Question

area of a triangle two sides of which are 18 cm and 12 cm

and the perimeter is 44 cm.​

Answer 1

Answer:

≈ 83.9 cm²

Step-by-step explanation:

Given:

• P = a+b+c = 44 cm
• a = 18 cm
• b = 12 cm

Then:

• c= P -(a+b) = 44 -(18+12) = 14 cm

Area of  the triangle is found by using Herons formula:

• A = √s(s-a)(s-b)(s-c),

where s = P/2 = 44/2 = 22 cm

• A = √22(22-18)(22-12)(22-14) = √22*4*10*8= √7040 ≈ 83.9 cm²

Answer 2

Answer:

$\huge \boxed{\mathrm{83.9 \ cm^2}}$

Step-by-step explanation:

Two sides of the triangle are given.

The perimeter is given.

We need to solve for the third side.

$P=a+b+c$

$P= \sf perimeter$

$a,b,c= \sf side \ lengths$

$44=18+12+c$

$c=14$

The measure of the third side is 14 cm.

When three sides of the triangle are given, we can solve for the area using Heron’s formula.

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

$s=\sf semi \ perimeter$

$\displaystyle s=\frac{P}{2} =\frac{44}{2} =22$

Plugging in the values and evaluating.

$A=\sqrt{22(22-18)(22-12)(22-14)}$

$A = 83.904708...$

The area of the triangle is approximately 83.9 cm².