Question

11. Find the area of a triangle having perimeter 32cm. One side of its side is equal to
11cm and difference of the other two is 5cm. give me answer i will mark brainlist ​

Answer 1

Step-by-step explanation:

perimeter=32

S1=11

let S2=x & s3=Y

so, X-Y=5(given)

so, x+y+11=32

then, x+y=21......(1)

X-Y=5.......(2)

by solving the above eqn. we get,

x=13 &y=8

area= 1/2(b*h)

= 0.5*8*11

=44 ans....

Answer 2

Given :-

Perimeter of ∆ = 32cm

1st side = 11cm

Difference of other two sides is 5cm

Let,

2nd and 3rd sides be x and y respectively.

Now,

x - y = 5 .......(1) (given)

(Perimeter = Sum of all sides)

x + y + 11 = 32

x + y = 32 - 11

x + y = 21 .........(2)

Solving (1) and (2),

(using elimination method)

x + y = 21

x - y = 5

_______

2y = 16

y = 16/2

y = 8

Now, y in (2)

x + y = 21

x + 8 = 21

x = 21 - 8

x = 13

•°• Other 2 sides of the ∆ are 8cm and 13cm.

Using Heron's formula to find area :-

= √s(s - a)(s - b)(s - c)

Here,

s = (a + b + c)/2

s = 32/2

s = 16

Substituting values in formula :-

= √(16(16- 11)(16 - 13)(16 - 8)

= √16(5)(3)(8)

= √2 x 2 x 2 x 2 x 5 x 3 x 2 x 2 x 2

= 2 x 2 x 2√2 x 3 x 5

= 8√30

= 43.4cm^2

•°• Area of ∆ is 43.4cm^2.