Question

The sides of a triangle are 10 , 12,15 cm . Find its area . using herons formulah

Answer 1

Given :

• Sides of the triangle are 10, 12 and 15 cm

To Find :

• Area of the triangle

Solution :

Heron gave the formula for finding the area of the ∆ in terms of it's three sides.

$\sf \: Heron's \: Formula$

If a, b and c denotes the lengths of the sides of a triangle ABC. Then,

$\sf \: Area \: of \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}$

where, $\sf \: s = \dfrac{a + b + c}{2}$ is the Semi-Perimeter of the triangle.

According to the Question :

$\sf \: s = \dfrac{10 + 12 + 15}{2}$

$\longrightarrow \sf \: s = \dfrac{37}{2} = 18.5$

Now, Area :

$\sf \: \triangle = \sqrt{s(s - a)(s - b)(s - c)}$

$\sf \longrightarrow\: \sqrt{18.5(18.5- 10)(18.5 - 12)(18.5 - 15)}$

$\sf \longrightarrow\: \sqrt{18.5 \times 185 \times 222 \times 277.5}$

$\sf \longrightarrow\:59.8117 \: {cm}^{2}$

$\therefore$ Area of the triangle is 59.8117 cm².

Answer 2

Answer:

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given ;

side of ∆ABC is 10,12,15

AB= 10 cm ,AC = 12 cm and BC = 15 cm

to find ;

area of ∆ ABC

solution

finding area of ∆ by using heron's formula

where the side of ∆ABC is 10,12,15

$s = \frac{a + b + c}{2} \\ \\ s = semiperimeter$

according to question

$s = \frac{10 + 12 + 15}{2} \\ \\ s = \frac{37}{2} \\ s = 18.5$

now area of ∆

$= > \sqrt{s(s - a(s - b)(s - c)} \\ \\ = > \sqrt{18.5(18.5 - 10)(18.5 - 12)(18.5 - 15)} \\ \\ = > \sqrt{18.5 \times 185 \times 222 \times 227.5} \\ = > 59.8117cm ^{2}$