Question

Find area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.please explain clearly.

Answer 1

HERE IS YOUR ANSWER..⬇⬇

$\underline{SOLUTION}$ ➡

We know that,

In an isosceles triangle two sides are equal.

Given,

Base of the isosceles triangle , b = 6 cm

Perimeter of the isosceles triangle = 16 cm

Let,

Equal sides of the isosceles triangle = a

A.T.Q.,

Perimeter = 16

=> 2a + b = 16

=> 2a + 6 = 16

=> 2a = 16 - 6

=> 2a = 10

=> a = 10/2

=> a = 5

•°• Equal sides of the isosceles triangle = 5 cm

Now,

We know that ,

$Area \: \: of \: an \: isosceles \: triangle \: \\ = \frac{b}{4} \sqrt{4a {}^{2} - b {}^{2} }$

$Area \: \: of \: \: the \: isosceles \: triangle \\ = \frac{b}{4} \sqrt{4a {}^{2} - b {}^{2} } \\ \\ = (\frac{6}{4} \sqrt{4 \times (5) {}^{2} - (6) {}^{2} } )\: \: cm {}^{2} \\ \\ = (\frac{3}{2} \sqrt{4 \times 25 - 36} )\: \:cm {}^{2}\\ \\ = (\frac{3}{2} \sqrt{100 - 36}\:) \:cm {}^{2} \\ \\ = (\frac{3}{2} \sqrt{64}\: )\: cm {}^{2}\\ \\ =( \frac{3}{2} \times 8 ) \: \:cm {}^{2} \\ \\ = (3 \times 4 \:) \: cm{}^{2}\\ \\ = 12 \: \: cm {}^{2}$

$\underline{ANSWER}$ ➡

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

✝✝..HOPE IT HELPS YOU..✝✝

Answer 2

It is given that

Base = 6 cm

Perimeter = 16 cm

Consider ABC as an isosceles triangle in which

AB = AC = x

So BC = 6 cm

We know that

Perimeter of △ABC = AB + BC + AC

Substituting the values

16 = x + 6 + x

By further calculation

16 = 2x + 6

16 – 6 = 2x

10 = 2x

So we get

x = 10/2 = 5

Here AB = AC = 5 cm

BC = ½ × 6 = 3 cm

In △ABD

AB2 = AD2 + BD2

Substituting the values

52 = AD2 + 32

25 = AD2 + 9

By further calculation

AD2 = 25 – 9 = 16

So we get

AD = 4 cm

Here

Area of △ABC = ½ × base × height

Substituting the values

= ½ × 6 × 4

= 3 × 4

= 12 cm2

${\fcolorbox{blue}{black}{\blue{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:HumanIntelligence\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}$