Question

13. Find the area of the shaded region in each of the following figures, in which the dimensions
given are in centimetres.
please solve ​

Answer 1

Answer :-

• Area of shaded portion = 26 cm²

Given :-

• AB = 8 cm
• AC = 15 cm
• DE = 4 cm

To Find :-

Area of shaded portion (ABDC) = ?

Explanation :-

In order to find the area of shaded portion first we need to find the are of both triangles.

As from the given figure, Δ ABC is a right angle triangle and thus we will use the Pythagoras theorem to find the length of 'BC'.

$\blue { \boxed { \bf \bigstar \: Pythagoras \: theorem = (B)^2 + (P)^2 = (H)^2 \; \bigstar }}$

Where,

• B = Base of triangle
• P = Perpendicular length of triangle
• H = Hypotenuse of the triangle

✦ In Triangle ABC,

$\sf : \; \implies (AC)^2 + (AB)^2 = (BC)^2$

$\sf : \; \implies (15)^2 + (8)^2 = (BC)^2$

$\sf : \; \implies 225 + 64 = (BC)^2$

$\sf : \; \implies 289 = (BC)^2$

$\sf : \; \implies BC = \sqrt{289}$

$\green { \sf : \; \implies BC = 17 \; cm \qquad \star}$

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Finding the area of both triangles,

$\pink { \boxed { \bf \bigstar \: Area \: of \; Triangle = \dfrac{1}{2} \times B \times H \; \bigstar }}$

Where,

• B = Base of triangle
• H = Height of triangle

✦ In Triangle ABC,

$\sf : \; \implies \dfrac{1}{2} \times AC \times AB$

$\sf : \; \implies \dfrac{1}{2} \times 15 \times 8$

$\sf : \; \implies 15 \times 4$

$\sf : \; \implies 60 \; cm^2 \qquad \star$

✦ In Triangle DBC,

$\sf : \; \implies \dfrac{1}{2} \times DE \times BC$

$\sf : \; \implies \dfrac{1}{2} \times 4 \times 17$

$\sf : \; \implies 2 \times 17$

$\sf : \; \implies 34 \; cm^2 \qquad \star$

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Finding area of shaded portion,

$\orange { \boxed { \bf \bigstar \: Area \: of \; shaded \; portion = Area \; of \; \triangle ABC - Area \; of \; \triangle DBC \; \bigstar }}$

$\sf : \; \implies 60 \; cm^2 - 34 \; cm^2$

$\sf : \; \implies (60-34) \; cm^2$

$\red { \underline { \boxed { \sf : \; \implies 26 \; cm^2 }}}$

Hence, the area of the shaded portion is 26 cm²

Answer 2

Answer:

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