Question

Abcd is a trapezium with ab parallel to dc ab jadu 18 cm bc is equal to 32 cm and distance between ab and bc is equal to 14 cm if the are of equal radius 10 centimetre with centre a b c and d don find the area of shaded

Answer 1

Answer:

From the GIVEN FIGURE, we have

AB = 18 cm , DC = 32 cm , Distance between AB  and DC(h) = 14 cm & radius of each circle(r) = 7cm

Since , AB ||DC

Therefore , ∠A + ∠D = 180° & ∠B + ∠C = 180°

[CO INTERIOR ANGLES are supplementary]

Area of sector =  (θ /360) ×πr²

Area of sector with ∠A & ∠D = (180 /360) × 22/7 × 7²

= ½ × 22 × 7 = 11 × 7 = 77 cm²

Similarly,  Area of sector with ∠B & ∠C = (180 /360) × 22/7 × 7²

= ½ × 22 × 7 = 11 × 7 = 77 cm²

Area of trapezium = ½ (sum of parallel sides) ×distance between Parallel sides(h)

Area of trapezium = ½ (AB + DC ) ×(h)

Area of trapezium = ½(18+32)×14

= ½(50)× 14 = 25 × 14 = 350 cm².

Area of shaded region = Area of trapezium - (Area of sector with ∠A & ∠D + Area of sector with ∠B & ∠C)

= 350 -(77+77) = 350 - 154 = 196 cm²

Area of shaded region = 196 cm²

Hence, the Area of shaded region is 196 cm².

Answer 2

Given

=> AB ll CD

=> AB = 18 cm

=> CD = 32 cm

=> Distance between AB and CD = 14 cm

Now

=> Area of ABCD = 1/2(32+18) ×14 cm^2

=> Area of ABCD ABCD = 350 cm^2

Now

Circle are Drowned of Radius 10cm from A B, C, D

=> Area of Circle with 10cm = πr^2

$= \frac{22}{7} \times {10}^{2} {cm}^{2} \\ \\ = \frac{2200}{7} \: {cm}^{2}$

Hence Remaining Area

$= \: 350 {cm}^{2} - \frac{2200}{7} {cm}^{2} \\ \\ = 350 {cm}^{2} - 314.28 {cm}^{2} \\ \\ = 35.72 {cm}^{2} \: \: \: \: \: \: \: [ANSWER]$

_-_-_-_-_$BE \: \: BRAINLY$_-_-_-_-_