Question

1. Find the area of a triangle, two sides of which are 8 cm and 11 cm, and the perimeter
32 cm
​

Answer 1

Answer:

Given:-

Side a = 8 cm

Side b = 11 cm

Side c = ???

Perimeter = 32 cm

To Find:-

Area of triangle

Solution:-

To find the area, we have to firstly find the all sides of triangle.

so, we know that

⇒ a + b + c = 32 cm

⇒ 8 + 11 + c = 32 cm

⇒ 19 + c = 32

⇒ c = 32 - 19

⇒ c = 13 cm

Area of triangle = √ (s (s-a) (s-b) (s-c) )

Here s is the semi-perimeter.

Semi-perimeter = Perimeter/2 = 32/2= 16 cm

Putting all the values, we get

⇒ √ (s (s-a) (s-b) (s-c) )

⇒ √(16 × (16−8) × ( 16−11) × (16−13))

⇒ √ (16 × 8 × 5 × 3)

⇒ √8 × 2 × 8 × 5 × 3

⇒ √8 × 8 × 2 × 5 × 3

⇒ 8 √ 2 × 5 × 3

= 8√30 cm²

Hence, Area of triangle is 8√30 cm²

Answer 2

Refer The Attachment for Triangle.

Two Sides are=8 cm and 11 cm

Total Sides in Triangle=3

Perimeter Given=32 cm

Let Length of 3 sides=p cm

8+11+p=32

19+p=32

p=32-19

p=13

Length of 3 sides are=8 cm,11 cm,13 cm

As It is Scalene Triangle In Which Length of all sides are different.

Use Heron's Formula to Find Area of Triangle.

Let A=8 cm

B=11 cm

C=13 cm

Firstly Find( s)

$s = \frac{a + b + c}{2} \\ \\ s = \frac{8 + 11 + 13}{2} \\ \\ s = \frac{32}{2} \\ \\ s = 16$

We Got S=16

Now

$area = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{16(16 - 8)(16 - 11)(1 6 - 13)} \\ \\ = \sqrt{16 \times 8 \times 5 \times 3} \\ \\ \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 3} \\ \\ 2 \times 2 \times 2 \sqrt{2 \times 5 \times 3} \\ \\ 8\sqrt{30}$

$\underline{\mathbf{\huge{Area=8\sqrt{30}\:{cm}^{2}}}}$