Question

Find the area of shaded region in figure.
(1 Point)
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Answer 1

—› Your Answer :-

In the fig. given there are two triangles ∆ ABC and ∆ BDC,

where,

∆ BDC is a right angle triangle and ∆ ABC is equivalent triangle,

Now,

$\sf \: Area \: of \: equilateral \: ∆ \: ABC = \frac{ \sqrt{3} }{4} ( {side)}^{2} \\ \\ = \sf \frac { \sqrt{3} }{4} ( {10)}^{2} \\ \\ \sf \: = \frac{ \sqrt{3} }{4} \times 100 \\ \\ \sf \: = \frac{ \sqrt{3} }{ \cancel4} \times \cancel{100} \: ^{25} \\ \\ \sf \: = 25 \sqrt{3} \: {cm}^{2}$

Now,

In ∆ BDC,

h = 10 cm

b = ?

p = 8 cm

By applying Pythagoras theorem,

h² = p² + b²

(10)² = (8)² + b²

b² = (10)² - (8)²

b² = 100 - 64

b² = 36

b = √36

b = 6 cm

so,

DC = 6 cm

$\sf \: Area \: of \: right \: angle \: ∆ \: BDC= \frac{1}{2} \times b \times h \\ \\ \sf = \frac{1}{2} \times DC \times DB \\ \\ = \sf \: \frac{1}{ \cancel{2}} \times \cancel{6} \: ^{3} \times 8 \\ \\ \sf \: = 24 \: {cm}^{2}$

Now,

To find,

Area of shaded region = Area of ∆ ABC - Area of ∆ BDC

=> 25√3 - 24

=> 1√3

=> 1 × 1.732 cm²

=> 1.732 or 1.73 cm² (ans)