Question

# To construct a ΔPQR, in which QR= 5.5 cm, ∠Q=60∘ and PR - PQ = 2.5 cm, the steps of construction are given below. Complete the third step Here, PR > PQ, i.e., the side containing base angle is less than the third side. Steps of construction: Step 1: Draw the base QR = 5.5 cm Step 2: At the point Q, make an ∠XQR=60∘ Step 3 will be

A
Cut line segment QS = PR - PQ = 2 cm from the line QX extended on opposite side of linesegment QR
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B
Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR
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C
Cut line segment QS = PR - PQ = 2.5 cm from the ray QX
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D
Draw the perpendicular bisector LM of SR.
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Solution

## The correct option is B Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR Steps of construction: Step 1: Draw the base QR = 5.5 cm Step 2: At the point Q, make an ∠XQR=60∘ Step 3 : Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR. Join RS and draw the perpendicular bisector of RS. Step 4 : Let it meet QX at P and join PR. PQR is the required triangle.

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