Question

Find the area of triangle sides of which are 122m, 22m and 120m.

Answer 1

Answer:

HINT: Use Heron's Formula

Step-by-step explanation:

If you don't know Heron's formula ,here it is in brief :

Heron’s formula, formula credited to Heron of Alexandria (c. 62 CE) for finding the area of a triangle in terms of the lengths of its sides. In symbols, if a, b, and c are the lengths of the sides:

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

where s is half the perimeter, or (a + b + c)/2.

Answer 2

Given :-

• Sides of traingle are 122 m, 22 m and 120 m.

To Find :-

• The area of traingle = ?

Solution :-

$\frak {\red{Here}}\begin{cases} \sf{\pink{a = 122 \: m}}\\ \sf{\gray{b =22 \: m}}\\ \sf{\orange{c = 120 \: m}}\end{cases}$

☆ Semi Perimeter of traingle :

$\dashrightarrow\:\: \textsf{Semi Perimeter (s)= \dfrac{ \text{a + b + c}}{ \text{2}} }$

$\dashrightarrow\:\: \textsf{Semi Perimeter (s)= \dfrac{ \text{122+ 22+ 120}}{ \text2} }$

$\dashrightarrow\:\: \textsf{Semi Perimeter (s)= \dfrac{ \text{264}}{ \text2} }$

$\dashrightarrow\:\: \textsf{Semi Perimeter (s)} = \frak{132 \: cm }$

Now, let's calculate the the area of ∆ by using heron's formula :

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

$:\implies \sf Area_{\tiny \triangle} = \sqrt{132(132 - 122)(132- 22)(132 - 120)}$

$:\implies \sf Area_{\tiny \triangle} = \sqrt{132 \times 10 \times 110 \times 12}$

$:\implies \sf Area_{\tiny \triangle} = \sqrt{11 \times 12 \times 10 \times 11 \times 10 \times 12}$

$:\implies \sf Area_{\tiny \triangle} = 11 \times 12 \times 10$

$:\implies \underline{ \boxed {\frak{ Area_{\tiny \triangle} = 1320 \: {cm}^{2}}}}$

Therefore,Area of ∆ is 1320 cm².