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Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ =2cm.

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The steps mentioned below will be followed to construct the required triangle. Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠XQR. Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and...
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The steps mentioned below will be followed to construct the required triangle.

Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠XQR.

Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and PR − PQ = 2 cm). Join SR.

Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.

ΔPQR is the required triangle.

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Maths tutor with 4 years experience

The steps of construction for the required triangles are as follows:Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.Step...
read more
The steps of construction for the required triangles are as follows:
Step I: Draw line segment QR of 6 cm. At point Q draw an angle of 60o say XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended an opposite side of line segment XQ. (As PR> PQ and PR - PQ = 2cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR. PQR is the required triangle.
 
 
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