Question

27. Find the area of the shaded part in the following figure.
D
C
8 cm
10 cm
E
A
18 cm
IS​

Answer 1

Answer:

Area of shaded part = 156cm²

Step-by-step explanation:

In ΔCEB, ∠CEB = 90°

CE² + BE² = BC²-----(pythagoras theorem)

8² + BE² = 10²

BE² = 100 - 64

BE = √36

BE = 6cm

Area of figure = Area of rectangle - area of triangle

= l x b - 1/2 x b x h

= 18 x 10 - 1/2 x 6 x 8

= 180 - 24

=156cm²

Answer 2

Answer:

=${156cm}^{2}$

Step-by-step explanation:

Given:

CL = 10cm

CE = 8cm

∠CEL = 90°

AL = 18cm

To Find:

shaded part of the figure

Solution:

Using Pythagoras Theorem, ‎{${ c}^{2} = \sqrt{ {a}^{2} + {b}^{2} }$}

${10}^{2} = {8}^{2} + {b}^{2} \\ {10}^{2} - {8}^{2} = {b}^{2} \\ 100 - 64 = {b}^{2} \\ {b }^{2} = 36 \\ b = \sqrt{36} \\ b = 6 \\ EL = 6cm$

{Area of Right Angled Triangle= $\frac{ab}{2}$}

Area of the figure = Area of Rectangle ALCD - Area of △CEL

$= l \times b - \frac{ab}{2} \\ = 18 \times 10 - \frac{8 \times 6}{2} \\ = 180 - \frac{48}{2} \\ = 180 - 24 \\ = {156cm}^{2}$

Thus, the area of the shaded portion is ${156cm}^{2}$