
The length of the arc of a sector having central angle 90 degrees and radius 7cm is:
(a) 22cm
(b) 44cm
(c) 11cm
(d) 33cm
Answer
586.8k+ views
Hint: For finding the length of the arc of a circle we need to use the formula $R\theta $ , where R is the radius of the circle and $\theta $ is the angle subtended by the arc at the at the centre of the circle take in radians. So, for the above question R is 7 cm and convert $90{}^\circ $ to radians to get the value of $\theta $ .
Complete step-by-step answer:
Let us start by drawing a representative diagram of the situation given in the question for better understanding.
Now let us first convert $90{}^\circ $ to radians. We know that $180{}^\circ ={{\pi }^{c}}$ . If we divide both sides of this relation by 2, we get
$\dfrac{180{}^\circ }{2}=\dfrac{{{\pi }^{c}}}{2}$
$\Rightarrow 90{}^\circ =\dfrac{{{\pi }^{c}}}{2}$
So, $\theta $ for the above question is equal to $\dfrac{\pi }{2}$ .
Now we know that the length of the arc of a circle we need to use the formula $R\theta $ , where R is the radius of the circle and $\theta $ is the angle subtended by the arc at the at the centre of the circle taken in radians.
$arc\left( BC \right)=R\theta =7\times \dfrac{\pi }{2}$
Now we will put the value of $\pi =\dfrac{22}{7}$ .
$arc\left( BC \right)=7\times \dfrac{\pi }{2}=7\times \dfrac{22}{7\times 2}=11cm$
Therefore, the answer to the above question is option (c).
Note: If you want you can solve the question using the fact that the angle subtended by the whole circle at the centre is $360{}^\circ $ , so $90{}^\circ $ would be subtended by one fourth of the whole perimeter of the circle. Therefore, one fourth of the perimeter of the circle is the answer. Also, remember the value of $\pi $ you choose out of the two values $3,14\text{ and }\dfrac{22}{7}$ decides the complexity of the calculations.
Complete step-by-step answer:
Let us start by drawing a representative diagram of the situation given in the question for better understanding.
Now let us first convert $90{}^\circ $ to radians. We know that $180{}^\circ ={{\pi }^{c}}$ . If we divide both sides of this relation by 2, we get
$\dfrac{180{}^\circ }{2}=\dfrac{{{\pi }^{c}}}{2}$
$\Rightarrow 90{}^\circ =\dfrac{{{\pi }^{c}}}{2}$
So, $\theta $ for the above question is equal to $\dfrac{\pi }{2}$ .
Now we know that the length of the arc of a circle we need to use the formula $R\theta $ , where R is the radius of the circle and $\theta $ is the angle subtended by the arc at the at the centre of the circle taken in radians.
$arc\left( BC \right)=R\theta =7\times \dfrac{\pi }{2}$
Now we will put the value of $\pi =\dfrac{22}{7}$ .
$arc\left( BC \right)=7\times \dfrac{\pi }{2}=7\times \dfrac{22}{7\times 2}=11cm$
Therefore, the answer to the above question is option (c).
Note: If you want you can solve the question using the fact that the angle subtended by the whole circle at the centre is $360{}^\circ $ , so $90{}^\circ $ would be subtended by one fourth of the whole perimeter of the circle. Therefore, one fourth of the perimeter of the circle is the answer. Also, remember the value of $\pi $ you choose out of the two values $3,14\text{ and }\dfrac{22}{7}$ decides the complexity of the calculations.
Recently Updated Pages
A 5m wide cloth is used to make a conical tent of base class 10 maths CBSE

Briefly describe the stages in the clotting of blo class 11 biology CBSE

The magnification produced by a spherical mirror is class 10 physics CBSE

Explain the formation of potential barriers in a pn class 12 physics CBSE

A particle of mass m is moving along the side of a class 11 physics CBSE

The figure shows a rectangle ABCD with a semicircle class 10 maths CBSE

Trending doubts
Which places in India experience sunrise first and class 9 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Write the 6 fundamental rights of India and explain in detail

Difference Between Plant Cell and Animal Cell

What is the Full Form of ISI and RAW

Golden Revolution is related to AFood production BOil class 9 social science CBSE

