In the given figure,OP = PQ = 14 , cm and sem | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: $OP = 14 \, cm$ and $PQ = 14 \, cm$. Therefore,$OQ = OP + PQ = 14 + 14 = 28 \, cm$.The shaded region is bounded by three semicircular arcs:

Vedclass · Accepted Answer

$88$

Vedclass · Answer

(A) Given: $OP = 14 \, cm$ and $PQ = 14 \, cm$. Therefore,$OQ = OP + PQ = 14 + 14 = 28 \, cm$.The shaded region is bounded by three semicircular arcs:$1$. The arc of the semicircle with diameter $OQ$ (length = $\frac{1}{2} \pi 	imes 28 = 14\pi \, cm$).$2$. The arc of the semicircle with diameter $OP$ (length = $\frac{1}{2} \pi 	imes 14 = 7\pi \, cm$).$3$. The arc of the semicircle with diameter $PQ$ (length = $\frac{1}{2} \pi 	imes 14 = 7\pi \, cm$).The perimeter of the shaded area is the sum of the lengths of these three arcs:Perimeter = $14\pi + 7\pi + 7\pi = 28\pi \, cm$.Using $\pi = \frac{22}{7}$,we get:Perimeter = $28 	imes \frac{22}{7} = 4 	imes 22 = 88 \, cm$.

In the given figure,$OP = PQ = 14 \, cm$ and semicircles are drawn on $OP$,$PQ$,and $OQ$ as diameters. Then,the perimeter of the shaded area is . . . . . . $cm$.

Similar Questions

The circumference of a circle is $176\, m$. Then,its area is

If the sides of a triangle are doubled,its area

The length of a rectangular hall is $5\, m$ more than its breadth. The area of the hall is $750\, m^{2}$. The length of the hall is.........$m$

The area of a parallelogram is $72 \, cm^{2}$ and its altitude is twice the corresponding base. The length of the base is.....$cm$.

Inside a square plot,a circular garden is developed which exactly fits in the square plot and the diameter of the garden is equal to the side of the square plot which is $28\, m$. What is the area of the space left out in the square plot after developing the garden? (in $m^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module