10 cm
In the figure, seg QR is tangent to the circle with centre O. Point Q is the point of contact. Radius of the
circle is 10 cm. OR = 20 cm. Find the area of the shaded region. (ne = 3.14, V3 = 1.73)
Answers
The area of the shaded region is 34.17 cm^2
Step-by-step explanation:
Since line drawn from the centre of a circle to the tangent is perpendicular to the tangent.
∴ OQ perpendicular to QRQR
△OQR is a right angled triangle.
OR^2 =QR^2 + OQ^2
QR = √ OR^2 - OQ^2
QR = √ 20^2 - 10^2
QR = 10 √ 3
Area of △OQR = 1/2 × QR × OQ
Area of △OQR = 1/2 x 10 x 10√ 3
Area of △OQR = 86.5 cm^2
Let △QOR = θ
sin θ = QR / OR
Sin θ = 10 √ 3 / 20
θ = 60°
Area of shaded region = Area of △QOR + Area of sector OQT
Area of shaded region = (86.5 - 52.33)cm^2
Area of shaded region = 34.17 cm^2
Thus the area of the shaded region is 34.17 cm^2
Answer :-
The area of the shaded region is 34.17 cm^2
Step-by-step explanation:
Since line drawn from the centre of a circle to the tangent is perpendicular to the tangent.
∴ OQ perpendicular to QRQR
△OQR is a right angled triangle.
OR^2 =QR^2 + OQ^2
QR = √ OR^2 - OQ^2
QR = √ 20^2 - 10^2
QR = 10 √ 3
Area of △OQR =
1/2 × QR × OQ
Area of △OQR =
1/2 x 10 x 10√ 3
Area of △OQR = 86.5 cm^2
Let △QOR = θ
sin θ = QR / OR
Sin θ = 10 √ 3 / 20
θ = 60°
Area of shaded region = Area of △QOR + Area of sector OQT
Area of shaded region = (86.5 - 52.33)cm^2
Area of shaded region = 34.17 cm^2
Thus the area of the shaded region is 34.17 cm^2