Question

find the area of the shaded figure please it is urgent​

Answer 1

$\huge \mathfrak{solution}$

now area of the shaded portion

=area of trapezium (ADCB-EHGF)

now area of the trapezium ADCB is

$= \frac{1}{2} \times (AD + BC) \times height \\ = \frac{1}{2} \times (40 + 24) \times 25 \\ = 32 \times 25 \\ = 800 \: cm {}^{2}$

now area of the trapezium EHGF is

$= \frac{1}{2} \times (EH + GF) \times height \\ = \frac{1}{2} \times (30 + 16) \times 15 \\ = 23 \times 15\\ = 345 \: cm {}^{2}$

therefore area of the shaded portion is

$\large \implies area_{shaded} = (800 - 345) \: cm {}^{2} \\ \large \implies area_{shaded} = 455 \: cm {}^{2}$

hope this helps you

Answer 2

Answer:

Area of the shaded region is 455 cm^2.

Step-by-step-explanation:

Given,

Lengths of || sides of bigger trapezium are 24 cm & 40 cm. Lengths of || sides of smaller trapezium are 30 cm & 16 cm.

Also,

Height of the bigger and smaller trapeziums are 25 cm and 15 cm respectively.

From the properties of quadrilaterals :

• Area of trapezium = 1 / 2 x sum of parallel sides x height

Here, on observing :

= > Area of the shaded region = Area of bigger trapezium - area of smaller trapezium

= > Area of the shaded region = [ 1 / 2 x ( 24 cm + 40 cm ) x 25 cm ] - [ 1 / 2 x ( 16 cm + 30 cm ) x 15 cm ]

= > Area of the shaded region = [ 32 cm x 25 cm ] - [ 23 cm x 15 cm ] = 800 cm^2 - 345 cm^2

= > Area of the shaded region = 455 cm^2

Hence, area of the shaded region is 455 cm^2.