Question

find the area of an isosceles triangle whose one side is 7 m greater than its equal side and perimeter is 70 m

Answer 1

let the equal sides be x
given that
perimeter=70m
it's one side=7m
perimeter of triangle=sum of all sides
70 = x+x+7
70 =2x+7
63=2x
x=63/2
After that you will use heron's formula and than you will get the answer.

Answer 2

Answer:

$\large{\underline{\boxed{\purple{\bf Area = 98 \sqrt{5} \: m^2 }}}}$

Step-by-step explanation:

Let equal side of isosceles triangle be x m. Other side is 7 m greater than the equal side.

⇒ other side = x + 7 m

Perimeter of triangle = sum of all its sides.

⇒ x + x + x + 7 = 70

⇒ 3x + 7 = 70

⇒ 3x = 70 - 7

⇒ 3x = 63

⇒ x = $\sf \dfrac{63}{3}$

⇒ x = 21 m

So,

• a = x = 21 m
• b = x = 21 m
• c = x + 7 = 28 cm

Using heron's formula -

S = $\rm \dfrac{a + b + c}{2}$

S = $\rm \dfrac{21 + 21 + 28}{2}$

S = $\rm \dfrac{70}{2}$

S = 35

Now,

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

$\rm Area = \sqrt{35(35 - 21)(35 -21)(35 - 28)} \\ \\ \rm Area = \sqrt{35 *14 * 14 * 7}\\\\\rm Area = \sqrt{7 * 5 * 7 * 2 * 7 * 2 * 7}\\\\\rm Area = \sqrt{7^2* 7^2 * 2^2 * 5}\\\\ \rm Area = 98 \sqrt{5} \: m^{2}$

Hence, the area of isosceles triangle is 98√5 m².