Question

Inside a triangular garden there is a flower bed in the form of a similar triangle. Around the flower bed runs a uniform path of such a width that the sides of the garden are double of the corresponding sides of the flower bed. The areas of the path and the flower bed are in the ratio

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 3 : 1

Answer 1

Answer: (D)

3 : 1

Let the sides of the garden be a, b and c
$\therefore$ The sides of the flower bed will be $\frac{a}{2}$, $\frac{b}{2}$ and $\frac{c}{2}$
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$\therefore$ Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$
where, $s=\frac{a+b+c}{2}$
$=\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)}$
$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
and the area of the flower bed
$=\sqrt{s_1(s_1-\frac{a}{2})(s_1-\frac{b}{2})(s_1-\frac{c}{2})}$
where, $s_1=\frac{a+b+c}{4}$
$=\sqrt{\left(\frac{a+b+c}{4}\right)\left(\frac{b+c-a}{4}\right)\left(\frac{a+c-b}{4}\right)\left(\frac{a+b-c}{4}\right)}$
$=\frac{1}{16}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
Area of the path
$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$=\frac{1}{16}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$=\frac{3}{16}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}$
$\therefore$ Ratio of the area of the path to the area of the flower bed
$=\frac{3}{16}:\frac{1}{16}=3:1$