Question

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

Answer 1

Let the sides a, b and c
a=18cm
b=10c
Perimeter=42cm

a+b+c=42
18+10+c=42
28+c=42
c=42-28=14cm

S=42/2=21
Area of triangle=√s(s-a)(s-b)(s-c)
=21√11cm^2

Answer 2

Hello dear friend ...
Ur answer is .

$sol. \\ perimeter \: of \: the \: triangle \: = 42cm \\ let \: a = 18cm \: \: and \: \: b = 10cm \\ \\ c = (42 - 18 - 10)cm \: = 14 \\ \\ s = \frac{a + b + c}{2} = \frac{42}{2} = 21cm \\ \\ using \: \: herons \: formula \\ area \: \: of \: triangle \: \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ \sqrt{21(21 - 18)(21 - 14)(21 - 10)} \\ \\ = \sqrt{21 \times 3 \times 7 \times 11} = \sqrt{21 \times 21 \times 21 } \\ \\ = 21 \sqrt{11} {cm}^{2}$
Hope it's helps you.
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