Question

chord.
Q.28 Find the area of a triangle two sides of which are 18 cm and 10 cm
and the perimeter is 42 cm.​

Answer 1

Solution!!

The perimeter of the triangle and its two sides are given. We have to find the area of the triangle. We can find the third side of the triangle by the perimeter formula and hence we can apply the Heron's formula to get the area of the triangle.

Perimeter = 42 cm

Side a = 18 cm

Side b = 10 cm

Side c = ?

Perimeter = Sum of all sides of triangle

42 cm = 18 cm + 10 cm + Side c

42 cm = 28 cm + Side c

Side c = 42 cm - 28

Side c = 14 cm

We know all the sides now. Let's use the Heron's formula.

Semi-perimeter (s) = (a + b + c)/2

s = (18 cm + 10 cm + 14 cm)/2

s = 42 cm ÷ 2

s = 21 cm

Area = √[s(s - a)(s - b)(s - c)]

Area = √[21(21 - 18)(21 - 10)(21 - 14)]

Area = √[21(3)(11)(7)]

Area ≈ 69.65 cm²

Answer 2

Answer:

Given :

Two sides of triangle are 18cm amd 10cm.

Perimeter of Triangle is 42cm.

To Find :

Area of Triangle.

Solution :

a = 18cm

b = 10cm

c = ?

$\longmapsto\tt{a+b+c=42}$

$\longmapsto\tt{18+10+c=42}$

$\longmapsto\tt{28+c=42}$

$\longmapsto\tt{c=42-28}$

$\longmapsto\tt\bf{c=14cm}$

Now ,

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

$\longmapsto\tt{s=\dfrac{18+10+14}{2}}$

$\longmapsto\tt{s=\cancel\dfrac{42}{2}}$

$\longmapsto\tt\bf{s=21cm}$

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

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

$\longmapsto\tt{\sqrt{21(3)(11)(7)}}$

$\longmapsto\tt{\sqrt{7\times{3}\times{3}\times{11}\times{7}}}$

$\longmapsto\tt{7\times{3}\sqrt{11}}$

$\longmapsto\tt\bf{21\sqrt{11}{cm}^{2}}$

So , The Area of Triangle is 21√11 cm²