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. 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) ⟹25√34=12×5×(OM+OL+OM) ⟹OM+ON+OL=5√32 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√32 Thus, OL+OM+ON=5√32

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