Question

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

Answer 1

Answer:

Third side of the triangle =42-(10+18)=14

semi perimeter=42/2=21

area = √s (s-a)(s-b)(s-c)

√21(21-18) (21-10) (21-14)

√21×3×11×7=69.6491206 or 21√11

hope it helps

Answer 2

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...