Question

In a triangle \(ABC\), if \(a=6,\ b=5,\ c=9\), then the sum of the squares of the reciprocals of the altitudes of the triangle is

• A. \(\dfrac{71}{400}\)
• B. \(\dfrac{142}{300}\)
• C. \(\dfrac{80}{421}\)
• D. \(\dfrac{15}{157}\)

Hint:

Whenever altitudes appear in a triangle problem, express them using area formulas.

Answer 1

The Correct Option is A

Concept: For a triangle, \[ \Delta=\frac12 ah_a=\frac12 bh_b=\frac12 ch_c \] Hence, \[ h_a=\frac{2\Delta}{a},\qquad h_b=\frac{2\Delta}{b},\qquad h_c=\frac{2\Delta}{c} \] Therefore, \[ \frac1{h_a^2}+\frac1{h_b^2}+\frac1{h_c^2} = \frac{a^2+b^2+c^2}{4\Delta^2} \]

Step 1: Find the semi-perimeter.
\[ s=\frac{a+b+c}{2} \] \[ =\frac{6+5+9}{2}=10 \]

Step 2: Apply Heron's formula.
\[ \Delta=\sqrt{s(s-a)(s-b)(s-c)} \] \[ =\sqrt{10(10-6)(10-5)(10-9)} \] \[ =\sqrt{10\cdot4\cdot5\cdot1} \] \[ =\sqrt{200}=10\sqrt2 \] Thus, \[ \Delta^2=200 \]

Step 3: Substitute into the formula.
\[ \frac1{h_a^2}+\frac1{h_b^2}+\frac1{h_c^2} = \frac{6^2+5^2+9^2}{4(200)} \] \[ = \frac{36+25+81}{800} \] \[ = \frac{142}{800} \] \[ = \frac{71}{400} \] Hence, \[ \boxed{\frac{71}{400}} \]