In Fig. O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take $π = 3.142$ )

Open in App

Solution

the angle acb=90 (because semicircle)
therefore by pythagorous in triangle acb
ab=20
area of triangle=96 sq,cm
area of circle= 157.14 sq.cm
area of shaded region= area of circle-area of trianlgle=61.14sq.cm

Suggest Corrections

Similar questions

Q. In the following figure, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

Q. The given figure shows the center of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take

π = 3.142

)

ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take $π = 3.142$ and $\sqrt{3} = 1.732$ )

In the given figure

O

is centre of circular arc and

A O B

is straight line find the area of the shaded region. Given

A C = 12 c m

B C = 16 c m

Q. In the following figure, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π =3.142 and

\sqrt{3}

= 1.732).

Join BYJU'S Learning Program

In Fig. O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π=3.142)

the angle acb=90 (because semicircle) therefore by pythagorous in triangle acb ab=20 area of triangle=96 sq,cm area of circle= 157.14 sq.cm area of shaded region= area of circle-area of trianlgle=61.14sq.cm

In Fig. O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take $π = 3.142$ )

the angle acb=90 (because semicircle)
therefore by pythagorous in triangle acb
ab=20
area of triangle=96 sq,cm
area of circle= 157.14 sq.cm
area of shaded region= area of circle-area of trianlgle=61.14sq.cm