Question

The sides of a triangular field are 11 cm, 12 cm and 13 cm. If length of its shortest and longest altitude are $k_1\sqrt{105}$ cm and $k_2\sqrt{105}$ cm respectively, then the value of $\frac{k_1+k_2}{k_1{k}_2}$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

Let the sides of the triangle ABC be AB = a = 11 cm, BC = b = 12 cm and AC = c = 13 cm.
∴ Semi-perimeter, (s) = $\frac{a+b+c}{2}$
$=\frac{11+12+13}{2}$
= 18 cm

∴ Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
=$\sqrt{18\times (18-11)\times (18-12)\times (18-13)}$
=$\sqrt{18\times 7\times \times 6\times 5}$
=$6\sqrt{105}c{m}^2$
We know that the altitude corresponding to the smallest side is longest and the altitude corresponding to the largest side is shortest.
Thus, length of the shortest altitude, $k_1\sqrt{105}$ is corresponding to the side c = 13 cm and the longest altitude, $k_2\sqrt{105}$ is corresponding to the side a = 11 cm.

$\therefore$ Area of triangle $=\frac{1}{2}\times \text{Base}\times \text{Altitude}$
$\Rightarrow 6\sqrt{105}=\frac{1}{2}\times 13\times k_1\sqrt{105}$ (Taking base as c =13 and altitude as $k_1\sqrt{105}$ )
$\Rightarrow k_1=\frac{6\sqrt{105}\times 2}{13\times \sqrt{105}}=\frac{12}{13}$
$\Rightarrow 6\sqrt{105}=\frac{1}{2}\times 11\times k_2\sqrt{105}$ (Taking base as a =11 and altitude as $k_2\sqrt{105}$ )
$\Rightarrow k_2=\frac{12}{11}$
$\therefore \frac{k_1+k_2}{k_1{k}_2}=\frac{\frac{12}{13}+\frac{12}{11}}{\frac{12}{13}\times \frac{12}{11}}$
$=\frac{\frac{(12\times 11)+(12\times 13)}{13\times 11}}{\frac{12\times 12}{13\times 11}}$
$=\frac{12\times 24}{12\times 12}$
= 2
Hence, the correct answer is option (b).