Question

find the area of the triangle

Answer 1

hope so this helps...

Answer 2

Question :

• Find the area of a triangle , two sides of which are 18 cm and 10 cm and the perimeter is 42 cm. ( Take √11 = 3.32 )

Given :

• Perimeter = 42 cm
• Side A = 18 cm
• Side B = 10 cm

To Find :

• Side C
• Area of the triangle

Solution :

First we have to find side C of the triangle

$\tt perimeter = a + b + c \\ \\ \tt 42 = 18 + 10 + c \\ \\ \tt42 = 28 + c \\ \\ \tt c = 42 - 28 \\ \\ \tt c = 14 \: cm$

Semi-perimeter of the triangle

$\tt s = \frac{a + b + c}{2} \\ \\\tt s = \frac{18 + 10 + 14}{2} \\ \\ \tt s = 21$

Now area of triangle is

$\large\boxed{\tt Area = \sqrt{s(s - a)(s - b)(s - c)}} \\ \\ \tt \implies \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ \\ \tt \implies \sqrt{21 \times 3 \times 11 \times 7} \\ \\\tt \implies \sqrt{7 \times 3 \times 3 \times 11 \times 7} \\ \\ \tt \implies7 \times 3 \sqrt{11} \\ \\ \tt \implies21 \sqrt{11} \\ \\\tt \implies21 \times 3.32 \\ \\ \tt \implies69.72 {cm}^{2}$

$\large \underline{ \tt \blue{Area \: of \: triangle \: is \: 69.72 \: {cm}^{2} }}$