Question

In a triangle ABC, It is given that s-a =5 s-b = 3 s-c = 1 where s is semi perimeter and a,b,
c are sides of triangle . of area of triangle is 3 root 15 square units​

Answer 1

hope it helps you... thanks

Answer 2

Step-by-step explanation:

Given: In ∆ABC, s-a=5, s-b=3 and s-c=1, where s is semi-perimeter and a,b and C are sides of ∆ABC. Area of triangle is 3√15 square units.

To find: Find the sides of triangle.

Solution:

Formula to be used:

Heron's formula: If a,b and c are sides of a triangle, then area of triangle is given by

$\bf Ar.(∆ABC) = \sqrt{s(s - a)(s - b)(s - c)} \: sq - units$

where s is semi-perimeter of the triangle.

Step 1: Put the given values in Heron's formula.

$3 \sqrt{15} = \sqrt{s(5)(3)(1)}$

or

$3 \sqrt{15} = \sqrt{15s}$

or

$3 \sqrt{15} = \sqrt{s} \sqrt{15}$

cancel √15 from both sides.

$\sqrt{s} = 3$

squaring both sides

$s = 9$

Thus,

Semi-perimeter of triangle is 9 cm.

Step 2: Find the sides of triangle.

As,

$s - a = 5$

put value of s

$9 - a = 5$

or

$a = 9 - 5$

Thus,

$\bf \red{ a = 4}$

Like the same way,

$s - b = 3 \\ b = 9 - 3$

$\bf \green{ b = 6 }$

and

$s - c = 1$

$c = 9 - 1$

$\bf \pink{c = 8}$

Therefore, sides of triangle are 4,6 and 8 cm.

Final answer:

Sides of triangle ABC are 4,6 and 8 cm.

Hope it will help you.

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