Question

find the area of a triangle, two sides of which are 8cm and 11cm and the perimeter is 34cm.​

Answer 1

★ Given :

• Sides of traingle are 8cm and 11cm and the perimeter is 34cm.

★ To find :

• Area of Traingle

★ Solution :

• In the question sides of traingle are given. Firstly By using the formula of Perimeter of traingle (a We can easily get the third side of the Traingle.

• After Getting the side of the triangle we will find the Area of Triangle using heron's Formula.

A = 11 cm,B = 8cm, C = ?

Let's Find out...!

✰ Perimeter = a + b + c ✰

➭34 = 11 + 8 + C

➭34 = 19 + C

➭C = 34 - 19

➭C = 15 cm

Third side of triangle is 13 cm.

✰ Side = Perimeter/2 ✰

➤ 34/2

➤ 17 cm

Now finding Area

✰ Area = √s (s - a)(s - b)(s - c) ✰

➤ √17 (17 - 11)(17 - 8)(17 - 13)

➤ √17 × 6 × 9 × 5

➤ 6√51cm²

Therefore

Area of Traingle is 6√51cm²

______________

Answer 2

ANSWER :

$\large\sf\implies6\sqrt{51}cm^2$

Solution :

Given :

• Two Sides of triangle are 8cm and 11cm respectively.
• Permiter of triange is 34cm.

To Find :

• Area of triangle.

Let's start ,

$\red{\bigstar}\boxed{\tt Perimeter \: of \; triangle\;=sum\; of\;all\;the\;sides}$

let the third side of triangle be x

So,

$\implies\sf8+11+x=34\\\\\implies19=x=34\\\\\implies\overline{\boxed{\pink{x=15}}}$

Now,

we can find the area of triangle using heron's formula.

$\red{\bigstar}\boxed{\sf Area_{triangle}=\sqrt{s(s-a)(s-b)(s-c)}}$

Let :

• a = 8 cm
• b = 11 cm
• c = 15 cm
• s = perimeter/2 = 34/2 = 17cm.

Substituting the values :

$\sf Area \;of \;triangle\; = {\sqrt{s(s-a)(s-b)(s-c)}}\\\\ \implies \sf\sqrt{17(17-8)(17-11)(17-15)}\\\\\implies\sf\sqrt{17(9)(6)(2)}\\\\\implies\sf \sqrt{17(108)}\\\\\implies\sf\sqrt{1836} \\\\\implies\sf6\sqrt{51} \\\\\implies\boxed{\sf42.84857 cm^2}$

Hence,

Area of Triangle is 6√51 cm².

${\fcolorbox{red}{blue}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: SugarCrash\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}$