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Question: Construct a triangle PQR in which \(QR = 6\,{\rm{cm}}\), \(\angle {\rm{Q = 60}}^\circ \) and \({\rm{...

Construct a triangle PQR in which QR=6cmQR = 6\,{\rm{cm}}, Q=60\angle {\rm{Q = 60}}^\circ and PRPQ=2cm{\rm{PR}} - {\rm{PQ}} = {\rm{2}}\,{\rm{cm}}.

Explanation

Solution

In this problem, to construct the triangle PQR use the method of construction of a triangle with one side, one angle and difference of other two sides.

Complete Step-by-step Solution
Given,
One side of the triangle is QR=6cmQR = 6\,{\rm{cm}}.
The angle Q is Q=60\angle {\rm{Q = 60}}^\circ .
The difference between PQ and PO is PRPQ=2cm{\rm{PR}} - {\rm{PQ}} = {\rm{2}}\,{\rm{cm}}.

The following are the steps to construct a triangle PQR.

1. Draw the baseline of triangle as QR=6cmQR = 6\,{\rm{cm}}.

2. Now, draw an angle of 6060^\circ from point Q. To make an angle of 6060^\circ , draw a semicircle from point Q and with the same radius of compass intersect this semicircle from point X.


3. Open the compass and fill the distance PRPQ=2cm{\rm{PR}} - {\rm{PQ}} = {\rm{2}}\,{\rm{cm}} and draw an arc from point Q at opposite side of ray QX.

4. Arc intersect ray QX at point D. join point R and D.

5. Now draw a perpendicular bisector of line RD, and extend the line up to ray QX. Mark point P where perpendicular bisector intersects ray QX.

6. Now join PR.


Hence, ΔPQR\Delta PQR is a required triangle.

Note: In such types of problems, while making bisectors of angles remember that radius of compass should be the same to bisect an angle and use a sharp pencil to draw points and to get accurate results.