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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.


  1. Data insufficient


Solution

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 =12× base × height

Thus Area of POR=12×PR×OM

Area of QOR=12×QR×ON

Area of POQ=12×PQ×OL

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

=12×PR×OM+12×QR×ON+12×PQ×OL

Area of an equilateral triangle =34×a2 , where 'a' is the length of 1 side
34×52=12×PR×(OM+OL+ON) (Since, PR = QR = PQ)
2534=12×5×(OM+OL+OM)
OM+ON+OL=532

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 =532
Thus, OL+OM+ON=532

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