Question

The perimeter of an isosceles triangle is 32cm.The ratio of the equal side to it's base is 3:2.Find the area of the triangl​

Answer 1

$\;\;\underline{\textbf{\textsf{ Given:-}}}$

• Perimeter of an isosceles triangle is = 32cm

• Ratio of the equal side to its base is = 3:2

$\;\;\underline{\textbf{\textsf{ To Find:-}}}$

• Area of the given triangle

$\;\;\underline{\textbf{\textsf{ Solution :-}}}$

Let the equal sides of the triangle be x cm and the base be y cm.

Given that,

• The perimeter is 32cm.

Then,

$\sf\dashrightarrow 2x + y= 32 \: \: .......(1)$

Again, it’s given that

• Ratio of the equal side to its base is = 3:2

Therefore,

$\sf \dashrightarrow x: y = 3 : 2 \\ \\ \sf\dashrightarrow \frac{x}{y} = \frac{3}{2} \\ \\ \sf \dashrightarrow x = \frac{3}{2} y \: \: \: .......(2)$

Putting the value of x from eq(2) in eq(1) :-

$\sf\dashrightarrow 2 \times \dfrac{3}{2} y + y = 32 \\ \\ \sf\dashrightarrow 4y = 32 \\ \\ \dashrightarrow \sf y = 8$

Hence,

• The base is of 8cm.

Again,

Putting the value of y in (2)

$\dashrightarrow \sf x = \dfrac{3}{2} \times 8 \\ \\ \dashrightarrow \sf x = 12$

• The equal sides are of 12cm

Now, using Pythagoras theorem :-

$\sf \dashrightarrow {k}^{2} + 4 {}^{2} = {12}^{2} \\ \\ \dashrightarrow \sf {k}^{2} = 144 - 16 \\ \\ \dashrightarrow \sf {k}^{2} = 128 \\ \\ \dashrightarrow \sf k = 8 \sqrt{2}$

Hence,

• The height is 8√2cm.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

A.T.Q :-

We are asked to find the area of the given triangle.

We know that ,

$\tt{\small{\boxed{\bold{\bold{\green{\sf{Area\:of\:Triangle=\dfrac{1}{2}\times{b}\times{h}}}}}}}}$

$\sf \dashrightarrow Area \: of \: triangle = \frac{1}{2} \times 8 \times 8 \sqrt{2} \\ \\ \sf\dashrightarrow Area \: of \: triangle = 4 \times 8 \sqrt{2} \\ \\ \sf \dashrightarrow Area \: of \: triangle = 32 \sqrt{2}$

$\;\;\underline{\textbf{\textsf{ Hence-}}}$

$\underline{\textsf{The area of the triangle is \textbf{ 32√2cm²}}}.$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer:

32√2 is ten correct answer