Question

find the area of the triangle with perimeter 22cm one side 9cm and difference of other two sides is 7cm

Answer 1

let the sides be x,y,z
x=9
y-z=7
x+y+z=22
9+y+z=22
y+z=13
on solving we get y=10 and z=3
so the semi perimeter=22/2
=11
area=√s(s-a)(s-b)(s-c)
=√11×2×1×8
√22×8
√176

Answer 2

Answer:

13.26 sq.cm.

Step-by-step explanation:

One side of triangle is 9 cm.

Let the other two sides be x and y

Perimeter of triangle = Sum of all sides

                                  =$9+x+y$

Since we are given that the perimeter is 22 cm

So, $9+x+y=22$ ---1

we are also given that difference of other two sides is 7cm

So, $x-y=7$  --2

Substitute value of x from 2 in 1

$9+7+y+y=22$

$16+2y=22$

$2y=6$

$y=3$

So, Substitute the value of y in 2

$x-3=7$

$x=10$

So, sides of triangle are 9,10,3

a=9

b=10

c=3

To calculate the area of given triangle we will use the heron's formula :

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

Where $s = \frac{a+b+c}{2}$

a,b,c are the side lengths of triangle  

Now substitute the values :

$s = \frac{9+10+3}{2}$

$s =11$

$Area = \sqrt{11(11-9)(11-10)(11-3)}$

$Area = 13.26$

Hence the area of given triangle is 13.26 sq.cm.