Question

its urgent,
solve it plzzzzzz....​

Answer 1

SOLUTION:-

Let ABC be a right angled isosceles ∆ in which ∠B= 90°

Let AB= BC= a & AC= b

Now,

AB² + BC² [Pythagoras theorem]

=) b²= a² + a²

=) b²= 2a²...........(1)

Let AEC be the equilateral ∆ described on diagonal AC.

Let BDC be the equilateral ∆ described on base BC.

Now,

Area of ∆AEC,

$= > \frac{ \sqrt{3} }{4} {b}^{2} \\ \\ = > {b}^{2} = \frac{4ar( \triangle \: AEC)}{ \sqrt{3} }$

Now,

Area of ∆BDC,

$= > \frac{ \sqrt{3} }{4} {a}^{2} \\ \\ = > {a}^{2} = \frac{4ar( \triangle BDC)}{ \sqrt{3} }$

Now,

From (1), we get

$\frac{4ar( \triangle \: AEC)}{ \sqrt{3} } = 2 \times \frac{4ar( \triangle \: </u><u>BDC</u><u>)}{ \sqrt{3} } \\ \\ = > ar( \triangle \: </u><u>BDC</u><u>) = \frac{1}{2} ar( \triangle \: </u><u>AEC</u><u>)$

Hope it helps ☺️