Question

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

Answer 1

Answer:

side = a (18cm)

second side = b(10cm)

third= c

a+b = 18+10= 28cm

perimeter of triangle = a+b+c = 42

C = 42cm-28cm= 14cm

s = a+b+c / 2 = 42/2 = 21

heron's formula,

area of triangle = A = ( s( s-a) (s-b) (s-c) )

a= (21(21-18) (21-10) (21-14))

a=(21*3*11*7)

a=21*21*11

a=21 √11cm²

Answer 2

Perimeter = Sum of all sides

➥ 42 = 18 + 10 + x     (where x is the other side)

➟ 42 = 28 + x

⇢ x = 42 - 28

⇒ x = 14 cm

Since we do not know the base and height of the triangle, we apply herons formula.

$\boxed{\sf{s = \dfrac{a + b + c}{2}}}$

$\sf{s = \dfrac{18+10+14}{2} = \dfrac{42}{2} = 21 cm}$

Where;

• s is the semi-perimeter
• a, b, c are the sides of the triangle.

$\bf{Area = \sqrt{s(s-a)(s-b)(s-c)} }\\\\\\\implies \sf{Area = \sqrt{21(21-18)(21-10)(21-14)} }\\\\\\\implies \sf{Area = \sqrt{21(3)(11)(7)} }\\\\\\\implies \sf{Area = \sqrt{21 \times 21 \times 11}}\\\\\\\therefore \boxed{\underline{\bf{Area = 21 \sqrt{11} \: cm^{2}}}}$

Know more:

When base and height of a triangle are given, we can directly use the formula 1/2 × Base × Height.

Area of Equilateral triangle = √3/4 × (side)²