Question

In the figure, diameter AB is 12cm long. AB is trisected at points P and Q. Find the area of the shaded region.​

Answer 1

Area of shaded region = 2 × ( area of semi circle AQ - area of semi circle AP)

Hope it helps you.

Answer 2

Given:

AB= 12cm

AP=PQ=QB (∵ P and Q trisect AB)

To find:

Area of the shaded area

Solution:

AB=12 cm

AP=PQ=QB=$\frac{12}{3}$cm

=4 cm

Area of the shaded region= (area of semi-circle AQ - area of semi-circle AP)+(area of semi-circle BP - area semi-circle BQ)

Now, the$area of semi-circle AQ=area of semi-circle BP$, since the diameters AQ= BP.

Similarly, the$area of semi-circle AP=area of semi-circle QB$, since the diameters AP=BQ.

So we can write,

Area of the shaded region=2× (area of semi-circle AQ - area of semi-circle AP)

AQ=(4+4)cm=8cm

AP=4cm

So,

Area of the shaded region=2×$(\frac{1}{2}\pi$×$(4)^2-\frac{1}{2}\pi$×$(2)^2)$

=2×$\frac{1}{2}$×$\frac{22}{7}$×$(16-4)$

=$\frac{22}{7}$×$12$

$=\frac{264}{7}$

=$37.7 cm^2$

Hence, the area of the shaded area is$37.7 cm^2$.