Question

1) sides of a triangle are 45cm, 39 cm, & 42 cm
find Area of triangle using heron's formula​

Answer 1

Given -

• Sides of triangle - 45cm, 39 cm and 42 cm

To find -

• Area of triangle

Formula used -

• Heron's formula

Solution -

In the question, we are provided with the length of the sides of a triangle, and we need to find the area of that Triangle. First of all, we will find the semi-perimeter and then we will use heron's formula to find the area of the Triangle. Let's do it!

So -

Side A = 45 cm

Side B = 39 cm

Side C = 42 cm

Finding semi-perimeter -

Semi-perimeter, s = $\sf \: \frac{sum \: of \: 3 \: sides}{2}$

Semi-perimeter, s = $\sf \frac{45 \: + \: 39 \: + \: 42}{2}$

Semi-perimeter, s = $\sf\dfrac{126}{2}$

Semi-perimeter, s = 63

Now -

We have obtained the semi-perimeter, now we will apply the heron's formula to find the area of the Triangle.

Heron's formula -

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

On substituting the values -

$\sf \sqrt{63(63 - 45)(63 - 39)(63 - 42)} \: \: {cm}^{2}$

$\sf \sqrt{63(18)(24)(21)} \: \: {cm}^{2}$

$\sf\sqrt{571536} \: \: {cm}^{2}$

$\bf \: 756 \: \: {cm}^{2}$

$\therefore$ The area of the Triangle us 756 cm²

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Answer 2

756 cm

$s = a + b + c \div 2$

45+39+42/2=63

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

$\sqrt{63(63 - 45)(63 - 39)(63 - 42)}$

$\sqrt{63 \times 18 \times 24 \times 21}$

$\sqrt{571536}$

${756 \: cm}^{2}$