Question

4. 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​

Answer 1

Question:-

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.

Answer:-

Let,

We know that,

Two equilateral triangles are always similar.

In Triangle ABC and Triangle DFC

$\implies$ $\dfrac{DE}{AB}$ = $\dfrac{8}{4}$ = 2

$\implies$ $\dfrac{EF}{BC}$ = $\dfrac{8}{4}$=2

$\implies$ $\dfrac{DF}{AC}$ =$\dfrac{8}{4}$ =2

Hence By SSS similarity

Triangle ABC ~ Triangle DFC

Now,

Diagnol of its square is $\sqrt{2}$ Of its side.

we know that,

All the sides of square are equal,

Let,

AB = BC = CD = DA = P cm

All angles of square are 90°.

so,

$\angle{C}$ = 90°

According to pythagoras therom

We know that

${Hypotenuse}^{2}$ = ${Height}^{2}$+${Base}^{2}$

Now ,

Substituting the values

${BD}^{2}$ = ${BC}^{2}$+${DC}^{2}$

${BD}^{2}$ =${P}^{2}$+${P}^{2}$

${BD}^{2}$ = ${2P}^{2}$

BD = $\sqrt{2}$ P

Hence

$\dfrac{Diagnol}{Side}$ = $\dfrac{BD}{DC}$

$\implies$ $\dfrac{\sqrt{2} P }{P}$

$\dfrac{ \sqrt{2} }{1}$

Diagnol = $\sqrt{2}$ Side.

Hence proved.

-Happies!!

Answer 2

Given :-

ABCD is a square whose one diagonal is AC = ΔAPC

ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD

To Find :-

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.

Solution :-

We know that,

$\underline{\boxed{\sf Area( \triangle BQC) =\dfrac{1}{2} Area( \triangle APC) }}$

Given that,

ABCD is a square whose one diagonal is AC = ΔAPC

ΔBQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

(Please refer to the following attachment for your reference)

According to the question,

ΔAPC and ΔBQC are both equilateral triangles.

∴ ΔAPC ~ ΔBQC (AAA similarity criterion)

$\sf \dfrac{Area(\triangle APC)}{Area( \triangle AQC)} = \bigg( \dfrac{AC^{2}}{BC^{2}} \bigg) = \dfrac{AC^{2}}{BC^{2}}$

Since,

$\sf Diagonal=\sqrt{2} \ side=\sqrt{2} \ BC=AC$

$\sf \bigg( \dfrac{\sqrt{2} BC}{BC} \bigg)^{2}=2$

Therefore,

$\longrightarrow \sf area(\triangle APC) = 2 \times area(\triangle BQC)$

$\longrightarrow \sf area(\triangle BQC) = \dfrac{1}{2} area(\triangle APC)$

Hence, proved!