Question

Using Heron's formule, find the area of an isosceles triangle whose perimeter is 16cm and base is 6 cm. with solution ​

Answer 1

Question :-

find the area of an isosceles triangle whose perimeter is 16cm and base is 6 cm ?

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Given :-

• perimeter :- 16cm
• base :- 6cm

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To find :-

• area of an isosceles triangle

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Formula :-

using heron's formula :-

$\rm \: area = \sqrt{s(s - a)(s - b)(s - c)}$

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solution :-

now Frist let's find the Equal sides so for that permeter of isosceles triangle is 16cm and the base of the triangle is 6cm then equal sides are (16-6/2) = 5cm

therefore the sides are 5cm,5cm and 6cm

now by using formula and by putting values we get,

$\rm :\longmapsto \: area = \sqrt{s(s - a)(s - b)(s - c)}$

$\rm :\longmapsto\: \sqrt{8 \times 2 \times 3 \times 3}$

$\rm :\longmapsto\: = 12sq.cm$

so therefore the area of the isosceles triangle is 12sq.cm

Answer 2

Explain:

Have Given

$\rm Perimeter \: of \: triangle = 16 \: cm$

$\rm Base = 6 \: cm$

We know that –

opposite side of a isosceles triangle are equal.

So, Let assume that opposite sides are x.

Perimeter of triangle is = side + side + side

$\rm \: 16 = x + x + 6$

$\rm 16 = 2x + 6$

$\rm 16 - 6 = 2x$

$\rm 10= 2x$

$\rm \frac{ \cancel{10}}{\cancel2} = x$

$\rm 5 = x$

$\rm or \: \: \: x = 5 \: cm$

Therefore sides are :-

$\rm a = 5 \: cm$

$\rm b = 5 \: cm$

$\rm c = 6 \: cm$

Now,

$\rm Semi \: perimeter (s) = \frac{a + b + c}{2}$

$= \frac{5 + 5 + 6}{6}$

$= \frac{16}{2}$

$\rm = 8 \: cm$

Now, according to the Heron's formule,

$\rm Area = \sqrt{s(s - a)(s - b)(s - c)}$

$= \sqrt{8(8 - 5)(8 - 5)(8 - 6)}$

$= \sqrt{8 \times 3 \times 3 \times 2}$

$= \sqrt{144}$

$\rm = 12 \: {cm}^{2}$

I hope it is helpful