Question

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

2 points

21√11 cmsq

21 cmsq

11cmsq

27 cmsq

​

Answer 1

Answer:

1st side=18cm

2nd side=10cm

perimeter=42cm

3rd side= 42-(18+10) cm

= 14 cm

u can find the area of triangle by the heron's formula..

Hope it helps..

Answer 2

$\bf{\underline{\underline{Question:-}}}$

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

$\bf{\underline{\underline{Given:-}}}$

• Three sides of ∆ are
• a = 18
• b = 10
• c = ?
• Perimeter of ∆ is 42cm

$\bf{\underline{\underline{Find:-}}}$

• Area of ∆ ?

$\bf{\underline{\underline{Solution:-}}}$

$\sf → 42 = a + b+c$

$\sf → 42 = 18+10+c$

$\sf → 42=28+c$

$\sf → 42-28=c$

$\sf → 14 = c$

Applying formula of semi perimeter

• S = ( a + b + c ) ÷ 2 where a , b , c are sides of ∆

$\sf → S = \dfrac{18+10+14}{2} \\ \sf→ S=\dfrac{42}{2} \\ \sf→ S=\dfrac{\cancel{42_{21}}}{\cancel{2}}$

$\sf→ S=21$

★ By Heron's Formula

$\bf → Area= \sqrt{s(s-a)(s-b)(s-c)}$

$\sf → \sqrt{21(21-18)(21-10)(21-14)}\\ \sf→ \sqrt{21(3)(11)(7)}\\ \sqrt{21(231)}\\ \sf→ \sqrt{4851}\\ \sf→ 21\sqrt{11}cm^2$

$\bf{\underline{\underline{Hence:-}}}$

• The area of ∆ is 21√11 cm²