Tag: circle measures

Questions Related to circle measures

A semicircle is drawn with $AB$ as its diameter. From $C$ a point on $AB$ a line perpendicular to $AB$ is drawn meeting the circumference of the semicircle at $D$. Given that $AC = 2\ cm$ and $CD = 6\ cm$ the area of the semicircle is :

  1. $\displaystyle 32\pi $

  2. $\displaystyle 50\pi $

  3. $\displaystyle 40\pi $

  4. $\displaystyle 36\pi $


Correct Option: B
Explanation:

Let O be the centre of the circle. Then, $OA = OB = OD= r$
Now, $OC = r - 2$
and $CD = 6$
Thus, in $\triangle ODC$
$OC^2 + CD^2 = OD^2$
$(r - 2)^2 + 6^2 = r^2$
$r^2 + 4 - 4r + 36 = r^2$
$4r = 40$
$r = 10$ $cm$
Area of semicircle $= \dfrac{\pi r^2}{2} = \dfrac{\pi (10)^2}{2} = 50 \pi$

If a wire is bent into the shape of a square, then the area of the square is $81\ cm^{2}$. When the same wire is bent into a semi-circular shape, then the area of the semi circle will be

  1. $22\ cm^{2}$

  2. $44\ cm^{2}$

  3. $77\ cm^{2}$

  4. $154\ cm^{2}$


Correct Option: C
Explanation:

Let the side of the square be $a$ cm

Thus, perimeter is $4a$ and area of the square is $a^2$
$\therefore a^2=81$
$\Rightarrow a=\sqrt{81}$
$\Rightarrow a=9$
Thus, perimeter $=4 \times 9$ $=36$ cm
Now, wire is bent into a semicircle.
Therefore, $2r+πr=36$
$⇒r(π+2)=36$ $
$⇒r=\cfrac{36}{\dfrac{22}{7}+2}$

$⇒r=\cfrac { 36 }{\dfrac{36}{7}}$
$⇒r=7$ cm

Area of semicircle $=\cfrac{1}{2}πr^2$
$=\cfrac{1}{2}×\cfrac{22}{7}×7×7$
$=77\ cm^2$

The inner circumference of a circular tracks is $220\ m$. The track is $7$ $m$ wide everywhere. Calculate the cost of putting up a fence along the outer circle at the rate of Rs. $2$ per metre. Use $\pi=\displaystyle\frac{22}{7}$

  1. Rs. $947$

  2. Rs. $726$

  3. Rs. $612$

  4. Rs. $528$


Correct Option: D
Explanation:

Circumference of inner side = $220 m$


$\Rightarrow 2\pi r=220\Rightarrow r=\dfrac { 220\times 7 }{ 44 } =35m$


Now, width of track  $=7m$

$\therefore $ Outer radius $=35+7 = 42 m$

Therefore outer circumference  $=2\pi R=2\times \dfrac { 22 }{ 7 } \times 42=264m$

$\therefore $ Cost of fencing $=$ Rs. $\left( 264\times 2 \right) $ $=$ Rs. $528$

The areas of two concentric circles forming a ring are 154 sq cm and 616 sq cm The breadth of the ring is

  1. $21 cm$

  2. $56 cm$

  3. $14 cm$

  4. $7 cm$


Correct Option: D
Explanation:

Breadth of the ring is equal to the difference between the radius of the outer circle and the radius of the inner circle
Given the area of outer circle=616$\displaystyle cm^{2}$
$\displaystyle \Rightarrow  \pi r _{2}^{2}=616 cm^{2}$
$\displaystyle \Rightarrow r _{1}^{2}=\frac{616\times 7}{22}=196$
$\displaystyle \therefore r _{1}=14 cm $
and the area of the inner circle $\displaystyle =154 cm^{2}$
$\displaystyle \Rightarrow \pi r _{2}^{2}= 154$
$\displaystyle \Rightarrow r _{2}^{2}=\frac{154\times 7}{22}=49$
$\displaystyle \therefore r _{2}=7 cm.$
$\displaystyle \therefore $ The required answer $\displaystyle =r _{1}-r _{2}=14-7=7 cm.$

If the difference between the circumference and radius of a circle is 37 cm, then the area of the circle is

  1. $111 cm^2$

  2. $148 cm^2$

  3. $259 cm^2$

  4. $154 cm^2$


Correct Option: D
Explanation:

Consider the difference between circumference and radius.


Taking$\pi $ as $3.14$ will give an approximate answer. we can do it by taking $22/7$ as wel


Let, the radius be$ r cm$


Circumference will be,

$2r=2\times 3.14r=6.28r$


ATQ,  $6.28r-1.r=37$

$5.28r=37$

 $r=\dfrac{37}{5.28}$

$r=$approx. $7 cm$ 

Area$=\pi {{r}^{2}}=3.14\times 7\times 7=154c{{m}^{2}}$


Hence, this is the answer.

If the circumference of a circle is reduced by 50%, then the area will be reduced by

  1. 50%

  2. 25%

  3. 75%

  4. 12.5%


Correct Option: C
Explanation:
Let the original radius be $r$.
So, the area of circle $=\pi r^2$             $....... (1)$

Since, the circumference of the circle is reduced by $50\%$.
It means that the radius of the circle is also reduced by $50\%$.

Then,
The new radius $=0.5r$

Therefore, the new area
$=\pi (0.5r)^2$
$=0.25\pi r^2$

Therefore, the required $\%$
$=\dfrac{\pi r^2-0.25\pi r^2}{\pi r^2}\times 100$
$=\dfrac{0.75\pi r^2}{\pi r^2}\times 100$
$=75\%$

Hence, this is the answer.

The diameter of semi circular field is $49$ m. What the cost of fencing the plot of Rs. $10$ per metre?

  1. Rs. $ 1259.3$

  2. Rs. $1260$

  3. Rs. $1250$

  4. None


Correct Option: A
Explanation:

The perimeter of a semi circle is $= \pi r+2r$

Here $r:$ radius of the semi circle
Given that $2r=49$
$\Rightarrow r= \dfrac { 49 }{ 2 } $
Perimeter of the semi circle $ = (\pi +2)r= (\pi +2)\dfrac { 49 }{ 2 } = 125.93$ m 
Cost of fencing the plot per metre $=$ Rs. $10$
Total cost of fencing $= 125.93\times10=$ Rs. $1259.3$.

Find the area of the semicircle whose radius is $4$ cm.

  1. $25.12$ sq. cm

  2. $15.12$ sq. cm

  3. $12.12$ sq.cm

  4. $21.12$ sq. cm


Correct Option: A
Explanation:

Given, radius $=4$ cm
We need area of semicircle with the given radius
Area of a semicircle $=$ $\dfrac{1}{2}\pi  r^2$
$=$ $\dfrac{1}{2}\pi\times  (4)^2$
$= 25.12$ sq. cm

Find the circumference of the semicircle whose diameter is $4\ cm.$

  1. $3.46\ cm$

  2. $5.89\ cm$

  3. $6.28\ cm$

  4. $10.28\ cm$


Correct Option: D
Explanation:

Diameter $= 2 \times$ radius
radius $= \dfrac{4}{2}=2$ $cm$
Circumference of the semicircle $= \pi r+ 2r$
= $3.14 \times  2 +4= 10.28$ $cm$

A window in the shape of a semicircle has a radius of $20\ cm$. Find the area of the semicircle.

  1. $128\ cm$

  2. $428\ cm$

  3. $628\ cm$

  4. $290\ cm$


Correct Option: C
Explanation:

Area of a semicircle $=\dfrac{1}{2}\pi r^2$
$=\dfrac{1}{2}\pi (20)^2$
$=628\ cm^2$