Question

plz solve it........​

Answer 1

Answer:

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First question

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}>21×b×h>21×6×8=24cm2

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

2nd question

In∠ADE

DE2=AD2+AE2

DE2=9+9=18

DE=3√2cm

In∠ABC

BC2=AB2+AC2

BC2=49+49=98

BC=7√2cm

BG+FC=7√2−3√2=4√2cm

BG=2√2=FC

h2=49−49/2=49/2cm

h=7/√2cm

Area of iso traingle =

1/2⋅7/12⋅7√2

49/2cm square

Area of Rectangle

3√2⋅2√2=12cm2

Area of shaaded region =49/2−12=25/2cm2