Advertisements
Advertisements
Question
Calculate the area of the shaded region, if the diameter of the semicircle is equal to 14 cm. Take `pi = 22/7`

Advertisements
Solution
The diameter of the semi-circle is 14 cm.
ED = AC = 14 cm
Therefore, AB = BC = AE = CD = 7 cm
Area of the shaded region = Area of semi-circle EFD [Area of rectangle AEDC – 2 quarter circle]
`= 1/2 pir^2 + [AE xx ED - 2 xx 1/4 pir^2]`
`= 1/2 pir^2 + AE xx AE - 1/2 pir^2`
= 7 x 14
`= 98 cm^2`

APPEARS IN
RELATED QUESTIONS
ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.

In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°. Calculate : ∠RNM

In the figure, given alongside, AB || CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°,
Calculate:
- ∠RPQ,
- ∠STP.

Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate: ∠AEF

In the given figure, AB is the diameter of a circle with centre O.
If chord AC = chord AD, prove that:
- arc BC = arc DB
- AB is bisector of ∠CAD.
Further, if the length of arc AC is twice the length of arc BC, find:
- ∠BAC
- ∠ABC

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°.
Calculate: ∠ADC
Also, show that the ΔAOD is an equilateral triangle.

In the given figure, O is the centre of the circle and ∠PBA = 45°. Calculate the value of ∠PQB.

