Question

the semi perimeter of a triangle having the length of its side as 20 CM 15 cm and 9cm

Answer 1

The semiperimeter of triangle is 22 cm

Solution:

Given that,

Triangle has length of its sides as 20 cm, 15 cm and 9 cm

We have to find the semiperimeter of triangle

For a triangle having sides of length a, b, and c, the semiperimeter of triangle is given as:

$s = \frac{1}{2}(a + b + c)$

From given information,

a = 20 cm

b = 15 cm

c = 9 cm

Substituting the values in above formula,

$s = \frac{1}{2}(20 + 15 + 9)\\\\s = \frac{1}{2}(44)\\\\s = 22$

Thus the semiperimeter of triangle is 22 cm

Learn more about semiperimeter

The semi-perimeter of triangle is 132cm the product of difference of semi-perimeter and its respective sides are 13200 cm (square). Find the area of triangle

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If a b c are the sides of a triangle and s be its semiperimeter and (s-a)=90 (s-b)=50 (s-c)=10 then find value of a

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Answer 2

Answer:

$\red {Semi \: perimeter = \frac{sum \:of \: the \: sides }{2} }\green {= 22 \:cm}$

Step-by-step explanation:

$Let \: a, b \:and \:c \: are \: three \:lengths \: of \: a \: Triangle$

$Given \: a = 20 \:cm, b = 15\:cm \:and c = 9 \:cm$

$Semi \: perimeter = \frac{sum \:of \: the \: sides }{2} \\= \frac{20 + 15 + 9 }{3} \\= \frac{44}{2}\\= 22 \:cm$

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