If the sum of any two sides of a triangle exceeds the third by 6 cm, the area of the triangle is

Q: If the sum of any two sides of a triangle exceeds the third by 6 cm, the area of the triangle is

(d) $9 \sqrt{3} cm^2$ Let s be the semi-perimeter of a tringle of sides a cm, b cm and c cm. Then, 2s - a + b + c. It is given that $a+b=c+6, b+c=a+6$ and $c+a=b+6$ $\Rightarrow a+b - c=6, b+c - a=0$ and $c+a-b=0$ $\Rightarrow (a+b+c)-2 c=6,(a+b+c) - 2 a=6$ and $(a+b+c)-2 b=6$ $\Rightarrow 2 s-2 c=6,2 s-2 a=6$ and $2 s - 2 b=6 \Rightarrow s-a=3, s-b=3$ and $s - c = 3$ Adding these three, we obtain $3 s-(a+b+c)=9 \Rightarrow 3 s-2 s=9 \Rightarrow s=9$ So , area $A$ of the triangle is given by $A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{9 \times 3 \times 3 \times 3} cm^2=9 \sqrt{3} cm^2$

M.C.Q

CBSE - English Medium

If the sum of any two sides of a triangle exceeds the third by 6 cm, the area of the triangle is

A$12 \sqrt{3} cm^2$
B$18 \sqrt{3} cm^2$
C$15 \sqrt{3} cm^2$
D$9 \sqrt{3} cm^2$

STD 9
Maths
Herons Formula
RD Sharma

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