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Exercise - 11.2

Constructions

**Question-1 :-** Construct a triangle ABC in which BC = 7cm, ∠B = 75° and AB + AC = 13 cm.

Steps of Construction : Step I: Draw a line segment BC of 7 cm. At point B, draw an angle of 75°, say XBC. Step II: Cut a line segment BD = 13 cm (that is equal to AB + AC) from the ray BX. Step III: Join DC and make an angle DCY equal to BDC. Step IV: Let CY intersect BX at A. ∆ABC is the required triangle.

**Question-2 :-** Construct a triangle ABC in which BC = 8cm, ∠B = 45° and AB – AC = 3.5 cm.

Steps of Construction : Step I: Draw the line segment BC = 8 cm and at point B, make an angle of 45°, say XBC. Step II: Cut the line segment BD = 3.5 cm (equal to AB − AC) on ray BX. Step III: Join DC and draw the perpendicular bisector PQ of DC. Step IV: Let it intersect BX at point A. Join AC. ∆ABC is the required triangle.

**Question-3 :-** Construct a triangle PQR in which QR = 6cm, ∠Q = 60° and PR – PQ = 2cm.

Steps of Construction : 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. ∆PQR is the required triangle.

**Question-4 :-** Construct a triangle XYZ in which ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11 cm.

Steps of Construction : Step I: Draw a line segment AB of 11 cm. (As XY + YZ + ZX = 11 cm) Step II: Construct an angle, PAB, of 30° at point A and an angle, QBA, of 90° at point B. Step III: Bisect PAB and QBA. Let these bisectors intersect each other at point X. Step IV: Draw perpendicular bisector ST of AX and UV of BX. Step V: Let ST intersect AB at Y and UV intersect AB at Z. Join XY, XZ. ∆XYZ is the required triangle.

**Question-5 :-** Construct a right triangle whose base is 12cm and sum of its hypotenuse and other side is 18 cm.

Steps of Construction : Step I: Draw line segment AB of 12 cm. Draw a ray AX making 90° with AB. Step II: Cut a line segment AD of 18 cm (as the sum of the other two sides is 18) from ray AX. Step III: Join DB and make an angle DBY equal to ADB. Step IV: Let BY intersect AX at C. Join AC, BC. ∆ABC is the required triangle.

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