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**Question 1. Construct a triangle ABC in which BC = 7 cm, âˆ B = 75Âº and AB + AC = 13 cm. Solution: Steps of construction:**

I . Draw a ray BX.

II. From cut off = 7 cm.

III. At B, construct âˆ CBY = 75Âº.

IV. From cut off BD = 13 cm (= AB + BC)

V. Join D and C.

VI. Bisect DC such that the bisector of DC meets BD at A.

VII.Join AC.

Thus, Î”ABC is the required triangle.

**Question 2. Construct a triangle ABC in which BC = 8 cm, âˆ B = 45Âº and AB âˆ AC = 3.5 cm. Solution: Steps of construction: **

I . Draw a ray

II. From , cut off BC = 8 cm.

III. Construct âˆ CBY = 45Âº.

IV. From cut off = 3.5 cm.

V. Join D and C.

VI. Draw PQ, perpendicular bisector of DC, which intersects at A.

VII. Join AC.

Thus, ABC is the required triangle.

**Question 3. Construct a triangle PQR in which QR = 6 cm, âˆ Q = 60Âº and PR âˆ PQ = 2cm. Solution: Steps of construction:**

I . Draw a ray .

II. From , cut off QR = 6 cm.

III. Construct a line YQY' such that âˆ RQY = 60Âº.

IV. Cut off QS = 2 cm (from QY').

V. Join â€˜Sâ€™ and â€˜Râ€™.

VI. Draw MN, perpendicular bisector of SR, which intersects QY at P.

VII. Join P and R.

Thus, PQR is the required triangle.

**Question 4. Construct a triangle XYZ in which âˆ Y = 30Âº , âˆ Z = 90Âº and XY + YZ + ZX = 11 cm. Solution: Steps of construction:**

I . Draw a line segment AB = 11 cm = (XY + YZ + ZX)

II. Construct âˆ BAP = 30Âº = âˆ Y

III. Construct âˆ ABQ = 90Âº = âˆ Z

IV. Draw , the bisector of âˆ BAP.

V. Draw , the bisector of âˆ ABQ, such that and intersect each other at X.

VI. Draw perpendicular bisector of AX, which intersects AB at Y.

VII. Draw perpendicular bisector of XB, which intersects AB at Z.

VIII. Join XY and XZ.

Thus, XYZ is the required triangle.

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

I . Draw = 12 cm.

II. Construct âˆ CBY = 90Âº.

III. From cut off BX = 18 cm.

IV. Join CX.

V. Draw PQ, the perpendicular bisector of CX, such that PQ meets BX at A.

VI. Join AC.

Thus, ABC is the required triangle.

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