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

Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that the rhombus, inscribed in a circle, is a square.
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, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°.
Calculate:
- ∠EBA,
- ∠BCD.

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.
Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
The following figure shows a circle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, find the perimeter of the cyclic quadrilateral PQRS.

In the figure, given below, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of lengths 24 cm and 18 cm respectively.

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 Fig, Chord ED is parallel to the diameter AC of the circle. Given ∠CBE = 65°, Calculate ∠ DEC.

In the given figure, AC is the diameter of the circle with center O.
CD is parallel to BE.
∠AOB = 80° and ∠ACE = 20°
Calculate:
- ∠BEC
- ∠BCD
- ∠CED

