Question

the base of an isosceles triangle is 24 cm and its area is 60sqcm . find its perimeter herons formula

Answer 1

Suppose the two other sides are 'a'.

Then, the semiperimeter (s) = (24 + a +a)/2

(s) = (12 + a) cm.

Now, by Heron's Formula =>

Area of the triangle =
$\sqrt{s(s - a)(s - b)(s - c)}$
v
$\sqrt{(12 + a)(12 + a - 24)(12 + a - a)(12 + a - a)}$
$\sqrt{(a + 12)(a - 12)(12)(12)}$
But the area is given 60cm*2. So,

$\sqrt{(a + 12)(a - 12)(12)(12)} = 60$
$( {a}^{2} - 144)(144) = 3600$
$( {a}^{2} - 144) = 25$

${a}^{2} = 144 + 25$
${a}^{2} = 169$
$a = 13cm$


So, the perimeter of the Triangle = 24 + 13 +13

Perimeter = 50 cm........ans.

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Answer 2

Step-by-step explanation:

Suppose the two other sides are 'a'.

Then, the semiperimeter (s) = (24 + a +a)/2

(s) = (12 + a) cm.

Now, by Heron's Formula =>

Area of the triangle =

\sqrt{s(s - a)(s - b)(s - c)}s(s−a)(s−b)(s−c)

v

\sqrt{(12 + a)(12 + a - 24)(12 + a - a)(12 + a - a)}(12+a)(12+a−24)(12+a−a)(12+a−a)

\sqrt{(a + 12)(a - 12)(12)(12)}(a+12)(a−12)(12)(12)

But the area is given 60cm*2. So,

\sqrt{(a + 12)(a - 12)(12)(12)} = 60(a+12)(a−12)(12)(12)=60

( {a}^{2} - 144)(144) = 3600(a2−144)(144)=3600

( {a}^{2} - 144) = 25(a2−144)=25

{a}^{2} = 144 + 25a2=144+25

{a}^{2} = 169a2=169

a = 13cma=13cm

So, the perimeter of the Triangle = 24 + 13 +13

Perimeter = 50 cm........ans.

Please mark as brainliest if it was helpful.