In a triangle, the average of any two sides is 6 cm more than half of the third side. The area of the triangle is

Q: In a triangle, the average of any two sides is 6 cm more than half of the third side. The area of the triangle is

(d) $36 \sqrt{3} cm^2$ Let the lengths of three sides of the triangle be $a, b, c$ (in cms). It is given that $\quad$$\frac{a+b}{2}=\frac{c}{2}+6, \frac{b+c}{2}=\frac{a}{2}+6$ and $\frac{c+a}{2}=\frac{b}{2}+6$ $\begin{array}{ll}\Rightarrow & a+b-c=12, b+c-a=12 \text { and } c+a-b=12 \\ \Rightarrow & (a+b+c)-2 c=12,(b+c+a)-2 a=12 \text { and }(c+a+b)-2 b=12 \\ \Rightarrow & 2 s-2 c=12,2 s-2 a=12 \text { and } 2 s-2 b=12, \text { where } 2 s=a+b+c \\ \Rightarrow & s-c=6, s-a=6 \text { and } s-b=6 \\ \Rightarrow & (s-a)+(s-b)+(s-c)=6+6+6 \Rightarrow 3 s-(a+b+c)=18 \Rightarrow 3 s-2 s=18 \Rightarrow s=18\end{array}$ So, area $A$ of the triangle is given by $A=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{18 \times 6 \times 6 \times 6} cm^2=36 \sqrt{3} cm^2$

M.C.Q

CBSE - English Medium

In a triangle, the average of any two sides is 6 cm more than half of the third side. The area of the triangle is

A$64 \sqrt{3} cm^2$
B$48 \sqrt{3} cm^2$
C$72 \sqrt{3} cm^2$
D$36 \sqrt{3} cm^2$

STD 9
Maths
Herons Formula
RD Sharma

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