Question

The semi perimeter of a triangle having the length of its sides 20cm, 15cm and 9cm is
(i) 44
(ii) 21
(iv) 22
(iv) none of these
​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Semi-Perimeter of a ∆ :- In geometry, the semiperimeter of a polygon is half its perimeter , or, half of sum of all three sides in case of ∆.

Given That :-

→ Side 1 = S1 = 20cm.

→ Side 2 = S2 = 15cm.

→ Side 3 = S3 = 9cm.

So,

→ Perimeter of given ∆ = sum of all sides = 20 + 15 + 9 = 44cm.

Therefore,

→ semi - Perimeter = (Perimeter)/2

→ semi - Perimeter = (44/2)

→ semi - Perimeter = 22cm. (iv). (Ans.)

_______________________

Extra :-

Use of semi - Perimeter is to find Area of ∆ with Heron's formula.

Lets Try to Find Area Also Now....

→ Area of ∆ = √[s * (s - a) *(s - b) * (s - c)] ,

where,

→ s = semi - Perimeter of ∆ .

→ a , b & c are three sides of ∆.

now, we have ,

→ Side 1 = a = 20cm.

→ Side 2 = b = 15cm.

→ Side 3 = c = 9cm.

→ semi - perimeter = s = 22cm.

Therefore ,

→ Area of given ∆ = √[22 * (22-20) * (22-15) * (22-9)]

→ Area = √[22 * 2 * 7 * 13]

→ Area = √[2 * 2 * 11 * 7 * 13]

→ Area = 2√[ 11 * 7 * 13]

→ Area = 2√(1001) cm². (Ans.)

________________________

Answer 2

${ \huge{ \bold{ \underline{ \red{Question:-}}}}}$

▪ The semi perimeter of a triangle having the length of its sides 20cm, 15 cm and 9 cm is

${ \huge{ \bold{ \underline{ \red{ Solution:-}}}}}$

${ \bold{ \underline{ \pink{ \bigstar{ \: \: Semi \: perimeter}}}}}$

In any triangle, the distance around the boundary of the triangle from a vertex to the point on the opposite edge touched by an excircle equals the semiperimeter. In geometry, the semiperimeter of a polygon is half its perimeter .

${ \bold{ \underline{ \blue{Given-}}}}$

▪ the length of the sides of triangle are given as follows-

¤ side 1 = 20 cm

¤ side 2 = 15 cm

¤ side 3 = 9 cm

${ \boxed{ \bold{ \red{perimeter = sum \: of \: all \: 3 \: sides}}}}$

▪ perimeter = 20 cm + 15 cm + 9 cm

= 44 cm

${ \boxed{ \bold{ \red{semi \: perimeter = \frac{perimeter}{2} }}}}$

${ \bold{ \implies{semi \: perimeter = \frac{44 \: cm}{2} }}}$

thus,

${ \boxed{ \bold{ \pink{ \: semi \: perimeter = 22 \: cm \: }}}}$

hence, option (iii) is correct