1

Question

In the given figure, a semicircle is drawn on the hypotenuse of the right-angled triangle. Another arc of radius 30 cm is drawn passing through A and C with O as the center. What is the area of the shaded region? (Take π as 3.14 and √3 = 1.732)

Open in App

Solution

The correct option is **C** 271.95 sq cm

Area of the shaded region = Area of semicircle APC - Area of segment AQC

Area of semicircle = πr22=3.14×15×152=353.25sqcm

Area of segment = Area of sector OAQC - Area of triangle OAC

Consider triangle OAC

Since all sides are equal to 30 cm in the triangle, it is an equilateral triangle.

Hence, each angle is equal to 60 degrees.

Hence ∠OAC=∠AOC=∠ACO=60∘.

Now, let the height be h.

sin60∘=h30

√32=h30

h=15√3

Now, area of triangle OAC is 12×base×h

=12×30×15√3=225√3sq.cm=389.7sq.cm

Area of the sector OAQC = 60360×π×r2=16×3.14×30×30=471sq.cm.

Hence, area of segment AQC = 471 - 389.7 = 81.3 sq cm

Hence, area of the shaded region = Area of semicircle - area of segment AQC

= 353.25 - 81.3 = 271.95 sq cm.

Area of the shaded region = Area of semicircle APC - Area of segment AQC

Area of semicircle = πr22=3.14×15×152=353.25sqcm

Area of segment = Area of sector OAQC - Area of triangle OAC

Consider triangle OAC

Since all sides are equal to 30 cm in the triangle, it is an equilateral triangle.

Hence, each angle is equal to 60 degrees.

Hence ∠OAC=∠AOC=∠ACO=60∘.

Now, let the height be h.

sin60∘=h30

√32=h30

h=15√3

Now, area of triangle OAC is 12×base×h

=12×30×15√3=225√3sq.cm=389.7sq.cm

Area of the sector OAQC = 60360×π×r2=16×3.14×30×30=471sq.cm.

Hence, area of segment AQC = 471 - 389.7 = 81.3 sq cm

Hence, area of the shaded region = Area of semicircle - area of segment AQC

= 353.25 - 81.3 = 271.95 sq cm.

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program