M.C.Q (1 Marks)

MCQ 11 Mark

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii $24\ cm$ and $7\ cm$ is:

A
$31\ cm$
B
$25\ cm$
C
$62\ cm$
✓
$50\ cm$

Answer

Correct option: D.

$50\ cm$

Let $r_1 = 24\ cm$ and $r_2 = 7\ cm$
$\therefore$ Area of first circle $=\pi\text{r}^2_1=\pi(24)^2=576\pi\text{ cm}^2$
and area of second circle $=\pi\text{r}^2_1=\pi(7)^2=79\pi\text{ cm}^2$
According to the given condition$,$
Area of circle $=$ Area of first circle $+$ Area of second circle
$\therefore\ \ \pi\text{R}^2=576\pi+49\pi [$where$, R$ be radius of circle$]$
$\Rightarrow\ \ \text{R}^2=625$ $\Rightarrow\ \text{R}=25\text{cm}$
$\therefore$ Diameter of a circle $= 2R = 2 \times 25 = 50\ cm$

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MCQ 21 Mark

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is :

A
$22:07$
✓
$14:11$
C
$07:22$
D
$11:14$

Answer

Correct option: B.

$14:11$

Let radius of circle be $ r$ and side of a square be a.
According to the given condition,
perimeter of a circle $=$ Perimeter of a square
$\therefore\ \ 2\pi\text{r}=4\text{a}$
$\Rightarrow\ \text{a}=\frac{\pi\text{r}}{2}\ \dots(\text{i})$
Now, $\frac{\text{Area of circle}}{\text{Area of square}} $
$=\frac{\pi\text{r}^2}{(\text{a}^2)}$ $=\frac{\pi\text{r}^2}{\Big(\frac{\pi\text{r}}{2}\Big)^2}\ \ \ [\text{from Eq. (i)}]$
$=\frac{\pi\text{r}^2}{\frac{\pi^2\text{r}^2}{4}}=\frac{4}{\pi}$
$=\frac{4}{\frac{22}{7}}=\frac{28}{22}=\frac{14}{11}$

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MCQ 31 Mark

If the sum of the circumferences of two circles with radii $R1$ and $R2$ is equal to the circumference of a circle of radius $R,$ then

✓
$R1 + R2 = R.$
B
$R1 + R2 > R.$
C
$R1 + R2 < R.$
D
Nothing definite can be said about the relation among $R1, R2$ and $R.$

Answer

Correct option: A.

$R1 + R2 = R.$

According to the given condition,
Circumference of circle $=$ Circumference of first $+$ Circumference of second circle
$\therefore\ \ 2\pi\text{R}=2\pi\text{R}_1+2\pi\text{R}_2$
$\Rightarrow\ \ \text{R}=\text{R}_1+\text{R}_2$

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MCQ 41 Mark

If the circumference of a circle and the perimeter of a square are equal, then:

A
Area of the circle $=$ Area of the square.
✓
Area of the circle $>$ Area of the square.
C
Area of the circle $<$ Area of the square.
D
Nothing definite can be said about the relation between the areas of the circle and square.

Answer

Correct option: B.

Area of the circle $>$ Area of the square.

According to the given condition,
Circumference of a circle $=$ Perimeter of square
$2\pi\text{r}=4\text{a}$
$[$where$, r$ and $a$ are radius of circle and side of square respectively$]$
$\Rightarrow\ \frac{22}{7}\text{r}=2\text{a}\Rightarrow11\text{r}=7\text{a}$
$\Rightarrow\ \text{a}=\frac{11}{7}\text{a}\Rightarrow\text{r}=\frac{7\text{a}}{11}\ \dots(\text{i})$
Now$,$ area of circle$,\text{A}_1=\pi\text{r}^2$
$=\pi\Big(\frac{7\text{a}}{11}\Big)^2=\frac{22}{7}\times\frac{49\text{a}^2}{121}\ \ [\text{from Eq. (i)}]$
$=\frac{14\text{a}^2}{11}\ \dots(\text{ii})$
and area of square$, A_2 = (a)_2 ...(iii)$
From $Eqs. (ii)$ and $(iii), \text{A}_1=\frac{14}{11}\text{A}_2$
$\therefore\ \ \text{A}_1>\text{A}_2$
Hence$,$ Area of the circle $>$ Area of the square.

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MCQ 51 Mark

It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters $16m$ and $12m$ in a locality. The radius of the new park would be:

✓
$10m$
B
$15m$
C
$20m$
D
$24m$

Answer

Correct option: A.

$10m$

Area of first circular park, whose diameter is $16m$
$=\pi\text{r}^2=\pi\Big(\frac{16}{2}\Big)^2=64\pi\text{m}^2$
$\Big[\because\text{r}=\frac{\text{d}}{2}=\frac{16}{2}=8\text{m}\Big]$
Area of second circular park, whose diameter is $12m$
$=\pi\Big(\frac{12}{2}\Big)^2=\pi(6)^2=36\pi\text{m}^2$
$\Big[\because\text{r}=\frac{\text{d}}{2}=\frac{12}{2}=6\text{m}\Big]$
According to the given condition,
Area of single circular park $=$ Area of first circular park $+$ Area of second circular park
$\pi\text{R}^2=64\pi+36\pi\ [\because R$ be the radius of single circular park$]$
$\Rightarrow\ \ \pi\text{R}^2=100\pi$
$\Rightarrow\text{R}^2=100$
$\therefore\ \ \text{R}=10\text{m}$

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MCQ 61 Mark

If the sum of the areas of two circles with radii $R_1$ and $R_2$ is equal to the area of a circle of radius $R, $ then :

A
$\text{R}_1+\text{R}_2=\text{R}$
✓
$\text{R}_1^2+\text{R}_2^2=\text{R}^2$
C
$\text{R}_1+\text{R}_2<\text{R}$
D
$\text{R}_1^2+\text{R}_2^2<\text{R}^2$

Answer

Correct option: B.

$\text{R}_1^2+\text{R}_2^2=\text{R}^2$

According to the given condition,
Area of circle $=$ Area of first circle $+$ Area of second circle
$\therefore\ \ \pi\text{R}^2=\pi\text{R}_1^2+\pi\text{R}^2_2$
$\Rightarrow\ \ \text{R}^2=\text{R}^2_1+\text{R}^2_2$

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MCQ 71 Mark

The area of the square that can be inscribed in $a$ circle of radius $8\ cm$ is:

A
$256\text{cm}^2$
✓
$128\text{cm}^2$
C
$64\sqrt{2}\text{cm}^2$
D
$64\text{cm}^2$

Answer

Correct option: B.

$128\text{cm}^2$

Given$,$ radius of circle$, r = OC = 8\ cm.$
$\therefore$ Diameter of a circie$, = AC = 2 \times 8 = 16\ cm$
Which is equal to the diagonal of a square.
Let side of square be $x.$
In right angled $\triangle\text{ABC}, AC^2 = AB^2 + BC^2\ [$by Phthagoras theorem$]$

$\Rightarrow (16)^2 = x^2 + x^2$
$\Rightarrow 256 = 2x^2$
$\Rightarrow x^2 = 128$
$\therefore$ Area of square $= x^2 = 128\ cm^2$
Alternate Answer
Radius of circle $(r) = 8\ cm$
Diameter of circle $(d) = 2r = 2 \times 8 = 16\ cm$
Since$,$ square inscribed in circle$,$
$\therefore$ Diagonal of the squre $=$ Diameter of circle
Now$,$ Area of square $=\frac{(\text{Diagonal})^2}{2}=\frac{(16)^2}{2}=\frac{256}{2}=128\text{cm}^2$

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MCQ 81 Mark

The radius of a circle whose circumference is equal to the sum of the circumferences of two circles of diameters $36\ cm$ and $20\ cm$ is:

A
$56\ cm$
B
$42\ cm$
✓
$28\ cm$
D
$16\ cm$

Answer

Correct option: C.

$28\ cm$

$\because$ Circumference of first circle $=2\pi\text{r}=\pi\text{d}_1=36\pi\text{ cm} [$given$, d_1 = 36\ cm]$
and circumference of second circle $=\pi\text{d}_2=20\pi\text{ cm} [$given$, d_2 = 20\ cm]$
According to the given condition$,$
Circumference of circle $=$ Circumference of first circle $+$ Circumference of second circle
$\Rightarrow\ \ \pi\text{D}=36\pi+20\pi\ [$where$, D$ is diameter of a circle$]$
$\Rightarrow\ \ \text{D}=56\text{cm}$
So$,$ diameter of a circle is $56\ cm.$
$\therefore$ Required radius of circle $\frac{56}{2}=28\text{cm}$

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MCQ 91 Mark

The area of the circle that can be inscribed in a square of side $6\ cm$ is :

A
$36\pi\text{ cm}^2$
B
$18\pi\text{ cm}^2$
C
$12\pi\text{ cm}^2$
✓
$9\pi\text{ cm}^2$

Answer

Correct option: D.

$9\pi\text{ cm}^2$

Given, side of square $= 6\ cm$

$\therefore$ Diameter of a circie $, (d) = $ Side of square $= 6\ cm$
$\therefore$ Radius of a circle $(\text{r})=\frac{\text{d}}{2}=\frac{6}{2}=3\text{ cm}$
$\therefore$ Area of circle $=\pi(\text{r})^2$
$=\pi(3)^2=9\pi\text{ cm}^2$

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MCQ 101 Mark

Area of the largest triangle that can be inscribed in a semi $-$ circle of radius $r$ units is :

✓
$\text{r}^2\text{ sq. units}$
B
$\frac{1}{2}\text{ r}^2\text{ sq. units}$
C
$2\text{r}^2\text{ sq. units}$
D
$\sqrt{2}\text{ r}^2\text{ sq. units}$

Answer

Correct option: A.

$\text{r}^2\text{ sq. units}$

Take a point $C$ on the circumference of the semi $-$ circle and join it by the end points of diameter $A$ and $B.$

$\therefore \angle\text{C}=90^\circ\ [$by property of circle$]$
$[$angle in a semi $-$ circle are right angle$]$
So, $\Delta\text{ABC}=\frac{1}{2}\times\text{AB}\times\text{CD}$
$=\frac{1}{2}\times2\text{r}\times\text{r}$
$=\text{r}^2\text{ sq. units}$

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MATHS M.C.Q (1 Marks)

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