Question

In the given figure A ABC is an equilateral triangle of side 10 cm and
A DBC is right angled at D. If BD=8 cm and CD-6 cm. Find the area of the
Shaded portion.
​

Answer 1

Answer:

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 =

4

3

(side)

2

=

4

3

(10)

2

=43.30

So area of ΔABC is 43.30cm

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

Step-by-step explanation:

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Answer 2

Answer:

Area of equilateral triangle ABC=√3/4 a²

=√3/4X10X10

=√3X5X5

=25√3 cm²

area of BDC=1/2XbXh

=1/2X8X6

=24 cm²

area of shaded portion=25√3-24 cm²