Question

using heron's formula find the area of an isosceles triangle where one side is 7m greater than its equal side and perimeter is 70m ​

Answer 1

To Find :–

⠀⠀⠀⠀⠀⠀⠀⠀The Area of the triangle.

Given :–

• Perimeter of the triangle = 70 m

We Know :–

⠀⠀⠀⠀⠀⠀Perimeter of a triangle :-

$\boxed{\underline{\bf{P = a + b + c}}}$

Where :-

• a = Side of the triangle
• b = Side of the triangle
• c = Side of the triangle

⠀⠀⠀⠀⠀⠀Heron's formula :-

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

Where :-

• a = Side of the triangle
• b = Side of the triangle
• c = Side of the triangle
• s = Semi-perimeter

$\boxed{\bf{Semi-perimeter (s) = \dfrac{a + b + c}{2}}}$

Concept :–

⠀⠀⠀Let the equal side of the triangle be x m.

⠀⠀⠀ ⠀⠀ Now , According to the question :

The base is 7 m more than the equal side , so the equation formed :-

⠀⠀⠀⠀⠀⠀⠀b = (x + 7) m

Now by putting the values (in terms of x) in the formula for Perimeter of a triangle , we can find the three lengths of the triangle.

After finding the three sides , we can substitute the values in the heron's fornula, we can find the area of the Triangle.

Solution :-

⠀⠀⠀⠀The Sides of the triangle :-

Given :-

• Equal side (a = b) = x

• Base = (x + 7)

By using the formula and substituting the value in it, we get :-

$:\implies \bf{P = a + b + c} \\ \\ \\ :\implies \bf{70 = x + x + (x + 7)} \\ \\ \\ :\implies \bf{70 = 3x + 7} \\ \\ \\ :\implies \bf{70 - 7 = 3x} \\ \\ \\ :\implies \bf{63 = 3x} \\ \\ \\ :\implies \bf{\dfrac{63}{3} = x} \\ \\ \\ :\implies \bf{21 = x} \\ \\ \\ \therefore \purple{\bf{x = 21 m}}$

⠀⠀⠀⠀⠀⠀Hence, the value of x is 21 m.

Side of the triangle :-

• a = 21 m

• b = 21 m

• c = (x + 7)

⠀==> ⠀⠀(21 + 7) m

⠀==> ⠀⠀ 28 m

Hence, the three sides of the triangle are 21 m , 21 m and 28 m.

⠀⠀⠀⠀⠀⠀Area of the triangle :-

Semi-Perimeter :-

Given :-

• a = 21 m
• b = 21 m
• c = 28 m

Using the formula and substituting the values in it, we get :-

$:\implies \bf{s = \dfrac{a + b + c}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{21 + 21 + 28}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{42 + 28}{2}} \\ \\ \\ :\implies \bf{s = \dfrac{70}{2}} \\ \\ \\ :\implies \bf{s = 35} \\ \\ \\ \therefore \purple{\bf{s = 35 m}}$

Hence, the semi-perimeter is 35 m.

Area :-

Given :-

• a = 21 m
• b = 21 m
• c = 28 m
• s = 35 m

Using the formula and substituting the values in, it we get :-

$:\implies \bf{A = \sqrt{s(s - a)(s - b)(s - c)}} \\ \\ \\ :\implies \bf{A = \sqrt{35(35 - 21)(35 - 21)(35 - 28)}} \\ \\ \\ :\implies \bf{A = \sqrt{35 \times 14 \times 14 \times 7}} \\ \\ \\ :\implies \bf{A = \sqrt{35 \times 1372}} \\ \\ \\ :\implies \bf{A = \sqrt{48020}} \\ \\ \\ :\implies \bf{A = 219.14 (approx.) } \\ \\ \\ \therefore \bf{A = 219.14 m^{2}}$

Hence, the area of the triangle is 219.14 m².