Given base QR = 6 cm
∠ Q = 60°
And PR – PQ = 2 cms.
Steps of construction:
i. Draw a base line QR of 6 cms.
ii. Construct ∠ Q = 60°.
a. With Q as centre and with any radius, draw another arc cutting the line QR at A.
b. With A as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point B.
c. Now join the ray QB which forms an angle of 60° with the line QR.
As, PR – PQ = 2
PR > PQ (negative)
So the ray QB will be below QR
iii. Extend the ray QB below QR. With Q as centre draw an arc with length 2 cms ( = PR- PQ given), such that it intersects ray QB at C below QR.
iv. Join CR and draw a perpendicular bisector for CR.
a. By drawing arcs on both sides of the line CR, with C and R as centers and with same lengths which is more than half length of CR. These arcs intersect at D and E on either side of line CR.
v. The perpendicular bisector for CR will intersect the ray QB at point P. Join PR.
Thus the formed triangle PQR is the required triangle.
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