Question

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

Answer 1

Answer:

Step-by-step explanation:

69.95

hope this helps

Answer 2

$\huge{\bold{\underline{\underline{Given:-}}}}$

• Side 1 of triangle = 18cm
• Side 2 of triangle = 10cm
• Perimeter of triangle = 42cm

$\huge{\bold{\underline{\underline{To\:Find:-}}}}$

• Area of triangle

$\huge{\bold{\underline{\underline{Solution:-}}}}$

Let's consider a ΔABC

Side 1 = AB = 18cm

Side 2 = BC = 10cm

Perimeter = 42cm

Side 3 = AC = ?

Let Side 3 be 'x'

A.T.Q.

Perimeter of triangle = Sum of all sides

42cm = 18cm + 10cm + X cm

42cm = 28cm + X cm

X cm = 42cm - 28cm

X cm = 14 cm

∴ Side Side 3 = AC = 14cm

◈ Area :-

A = 18cm

B = 10cm

C = 14cm

Semi perimeter = $\frac{A+B+C}{2}$

= $\frac{18+10+14}{2}$

= $\frac{42}{2}$

= 21 cm

Area =

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

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

$= \sqrt{21(3)(11)(7)}$

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

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

$= 21\sqrt{11}$

So, Area of triangle = 21√11 cm²