Question

7. Prove that the area of an equilateral triangle described on one side of a square is equal
to half the area of the equilateral triangle described on one of its diagonal s​

Answer 1

Step-by-step explanation:

The answer is in the attachment above..

Hope it helps ya!

Answer 2

To prove :-

Prove that the area of an equilateral triangle described on one side of a square is equal

to half the area of the equilateral triangle described on one of its diagonals.

Proof :-

In the given figure ,

Let ,ABCD be a square of side "a".

Hence, length of diagonal is √2 a

Let a point" E " ,join AE and BE.

again let a point "F" ,join CF and BF.

Length of side ∆ ABE = a

Length of one side of ∆DBF = √2 a

We know that ,the equilateral triangle have side angles are 60° .

Hence,All these type of triangle are similar.

Therefore,

The ratio of the area of ∆ABE and ∆DBF

is equal to the square of the ratio of corresponding sides of ∆ ABE and ∆DBF.

$\therefore \: \frac{area \: of \: \triangle \: ABE}{area \: of \: \triangle \: DBF} = { \bigg( \frac{a}{ \sqrt{2}a} \bigg) }^{2} \\ \\ \implies \: \frac{ {a}^{2} }{2 {a}^{2} } \\ \\ \implies \: \frac{1}{2}$

Hence, proved.

: )