Question

In the given figure, AABC is equilateral triangle with side 10 cm and ADBC is right angled
at D. If BD = 6 cm, find the area of the shaded region (√3 = 1.732).​

Answer 1

Given side of Equilateral ∆ = √3/4a²

a= 10 cm

Area = √3/4 ×10 ×10 cm2

Area = 100√3/4 cm² = 43.3

Now In ∆ BDC

/_ D = 90°

BC² = BD² + CD²

100 = 36 + CD²

CD = √64 = 8

Area of ∆ BDC =

$\begin{gathered} > \large \: \frac{1}{2} \times b \: \times h \: \\ \\ > \frac{1}{2} \times 6 \times 8 = 24 \: {cm}^{2} \end{gathered}$

Area of shaded region = 43.3 - 24 =19 .3 cm²

${ \pink{\huge{ \underline{thank \: - u \: }}}}$

Answer 2

Answer:

Given side of Equilateral ∆ = √3/4a²

a= 10 cm

Area = √3/4 ×10 ×10 cm2

Area = 100√3/4 cm² = 43.3

Now In ∆ BDC

/_ D = 90°

BC² = BD² + CD²

100 = 36 + CD²

CD = √64 = 8

Area of ∆ BDC =

\begin{gathered}\begin{gathered} > \large \: \frac{1}{2} \times b \: \times h \: \\ \\ > \frac{1}{2} \times 6 \times 8 = 24 \: {cm}^{2} \end{gathered} \end{gathered}

>

2

1

×b×h

>

2

1

×6×8=24cm

2

Area of shaded region = 43.3 - 24 =19