arc is

${5\pi}$

2108064_536031_ans_b12040c65ced4c67b3763c330794e049.png

A show-piece, as shown in the figure, is made of a cube and a hemisphere. If the measure of the total surface area of the cube is represented by

$A$ , the curved surface area of the hemisphere is represented by

$B$ and the area of the base of the hemisphere is represented by

$C$ , then ................ is true for the total surface area of the show-piece.

0%
$A+B+C$
0%
$A+B-C$
0%
$B+C-A$
0%
$A+C-B$

Explanation

We know that the total surface area of the showpiece is obtained by adding the surface area of every part.

The surface area of the cube and hemisphere will include the area of the base of the hemisphere which is extra.

So, to get the total surface area, we need to subtract the area of the base of the hemisphere from that of the hemisphere.

$\therefore$ Total surface Area

$=$ TSA of cube

$+$ CSA of hemisphere

$-$ (Area of base of the hemisphere)

$=A+B-C$

Area of figure as shown

0%
$12\ \text{m}^{2}$
0%
$13\ \text{m}^{2}$
0%
$14\ \text{m}^{2}$
0%
$15\ \text{m}^{2}$

Explanation

Area of rectangle

$( ABCD)=5\times 1=5m^{2}$

Area of rectangle

$( EFGH)=7\times 1=7m^{2}$

Total =

$7+5=12m^{2}$

1087373_923779_ans_461465a590584012a7d1b4fa0042211c.png

In the given figure,

$ABC$ is a quadrant of a circle of radius

$14\ cm$ and a semicircle is drawn with

$BC$ as diameter. Find the area of the shade region.

0%
$98cm^2$
0%
$38cm^2$
0%
$48cm^2$
0%
$58cm^2$

Explanation

Area of shaded region=area of semicircle of diameter BC-{area of quadrant of radius AB/AC- area of

$\triangle$ ABC}

$\because$ BC is hypotenuse of right angle

$\triangle ABC$

here

$AB=BC=14$

So,

$BC=14\sqrt2=2\times radius\Rightarrow radius=7\sqrt2$

So, Area of semicircle of diameter BC

$=\dfrac{\pi r^2}{2}$

$=\dfrac{1}{2}\times\dfrac{22}{7}\times(7\sqrt2)^2$

$=154cm^2$

Area of quadrant of radius AB/AC

$=\dfrac{\pi r^2}{4}$

$=\dfrac{1}{4}\times\dfrac{22}{7}\times14\times14$

$=154cm^2$

Area of

$\triangle ABC=\dfrac{1}{2}\times h\times b$

$=\dfrac{1}{2}\times14\times14=98cm^2$

Now, area of shaded region

$=154-\{154-98\}=98cm^2$

Hence, area of shaded region

$=98cm^2$

From the above figure area of shade region is

0%
$42\ cm^{2}$
0%
$196\ cm^{2}$
0%
$154\ cm^{2}$
0%
$350\ cm^{2}$

Explanation

Area of

$ABCD$

$={14}^{2}=196$ sq.cm

Area of

$APBA$

$\dfrac{\pi{d}^{2}}{{8}}=\dfrac{\pi{(14)}^{2}}{8}$ sq.cm

$=76.97$ sq.cm

Area of

$CPDC$

$=$ Area of

$APBA$

$=76.97$ sq.cm

$\therefore$ Area of the shaded region

$=196-\left(2\times76.97\right)$ sq.cm

$=42.06$ sq.cm

$\sim 42$ sq.cm

$Note : \left[sq.cm={cm}^{2}\right]$

1390645_1126001_ans_d290631b26a445a5b2bfa4554318984e.png

The area of small square is

$100\ sq.\ cm$ . The area of the large square in

$sq.\ cm$ is

0%
$144$
0%
$165$
0%
$200$
0%
$400$

Explanation

Small square area

$=100cm^2$

$s^2=100\ cm^2$

$s=10\ cm$

Diagonal of square

$=10\sqrt{2}$

Half the diagonal length

$=$ radius of circle

$=\dfrac{10\sqrt{2}}{2}$

$=5\sqrt{2}$

Diameter of circle

$=$ length of side of big square=diagonal of smaller square =

$10\sqrt{2}$

Area of big square

$=(10\sqrt{2})^2$

$=200\ cm^2$ .

If the perimeter and the area of a circle are numerically equal, then the radius of the circle is

0%
$2\ units$
0%
$p\ units$
0%
$4\ units$
0%
$7\ units$

Explanation

Given,
Numerically perimete of the circle is equal to area of circle

$\Rightarrow 2\pi r=\pi r^2$

$\Rightarrow 2=r$

$\therefore r=2$

In the given figure

$ABCDE$ is a pentagonal park in which

$DE=DC,AB=BC=CE=EA=25m$ and its total height is

$41m$ . Find the area of the park

0%
$225cm^2$
0%
$525cm^2$
0%
$625cm^2$
0%
$825cm^2$

Explanation

Height of triangle

$=41-25 =16cm$

Area of park

$=$ area of square

$ABCE \,+$ area of

$\Delta DEC$

$= 25\times 25+\dfrac{1}{2}\times 25 \times 16$

$=625+200$

$=825cm^{2}$

The length of the are that subtends an angle

${30}^{o}$ at the centre of a circle of radius

$4\ cm$ :

0%
$\dfrac {2\pi}{3}\ cm$
0%
$\dfrac {4\pi}{8}\ cm$
0%
$\dfrac {5\pi}{4}\ cm$
0%
$\dfrac {\pi}{3}\ cm$

Explanation

Length of arc

$= (\dfrac{\theta }{360}\times 2\pi r)\,cm$

$\therefore$ Length

$= (\dfrac{30}{360}\times 2\times \dfrac{22}{7}\times 4)\,cm$

$\therefore$ Length

$= (\dfrac{1}{6}\times \dfrac{22\times 4}{7})\,cm$

$\therefore$ Length

$= \dfrac{44}{21}\,cm$

$\therefore$ Length

$= 2.09\,cm$

$= (\dfrac{2\pi }{3})\,cm$

Find the area of the shaded region.

0%
$5634cm^2$
0%
$1004cm^2$
0%
$1034cm^2$
0%
$1134cm^2$

Explanation

$r+2r+r=42$

$4r=42$

$r=\dfrac{21}{2}cm$

Area of shaded region =

$4\times \dfrac{\pi r^2}{2}+(2r)(2r)$

$=2\pi r^2+4r^2$

$2\times \dfrac{22}{7}\times \dfrac{21}{2}\times \dfrac{21}{2}+4\times\dfrac{21}{2}\times \dfrac{21}{2}$

$=33\times 21+21\times 21$

$=54\times 21$

$=1134 cm^2$

1371941_1169930_ans_8b3d1e851dfa44f099db9dae2797e9f2.png

A chord of a circle of radius

$6\ cm$ subtends a right angle at centre. The area of the minor segment

$(\pi=3.14)$ is____.

0%
$9.82\ cm^{2}$
0%
$10.14\ cm^{2}$
0%
$10.26\ cm^{2}$
0%
$12.35\ cm^{2}$

Explanation

$\text{Area of the minor segment}$

$= \text{Area}(Sector\; OACB) - \text{Area}(\triangle AOB)$

$= \dfrac{90^{\circ}}{360^{\circ}} \times \pi \times r^2 - \dfrac{1}{2} \times OA\times OB$

$= \dfrac{90^{\circ}}{360^{\circ}} \times 3.14 \times6^2 - \dfrac{1}{2} \times 6\times 6$

$= 28.26 - 18$

$= 10.26 \text{ cm}^2$

Option C is correct.

The area of shaded region of the figure given below is __________. [Take

$\pi = \dfrac {22}{7}$ and

$\sqrt {3} = 1.732]$ .

0%
$4.28\ cm^{2}$
0%
$5.76\ cm^{2}$
0%
$3.27\ cm^{2}$
0%
$18.84\ cm^{2}$

Explanation

Area of sector

$OAB = \dfrac{\theta}{360^{\circ} } \times \pi r^2$

where

$\theta = \angle AOB$

Since

$\Delta AOB$ is an equilateral triangle

$\therefore \angle AOB = 60^{\circ}$

$\therefore$ Area of sector

$OAB = \dfrac{60^{\circ}}{360^{\circ}} \times \pi \times 6^2$

$= \dfrac{1}{6} \times \dfrac{22}{7} \times 6^2$

$= 18.85 cm^2$

Area of equilateral triangle

$OAB$

$= \dfrac{\sqrt{3}}{4} a^2$

$= \dfrac{\sqrt{3}}{4} \times 6 \times 6$

$= 9 \sqrt{3} cm^2$

$= 15.58 cm^2$

Area of shaded region

$= (18.85 - 15.58) cm^2$

$= 3.27 cm^2$

Answer:

$C) \, 3.27 cm^2$

2124125_1170899_ans_2f4558b21e504908b0fa327ae803bb39.png

If three circles of radius

$a$ each, are drawn such that each touches the other two, then the area included between them is equal to

$\frac{4}{25}a^2$ .

0%
True
0%
False

Five identical rectangles are placed inside a square with side

$24 cm$ as shown in the diagram. What is the area of one rectangle ?

0%
$12 \mathrm { cm } ^ { 2 }$
0%
$16 \mathrm { cm } ^ { 2 }$
0%
$24 \mathrm { cm } ^ { 2 }$
0%
$32 \mathrm { cm } ^ { 2 }$

Explanation

Consider vertical line, let

$b$ &

$h$ be breadth and height of rectangle then,

$24=b+h+h+b$ (vertically)

$b+h=12 \ \ \ \ \ \ ....(1)$

$24=h+(h-b)+b+h$ (horizontally)

$3h=24$

$h=8$

Substituting

$h=8$ in equation

$(1)$

$b=4$

Area

$= bh=32 \ cm^2$ .

In the following figure

$A B = B C$ and

$A C = 84 \mathrm { cm } .$ The radius of the smallest circle is

$14\mathrm { cm } .$

$B$ is the centre of the largest semicircle. What is the area of the shaded region.

0%
$335 \mathrm { cm } ^ { 2 }$
0%
$770 \mathrm { cm } ^ { 2 }$
0%
$840 \mathrm { cm } ^ { 2 }$
0%
$650 \mathrm { cm } ^ { 2 }$

Explanation

REF.Image

AC = 84 cm

AB = BC =

2AB = 84

AB = 42

Area of circle (semi) whose diameter is AB.

$A_{1}=\pi R^{2}/2$

$=\pi (21)^{2}/2=\frac{22}{7}\times 22\times 21\times \frac{1}{2}$

$= 11\times 3\times 21=33\times 21$

$= 693 cm^{2}$

Area of circle (semi) whose diameter is BC

$A_{2}=\frac{\pi R^{2}}{2}$

$= 693 cm^{2}$

Area of circle whose diameter is AC

$A=\frac{\pi R^{2}}{2}=\frac{22}{7}\times 42\times 42\times \frac{1}{2}$

$A=\frac{5544}{2}cm^{2}= 2772 cm^{2}$

Area of smallest circle

$a=\pi r^{2}=\frac{22}{7}\times 14\times 14=616 cm^{2}$

Area of the shaded region =

$A-(A_{1}+A_{2}+a)$

$=2772-(693+693+616)$

$=2772-2002$

$770 cm^{2}$

1200710_1294152_ans_fcb8fd2efe9b403b900cdd5152606888.jpg

The geometric shape that has no corners is _____.

0%
square
0%
circle
0%
rectangle
0%
hexagon

The region between an arc and its corresponding chord is called _____.

0%
Area of sector
0%
Segment
0%
Sector
0%
Semicircle

If chord of a circle of radius 28 cm make an angle of

$90^{\circ}$ at the centre, then the area of the major segment is

0%
392 $\text{cm}^{2}$
0%
1456 $\text{cm}^{2}$
0%
1848 $\text{cm}^{2}$
0%
2240 $\text{cm}^{2}$

Explanation

Chord of a circle of radius 28 cm makes an angle of

$90^{\circ}$ from the center.

Area of circle =

$\pi r^2$

Area of shaded part =

$\dfrac{3}{4}\pi r^2$

$= \dfrac{3}{4}\times\dfrac{22}{7}\times 28^2 cm^2$

$= 3\times 22\times 28 cm^2$

$= 66\times 28 cm^2$

$= 1848 cm^2$

Area of triangle =

$\dfrac{1}{2}\times b\times h$

$= \dfrac{1}{2}\times 28\times 28 cm^2$

$= 14\times 28 cm^2$

$= 392 cm^2$

So, Area of major segment = 1848 + 392

$= 2240 cm^2$

State whether following statement is true or false:

The nearer a chord to the centre of a circle, the longer is its length.

0%
True
0%
False

In a circle of radius

$\displaystyle 14$ cm an arc subtends an angle of

$\displaystyle 36^0$ at the centre. The length of the arc is

0%
$\displaystyle 6.6$ cm
0%
$\displaystyle 7.7$ cm
0%
$\displaystyle 8.8$ cm
0%
$\displaystyle 9.1$ cm

Explanation

$\displaystyle \theta = \left( 36 \times \frac{\pi}{180}\right)^c = \left( \frac{\pi}{5} \right)^c$ and

$\displaystyle r = 14$ cm.

$\displaystyle \therefore \ \ l = r \theta = \left( 14 \times \frac{\pi}{5} \right)$ cm =

$\displaystyle \left( 14 \times \frac{22}{7} \times \frac{1}{5} \right)$ cm =

$\displaystyle \frac{44}{5}$ cm =

$\displaystyle 8.8$ cm.

Is it true to say that area of a segment of a circle is less the area of its corresponding sector ? Why ?

0%
True
0%
False

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 4 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Areas Related To Cricles Quiz 4 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.