Question

find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer 1

You can satisfy for this answer.. i know

Answer 2

1st side = 18cm
2nd
$\sqrt{21(3)(11)(7)}$
$\sqrt{21(3)(7)(11)}$
side = 10cm
3rd side =?
Therefore 1st + 2nd+ 3rd side = perimeter
Therefore 18+10+3rd =42
3rd=42-28
3rd=14
Therefore area =
$\sqrt{s(s - a)(s - b)(s - c)}$
Where s= semi perimeter = perimeter /2=42/2
=21
Therefore
$\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$
$\sqrt{21(3)(11)(7)}$
$3 \times 7 \sqrt{11}$
$21 \sqrt{11}$
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