Question

find the area of the triangle whose sides are 13cm, 20cm, 21cm ​

Answer 1

Solution :

Let a, b and c are the sides of the triangle,and s = semi-perimeter i.e. half the perimeter of the triangle.

Let's take a = 13 cm, b = 20 cm, c = 21 cm

$\star \: \boxed{ \sf \: s = \dfrac{a + b + c}{2} }$

$\longrightarrow \sf \: \dfrac{13 + 20 + 21}{2}$

$\longrightarrow \sf \: \cancel\dfrac{54}{2}$

$\longrightarrow \red{\sf \: 27}$

Now,

• s - a = (27 - 13) = 14 cm

• s - b = (27 - 20) = 7 cm

• s - c = (27 - 21) = 6 cm

Therefore,area of the triangle :

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

$\longrightarrow \sf \: \sqrt{27 \times 14 \times \times 7 \times 6 }$

$\longrightarrow \sf \: \sqrt{3 \times 3 \times 3 \times 2 \times 7 \times 7 \times 2 \times 3 }$

$\longrightarrow \sf \: \sqrt{3 \times 3 \times 2 \times 7 }$

$\longrightarrow \red{\sf \: 126 {cm}^{2} }$

$\therefore$ Area of the triangle is 126 cm².

Answer 2

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$\huge\tt{GIVEN:}$

• A Triangle whose sides are 13 cm, 20 cm, 21cm

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$\huge\tt{TO~FIND:}$

• It's area

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$\huge\tt{FORMULA~USED:}$

• s = a+b+c/2 , where s is semi-perimeter of ∆
• Heron's formula= √{s(s-a)(s-b)(s-c)}

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$\huge\tt{SOLUTION:}$

Firstly, find the value of S here,

↪S = (13+20+21)/2

↪S = 54/2

↪S = 27

Now,

Area of triangle would be,

↪Area of ∆ = √ s(s-a)(s-b)(s-c)

↪Area of ∆ =√ 27(27-13)(27-20)(27-21)

↪Area of ∆ =√ 27(14)(7)(6)

↪Area of ∆ =√3×3×3×2×7××7×2×3

↪Area of ∆ =√3×3×2×7

↪Area of ∆ =126cm²

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