Question

find the area of triangle whose perimeter is 32 cm.one side of its equal to 11 cm and difference of other two is 5cm​

Answer 1

Step-by-step explanation:

a + b + c = 32 cm

a = 11 cm

b - c = 5 cm

So , b + c = 21 cm

2b = 26 cm

b = 13 cm

c = 8 cm

S = a+b+c/2 = 32/2 = 16 cm

Area of triangle

0 = √S(S-a)(S-b)(S-c)

= √16(16-11)(16-13)(16-8)

= √16*5*3*8

= √16*5*3*4*2

= 4*2√5*3*2

= 8√30 cm²

Answer 2

Step-by-step explanation:

perimeter of the triangle=a+b+c

whare a,b,c are sides of a triangle

a=11cm

b=x

c=x+5

perimeter of the triangle=11+x+x+5=

11+2x+5=16+2x=32

☆16+2x=32

2x=16

x=8

let the side b length be x

☆a=11 cm

☆b=x=8 cm

☆c=x+5=8+5=13 cm

area of the triangle=bh/2

where b=base;h=height

By using herons formula,

$area \: of \: the \: triangle = \\ \sqrt{s(s - a)(s - b)(s - c)}$

where s=semi perimeter=(a+b+c)/2=(8+13+11)/2=

32/2=16

a,b,c are the sides of the triangle

area of the triangle=(s(s-a)(s-b)(s-c))^1/2

=(16(16-11)(16-8)(16-13))^1/2

=(16(5)(8)(3))^1/2

=(80×24)^1/2

=1920^1/2

=43.8178046004 cm^2

Hope it helps you.

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