Listen NCERT Audio Books to boost your productivity and retention power by 2X.
In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded region. [Use π = 3.14.]
Here in order to find the area of the shaded region (region excluding the area of segment AC and quadrant OCD) can be calculated by subtracting the area of triangle and quadrant OBD from the area of the circle.
Given AC = 24 cm, AB = 7 cm and ∠BOD = 90°
Here we see that the triangle ACB is right angle triangle, therefore, we can apply Pythagoras theorem i.e.
BC2 = AC2 + AB2
⇒ BC2 = 242 + 72 (putting the given values)
⇒ BC2 = 576 + 49
⇒ BC2 = 625
BC = √625
∴ BC = 25 cm
Area of ∆ACB = 1/2×AB×AC (Area of triangle = 1/2×base×height)
On putting values we get,
⇒ Area of ∆AED = 7×12
∴ Area of ∆AED = 84 cm2→ eqn1
Area of circle = πR2 (R = radius of circle)
⇒ R = 12.5 cm
∴ Area of circle = π× 12.52
⇒ Area of circle = 156.25×3.14 (put π = 3.14)
∴ Are of circle = 490.625 cm2→ eqn2
⇒ Area of quadrant OBD = 122.65625 cm2→ eqn3
Area of shaded region = Area of circle – Area of quadrant – Area of ∆
⇒ Area of shaded region = 490.625 – 84 – 122.65625 (from eqn1, 2 and 3)
⇒ Area of shaded region = 283.96875 cm2
Area of shaded region is 283.96875 cm2.
ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m.
Find the following:
(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.
A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take π = 3.14.]