Find the area of a triangle whose two sides are 18 cm and 10 cm and th

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

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

Answer»

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

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