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.

• A.
• B.
• C.
• D. Data insufficient

Answer 1

The correct option is 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 $△POR=\frac{1}{2}\times PR\times OM$

Area of $△QOR=\frac{1}{2}\times QR\times ON$

Area of $△POQ=\frac{1}{2}\times PQ\times OL$

Sum = Area of $△POR+$Area of $△QOR+$Area of $△POQ=$ Area of $△PQR$

$=\frac{1}{2}\times PR\times OM+\frac{1}{2}\times QR\times ON+\frac{1}{2}\times PQ\times OL$

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}\times a^2$ , where 'a' is the length of 1 side
$\frac{\sqrt{3}}{4}\times 5^2=\frac{1}{2}\times PR\times (OM+OL+ON)$ (Since, PR = QR = PQ)
$⟹25\frac{\sqrt{3}}{4}=\frac{1}{2}\times 5\times (OM+OL+OM)$
$⟹OM+ON+OL=5\frac{\sqrt{3}}{2}$

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.

Thus, the diagram will look as follows.

The answer can now be obtained in a single step.
OL = 0, OM =0 and ON is the height of an equilateral triangle $=5\frac{\sqrt{3}}{2}$
Thus, $OL+OM+ON=5\frac{\sqrt{3}}{2}$