Question

A square and an equilateral triangle have equal perimeter if the diagonal of the square is 12sqrt2cm find the area of triangle​

Answer 1

Given:

A square and an equilateral triangle with -

• Equal perimeter

Diagonal of the square -

• 12√2 cm

What To Find:

We have to find the -

• Area of the equilateral triangle

How To Find:

To find it, we have to -

• First, find the side of the square.
• Next, find the perimeter of the square.
• Then, find the side of the triangle.
• Finally, find the area of the triangle.

Solution:

• Finding the side of the square.

We know that -

$\sf \to Diagonal \: of \: square = s\sqrt{2}$

Where 's' is the side.

Substitute and solve,

$\sf \to 12\sqrt{2} = s\sqrt{2}$

Take √2 to LHS,

$\sf \to \dfrac{12\sqrt{2}}{\sqrt{2}} = s$

Cancel √2,

$\sf \to 12 \: cm = s$

• Finding the perimeter of the square.

We know that -

$\sf \to Perimeter = 4s$

Substitute and solve,

$\sf \to Perimeter = 4 \times 12$

Multiply 4 with 12,

$\sf \to Perimeter = 48 \: cm$

• Finding the side of the triangle.

We know that -

$\sf \to Perimeter \: of \: square = Perimeter \: of \: triangle$

We also know that -

$\sf \to Perimeter = 48 \: cm$

We know that -

$\sf \to 3s = 48 \: cm$

Take 3 to RHS,

$\sf \to s = \dfrac{48}{3}$

Divide 48 by 3,

$\sf \to s = 16 \: cm$

• Finding the area of the triangle.

We know that -

$\sf \to Area \: of \: equilateral \: triangle =\dfrac{\sqrt{3}}{4}s^2$

Substitute the values and solve,

$\sf \to Area \: of \: equilateral \: triangle =\dfrac{\sqrt{3}}{4}(16^2)$

Find the square of 16,

$\sf \to Area \: of \: equilateral \: triangle =\dfrac{\sqrt{3}}{4} \times 256$

Cancel 256 and 64,

$\sf \to Area \: of \: equilateral \: triangle =\sqrt{3} \times 64$

Multiply,

$\sf \to Area \: of \: equilateral \: triangle =64\sqrt{3} \: cm^2$

Final Answer:

∴ Thus, the area of the triangle is 64√3 cm².

Answer 2

$let \: square \: has \: side \: _{a} \: and \: triangle \: _{t}$

$4a = 3t = a \frac{3}{4} diagonal \: of \: square = \sqrt{2a}$

$= \sqrt{2a} = 12 \sqrt{2} = a = 12 = t = \frac{4}{3} \times 12 = 16cm$

$area \: of \: equilateral \: triangle \: = \frac{ \sqrt{3} }{4} {t}^{2}$

$= \frac{ \sqrt{3} }{4} (256) = 64 \sqrt{3}$

$= 64 \sqrt{3}$