Triangle PQR is equilateral with sides of length 5 units. O is any point in the

Question: Triangle PQR is equilateral with sides of length 5 units. O is any point in the interior of triangle PQR. Segment OL, OM, and ON are perpendicular to PQ, PR and QR respectively. Find the sum of segments OL, OM and ON.

Options:

	A.	10√32
	B.	3√32
	C.	5√32
	D.	Data insufficient

Answer: Option C
: C

Conventional Method Join PO, OQ and OR

There are 3 triangles whose sum of areas = area of the equilateral triangle Area of a triangle

= \frac{1}{2} \times

base

\times

height Thus Area of

△ P O R = \frac{1}{2} \times P R \times O M

Area of

△ Q O R = \frac{1}{2} \times Q R \times O N

Area of

△ P O Q = \frac{1}{2} \times P Q \times O L

Sum = Area of

△ P O R +

Area of

△ Q O R +

Area of

△ P O Q =

Area of

△ P Q R

= \frac{1}{2} \times P R \times O M + \frac{1}{2} \times Q R \times O N + \frac{1}{2} \times P Q \times O L

Area of an equilateral triangle

= \frac{\sqrt{3}}{4} \times a^{2}

, where 'a' is the length of 1 side

\frac{\sqrt{3}}{4} \times 5^{2} = \frac{1}{2} \times P R \times (O M + O L + O N)

(Since, PR = QR = PQ)

⟹ 25 \frac{\sqrt{3}}{4} = \frac{1}{2} \times 5 \times (O M + O L + O M)

⟹ O M + O N + O L = 5 \frac{\sqrt{3}}{2}

Shortcut:- Assumption Note the most important word in the question-ANY. We can assume ANY point inside the triangle. We will consider a point infinitely close to P.