The sides of a triangle are 4 , cm, 5 , cm an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given sides of the triangle are $a = 4 \, cm, b = 5 \, cm, c = 6 \, cm$. Semi-perimeter $s = \frac{a + b + c}{2} = \frac{4 + 5 + 6}{2} = \frac{15}

Vedclass · Accepted Answer

$\frac{15}{4} \sqrt{7} \, cm^2$

Vedclass · Answer

(B) Given sides of the triangle are $a = 4 \, cm, b = 5 \, cm, c = 6 \, cm$. Semi-perimeter $s = \frac{a + b + c}{2} = \frac{4 + 5 + 6}{2} = \frac{15}{2} \, cm$. Using Heron's formula,Area $= \sqrt{s(s - a)(s - b)(s - c)}$. Area $= \sqrt{\frac{15}{2} \left( \frac{15}{2} - 4 ight) \left( \frac{15}{2} - 5 ight) \left( \frac{15}{2} - 6 ight)}$. Area $= \sqrt{\frac{15}{2} 	imes \frac{7}{2} 	imes \frac{5}{2} 	imes \frac{3}{2}}$. Area $= \sqrt{\frac{15 	imes 7 	imes 15}{16}} = \frac{15}{4} \sqrt{7} \, cm^2$.

The sides of a triangle are $4 \, cm, 5 \, cm$ and $6 \, cm$. The area of the triangle is equal to

Similar Questions

The area of a triangle whose vertices are $(a \cos \theta, b \sin \theta)$,$(-a \sin \theta, b \cos \theta)$,and $(-a \cos \theta, -b \sin \theta)$ is:

The area enclosed within the curve $|x| + |y| = 1$ is

$A$ triangular corner is cut from a rectangular piece of paper and the resulting pentagon has sides $5, 6, 8, 9, 12$ in some order. The ratio of the area of the pentagon to the area of the rectangle is

The area enclosed by the pair of lines $xy=0$,the line $x-4=0$ and $y+5=0$ is

The midpoints of the three sides of a triangle are $(1, 2)$,$(-1, 1)$,and $(0, 3)$. The area of this triangle is (in sq. units):

Vedclass Test Series

Exam Paper Generator

Online Exam Module