Question

In the following figure, ABCD is a trapezium of area 24.5 cm² , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. ​

Answer 1

Answer:

The area of the shaded region = 14.875 cm².

Step-by-step explanation:

Given :

Area of trapezium ABCD, A = 24.5 cm²

AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle.

Area of the trapezium, A = ½ (sum of parallel sides) × perpendicualr distance between the parallel sides

A =  ½ (AD + BC) × AB  

24.5 = ½ (10 + 4) × AB

24.5 × 2 = 14 AB  

49 = 14 AB  

AB = 49/14  

AB = 7/2

AB = 3.5 cm

Radius of the quadrant of the circle ,r =  AB = 3.5 cm

Area of the quadrant of the circle = ¼ ×πr²

= (1/4) (22/7 x 3.5 x 3.5)  

= 9.625 cm²

Area of the quadrant of the circle,ABE = 9.625 cm²

 

Area of the shaded region = Area of the trapezium,ABCD  - Area of the quadrant of the circle,ABE

= 24.5 - 9.625

Area of the shaded region = 14.875 cm²

Hence, the area of the shaded region = 14.875 cm².

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

$\huge\underline\mathcal\red{Solution}$

It is given that the area of trapezium ABCD is 24.5 cm^2 .

And AD || BC

<DAB = 90°

AD = 10 cm

BC = 4cm

As we know that

area of trapezium = 1/2 + (sum of parallel sides )× height

=> 24.5 = 1/2 ( AD + BC )× AB

=> 24.5 × 2 = (10+4) × AB

=> 49 = 14 × AB

=> AB = 49/14

=> AB = 3.5 cm

Now,

Area of quadrant

$= \frac{1}{4} \pi {r}^{2}$

$= \frac{1}{4}\pi {AB}^{2}$

$= \frac{1} {4} \times \frac{22}{7} \times 3.5 \times 3.5$

$= \frac{1}{2} \times 11 \times 0.5 \times 3.5$

$= 9.625$

Area of quadrant = 9.625 cm^2 .

Now ,

Area of shaded region = area of trapezium - area of quadrant

$= 24.5 - 9.625$

$= 14.875$

Hence , the area of shaded region is 14.875 cm^2 .