Properties of Triangles 2 Question 7
7. If $\Delta$ is the area of a triangle with side lengths $a, b, c$, then show that
$$ \Delta \leq \frac{1}{4} \sqrt{(a+b+c) a b c} $$
Also, show that the equality occurs in the above inequality if and only if $a=b=c$.
$(2001,6 M)$
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Answer:
Correct Answer: 7. 4 sq units
Solution:
- Given,
$$ \Delta \leq \frac{1}{4} \sqrt{(a+b+c) a b c} $$
$$ \begin{aligned} & \Rightarrow \quad \frac{1}{4 \Delta} \sqrt{(a+b+c) a b c} \geq 1 \\ & \Rightarrow \quad \frac{(a+b+c) a b c}{16 \Delta^{2}} \geq 1 \\ & \Rightarrow \quad \frac{2 s a b c}{16 \Delta^{2}} \geq 1 \\ & \Rightarrow \quad \frac{s a b c}{8 \cdot s(s-a)(s-b)(s-c)} \geq 1 \\ & \Rightarrow \quad \frac{a b c}{8(s-a)(s-b)(s-c)} \geq 1 \\ & \Rightarrow \quad \frac{a b c}{8} \geq(s-a)(s-b)(s-c) \end{aligned} $$
Now, put $s-a=x \geq 0, s-b=y \geq 0, s-c=z \geq 0$
$$ \begin{aligned} s-a+s-b & =x+y \\ 2 s-a-b & =x+y \\ c & =x+y \end{aligned} $$
Similarly, $a=y+z, b=x+z$
$\Rightarrow \quad \frac{(x+y)}{2} \cdot \frac{(y+z)}{2} \cdot \frac{(x+z)}{2} \geq x y z$
which it true.
Now, equality will hold, if $x=y=z$
$\Rightarrow a=b=c$
$\Rightarrow$ Triangle is equilateral.
