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In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. [Use π = 3.14.]
Here in order to find the area of the shaded region (region excluding the triangle) we have to subtract the area of the triangle from the area of the rectangle and then add the area of the semicircle.
Given AB = 20 cm, DE = 12 cm, AE = 9 cm and ∠AED = 90°
Here we see that the triangle AED is right angle triangle, therefore, we can apply Pythagoras theorem i.e.
AD2 = DE2 + AE2
AD2 = 122 + 92 (putting given values)
⇒ AD2 = 144 + 81
⇒ AD2 = 225
⇒ AD = √225
∴ AD = 15 cm
Area of ∆AED = 1/2×AE×DE
(Area of triangle = 1/2×base×height)
On putting values we get,
⇒ Area of ∆AED = 9×6
∴ Area of ∆AED = 54 cm2→ eqn1
Here radius of semicircle = BC/2 = 15/2
⇒ R = 7.5 cm
⇒ Area of semicircle = 1.07×56.25
∴ Area of semicircle = 88.3125 cm2→ eqn2
Area of rectangle = ℓ×b
(ℓ = length of rectangle, b = breadth of rectangle)
⇒ Area of rectangle = 20×15
(putting the values of ℓ & b)
∴ Area of rectangle = 300 cm2→ eqn3
Area of shaded region = Area of rectangle + Area of semicircle – Area of ∆
⇒ Area of shaded region = 300 + 88.3125 – 53 (from eqn1, eqn2, eqn3)
∴ Area of shaded region = 334.3125 cm2
Area of shaded region is 334.3125 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.]