Question

find the area of a triangle two sides of which are 8 CM 11 cm and the perimeter is 32 CM

Answer 1

Given:

AB = 8cm

AC = 11cm

Perimeter of triangle ABC = 32cm

To find

Area of triangle ABC

Solution:

Perimeter of a triangle = side + side + side

32cm = 8 + 11 + BC

32 - 19 = BC

13cm = BC

Now,

s of triangle ABC = a + b + c/2

= 8 + 11 + 13/2

= 32/2

= 16cm

Now, using heron's formula,

Area of triangle ABC = √s(s-a)(s-b)(s-c)

= √16(16-8)(16-11)(16-13)

= √16(8)(5)(3)

= √ 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5 x 3

Ans = 8√30cm²

Answer 2

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

_______________

${ \huge{ \bold{ \underline{ \underline{ \purple{Answer:-}}}}}}$

✒Here we have perimeter of the triangle = 32 cm ..

• a = 8
• b = 11

★ Refer to Attachment ★

✒Third side c = 32cm - (8+11)cm = 13 cm ..

✒Now ,

$\dashrightarrow\sf{2s=32\:(i.e.)=16cm}$

$\dashrightarrow\sf{s-a=(16-8)cm = 8cm}$

$\dashrightarrow\sf{s-b=(16-11)cm=5cm}$

$\dashrightarrow\sf{s-c=(16-13)cm=3cm}$

✒Therefore , area of triangle : -

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

$\dashrightarrow\sf{\sqrt{16\times{8}\times{5}\times{3}{cm}^{2}}}$

$\leadsto\sf{8\sqrt{30{cm}^{2}}}$