Question

23) Find the area of a triangle two sides of which are 18 cm and 10 cm and the
perimeter is 42 cm
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Answer 1

Answer:

area of triangle = 1/2 × b × h

1/2 × 18 × 10

=90

perimeter = sum of 3 side

42 = 1st side + 2nd side + 3rd side

42 = 18 + 10 + 3rd side

42 = 28 + 3rd side

42 -28 = 3rd side

14 = 3rd side

perimeter = sum of 3 sides

perimeter = 14 + 18 + 10

perimeter = 42

so , area of triangle is 90 and perimeter of triangle is 42

Answer 2

Answer:

$a = 18$

$b = 10$

$find \: c$

$perimeter = a + b + c$

$18 + 10 + c = 42$

$28 + c = 42$

$c = 42 - 28$

$c = 14$

$here \:$

$s \: = \frac{a + b + c}{2} \: = \frac{18 + 10 + 14}{2}$

$= \frac{42}{2 } = 21cm$

$area \: of \: triangle$

$\sqrt{s(s - a)(s - b)(s - c}$

$\sqrt{(21(21 - 18)(21 - 10)(21 - 14)}$

$\sqrt{21 \times 3 \times 11 \times 7}$

$\sqrt{7 \times 3 \times 3 \times 11}$

$3 \times 7 \times \sqrt{11} = 21 \sqrt{11} cm$