Question

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

Answer 1

Answer:

Given Information:-

• Sides of the ∆ :-

1. First Side = 8 centimeters
2. Second Side = 11 centimeters

• Perimeter of the ∆ = 32 centimeters

To Find:-

• Area of the triangle.

SOLUTION:-

Given in this question is a Scalene Triangle.

There is an appropriate formulae for finding the area of certain ∆(s).

Heron's Fomula:-

$\boxed{ \sqrt{s(s - a)(s - b)(s - c)} \ unit \ sq.}$

where,

• a, b, c = Sides of the ∆
• s = Semi perimeter of the ∆

Hence,

Third side will be:-

32 - (8 + 11)

= 32 - 19

= 13 cms

&

Semi Perimeter = 32/2 = 16 cms

_______...

Therefore, According to the Formula,

$= \sqrt{16(16 - 8)(16 - 11)(16 - 13)} \ {cm}^{2}$

(We have put the values of s, a, b, and c)

$= \sqrt{16(8)(5)(3)} \ {cm}^{2}$

(Result after Operation of {-} in brackets)

$= \sqrt{(40)(48)} \ {cm}^{2}$

(Multiplied,

• 16 with 3
• 8 with 5)

$= \sqrt{1920} \ {cm}^{2}$

(Multiplied 40 with 48)

$= 43.81 \ {cm}^{2}$

$\text{(Solution)}$

__________...

REQUIRED ANSWER:-

$\therefore$ Area of the triangle is 43.81 cm.

Answer 2

Answer:

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

Step-by-step explanation:

here we have perimeter of the triangle = 32cm, a = 8cm and b = 11cm,

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

So, 2s = 32 i.e, s = 16 cm,

s - a = ( 16-8) cm = 8cm

s - b = ( 16-11) cm = 5 cm

s - c = ( 16 -13) cm = 3 cm

$\sf{ \therefore \: \: \: \: area \: of \triangle \: = \sqrt{s(s - a)(s - b)(s - c)}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \sqrt{16 \times 8 \times 5 \ \times 3} \: \sf{cm}^{2} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 8\sqrt{30} \: \: \: \: \sf{cm}^{2}$