Question

1) Find the area of triangle whose sides are 22 cm,120cm and
122 cm by using heron's formula.​

Answer 1

Answer:

$\sf{Area \ of \ triangle \ is \ 1320 \ cm^{2}}$

Given:

$\textsf{Sides of triangle are 22 cm, 120 cm and 122 cm}$

To find:

$\sf{The \ area \ of \ the \ triangle.}$

Solution:

$\sf{By \ heron's \ formula,} \\ \\ \sf{s=\dfrac{a+b+c}{2}} \\ \\ \sf{\therefore{s=\dfrac{22+120+122}{2}}} \\ \\ \sf{\therefore{s=\dfrac{264}{2}}} \\ \\ \sf{\therefore{s=132}} \\ \\ \sf{A(\triangle)=\sqrt{s(s-a)(s-b)(s-c)}} \\ \\ \sf{\therefore{A(\triangle)=\sqrt{132(132-22)(132-120)(132-122)}}} \\ \\ \sf{\therefore{A(\triangle)=\sqrt{132\times110\times12\times10}}} \\ \\ \sf{\therefore{A(\triangle)=\sqrt{1742400}}} \\ \\ \sf{\therefore{A(\triangle)=1320 \ cm^{2}}} \\ \\ \purple{\tt{\therefore{Area \ of \ triangle \ is \ 1320 \ cm^{2}}}}$

Answer 2

AnswEr :-

• Area of the triangle is 1320cm².

Given :-

• Sides of a triangle as 22cm, 120cm and 122cm.

To Find :-

• Area of the triangle.

SoluTion :-

Here,

• a = 22

• b = 120

• c = 122

★ S = a + b + c /2

→ S = 22 + 120 + 122 / 2

→ S = 264 / 2

→ S = 132

By heron's formula

Area of ∆ :-

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

$\sqrt{132(132 - 22)(132 - 120)(132 - 122}$

$=> \sqrt{132 \times 110 \times 12 \times 10}$

$=> \sqrt{1742400}$

$=> \sqrt{1320 \times 1320}$

$= > 1320 {cm}^{2}$

Hence, the area of the traingle is 1320cm².

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