Question

In a given figure ABC is an equilateral triangle with 10 cm and DBC is right angle triangle with angle d =90 if BD is 6 cm then find the area of the shaded portion

Answer 1

Step-by-step explanation:

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Given, ΔABC is an equilateral triangle the length of whose side is equal to 10 cm, and ΔDBC is right-angled at D

and BD=8cm.

From figure:

Area of shaded region = Area of ΔABC− Area of ΔDBC.....(1)

Area of ΔABC:

$= Area = \frac{ \sqrt{3} }{4} (side {)}^{2} = \frac{ \sqrt{3} }{4} (10 {)}^{2} =43.30 \\ = area = > 43.3 {0cm}^{2}$

Area of right ΔDBC:

Area = 2/1 ×base×height...(2)

From Pythagoras Theorem:

Hypotenuse^2

= Base^2 + Height^2

BC^2 =DB^2 +Height^2

100−64=Height^2

36=Height^2 or Height =6

equation (2)⇒

Area = 2/1×8×6=24

So area of ΔDBC is 24cm^2

Equation (1) implies

Area of shaded region =43.30−24=19.30

Therefore, Area of shaded region =19.3cm^2

I hope this will be helpful...☃️