Question

7. Find the area of a triangle whose two sides are 8 cm and 11 cm and the perimeter is 32 cm. ​

Answer 1

Answer:

9th

Maths

Heron's Formula

Area of a Triangle

Find the area of a triangle...

MATHS

Find the area of a triangle , two sides of which are 8 cm and 11 cm and the perimeter is 32 cm.

MEDIUM

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ANSWER

Let a,b,c be the sides of the given triangle and 2s be its perimeter such that

a=8 cm, b=11 cm and 2s=32 cm

Now, a+b+c=2s

8+11+c=32

c=13

Therefore,

s−a=16−8=8,s−b=16−11=5,s−c=16−13=3

Hence, area of given triangle =

s(s−a)(s−b)(s−c)

=

16×8×5×3

=8

30

cm

2

Answered By

Answer 2

Given:

• Length of two sides of triangle are 8 cm and 11 cm.
• Perimeter of triangle = 32 cm

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To find:

• Area of triangle?

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☯ Let sides of triangle be,

• a = 8 cm

• b = 11 cm

• c = ?

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We know that,

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☯ Perimeter of a triangle is the sum of all three sides of a triangle.

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$:\implies\sf a + b + c = 32$

$:\implies\sf 8 + 11 + c = 32$

$:\implies\sf 19 + c = 32$

$:\implies\sf c = 32 - 19$

$:\implies{\boxed{\frak{\purple{c = 13\;cm}}}}\;\bigstar$

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Also,

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☯ Let s be the semi - perimeter of ∆.

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$:\implies\sf s = \dfrac{Perimeter}{2}$

$:\implies\sf s = \cancel{ \dfrac{32}{2}}$

$:\implies{\boxed{\frak{\purple{s = 16\;cm}}}}\;\bigstar$

Now, calculating Area of triangle,

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$\underline{\sf{\bigstar\; Using\; Heron's\;Formula\;:}}$

$\star\;{\boxed{\sf{\pink{Area = \sqrt{s(s - a)(s - b)(s - c)}}}}}$

$:\implies\sf \sqrt{16(16 - 8)(16 - 11)(16 - 13)}$

$:\implies\sf \sqrt{16 \times 8 \times 5 \times 3}$

$:\implies\sf \sqrt{1920}$

$:\implies{\boxed{\frak{\purple{8 \sqrt{30}\;cm^2}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;Area\;of\; \triangle\;is\; \bf{8 \sqrt{30}\;cm^2}.}}}$