Question

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Answer 1

Question :

The base of an isosceles triangle is 12 cm and its perimeter is 32 cm. Find its area.

Answer :

Given :

• Base of isosceles triangle = 12 cm
• Perimeter of isosceles triangle = 32 cm

To Find :

• Area of isosceles triangle = ?

Solution :

We know that two sides of an isosceles triangle are equal.

Let one side be "x", so another side will also be "x" and we are given base = 12 cm.

Now, we have perimeter = 32 cm and perimeter is sum of all sides.

: ⟹ x + x + 12 cm = 32 cm

: ⟹ 2x = 32 cm - 12 cm

: ⟹ 2x = 20 cm

: ⟹ x = $\sf \dfrac{20}{2}$ cm

: ⟹ x = 10 cm

Now, we have values of sides :

• First side, a = x = 10 cm
• Second side, b = x = 10 cm
• Third side, c = 12 cm

Now, by Heron's formula :

$\large \underline{\boxed{\bf{Area_{(triangle)} = \sqrt{s(s-a)(s-b)(s-c)}}}}$

Here :

• First side, a = 10 cm
• Second side, b = 10 cm
• Third side, c = 12 cm
• s is semi-perimeter = $\sf \dfrac{a + b + c}{2}= \dfrac{10 + 10 + 12}{2} = \dfrac{32 \: cm}{2}= 16 \: cm$

Now, By by filling values :

$\sf : \implies Area = \sqrt{16 \: cm(16 \: cm -10 \: cm)(16 \: cm -10 \: cm)(16 \: cm -12 \: cm)}$

$\sf : \implies Area = \sqrt{16 \: cm(6 \: cm)(6 \: cm)(4 \: cm)}$

$\sf : \implies Area = \sqrt{2304 \: cm^{4}}$

$\sf : \implies Area = \sqrt{(48 \: cm^{2})^{2}}$

$\sf : \implies Area = 48 \: cm^{2}$

$\large \underline{\boxed{\bf{Area_{(triangle)} = 48 \: cm^{2}}}}$

Hence, Area of given triangle is 48 cm².

Answer 2

Question:-

The Base Of An Isosceles ∆ is 12cm and it's perimeter is 32 cm . Find Its Area.

Answer:-

Given:-

Base Of An Isosceles ∆ is 12 cm

Perimeter is 32cm

Find:-

Area Of The Isosceles ∆

$\leadsto \: x + x + 12 = 32 \\\leadsto 2x + 12 = 32 \\ \leadsto2x = 32 - 12 \\\leadsto 2x = 20 \\ x = \frac{20}{2} \\ x = 10$

So, The 2Sides of The isosceles ∆ is 10

$\because \: we \: know \: that \: in \: isoscles \: traingle \: 2sides \: are \: equal$

So,

1st Side (a) = 10cm

2nd side (b) = 10cm

3rd Side (c) = 12cm

Now, We Use Heron's Formula

Firstly We Find The Semi-Perimeter (s)

$\leadsto \: \frac{a + b + c}{2} \\ \\ \leadsto \: \frac{10 + 10 + 12}{2} \\ \\ \leadsto \: \frac{32}{2} \\ \\ \leadsto \: 16 cm$

Now, By Putting The Values In Heron's Formula

$\leadsto \: area \: = \sqrt{16(16 - 10)(16 - 10)(16 - 12)} \\ \leadsto \: area \: = \sqrt{16(6)(6)(4)} \\ \leadsto \: area \: = \sqrt{2304cm ^{4} } \\ \leadsto \: area \: = \sqrt{(48cm ^{2})^{2} } \\ \leadsto \: area = \: 48cm ^{2}$

$so \: the \: Area \: of \: given \: triangle \: is \: 48cm ^{2}$