Jharkhand Board JAC Class 9 Maths Solutions Chapter 11 Constructions Ex 11.2 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 11 Constructions Ex 11.2

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Question 1.

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

Answer:

Steps of Construction:

Step 1: A line segment BC of 7 cm is drawn.

Step 2: At point B, an angle ∠XBC is constructed such that it is equal to 75°.

Step 3: A line segment BD =13 cm is cut on BX (which is equal to AB + AC).

Step 4: DC is joined.

Step 5: Draw perpendicular bisector of CD which meets BD at A.

Step 6: Join AC.

Thus, ∆ABC is the required triangle.

Question 2.

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

Answer:

Steps of Construction:

Step 1: A line segment BC = 8 cm is drawn and at point B, make an angle of 45° i.e. ∠XBC.

Step 2: Cut the line segment BD = 3.5 cm (equal to AB – AC) on ray BX.

Step 3: Join DC and draw the

perpendicular bisector PQ of CD.

Step 4: Let it intersect BX at point A. Join AC.

Thus, ∆ABC is the required triangle.

Question 3.

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

Answer:

Steps of Construction:

Step 1: A line segment QR = 6 cm is drawn.

Step 2: A ray QY is constructed making an angle of 60° with QR and YQ is produced backwards to form a line YY’.

Step 3: Cut off a line segment QS = 2 cm from QY’. RS is joined.

Step 4: Draw perpendicular bisector of RS intersecting QY at a point P. PR is joined. Thus, ∆PQR is the required triangle.

Question 4.

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

Answer:

Steps of Construction:

Step 1: A line segment PQ = il cm is drawn. (XY + YZ + ZX =11 cm)

Step 2: An angle, ZRPQ = 30° is constructed at point P and an angle ZSQP = 90° at point Q.

Step 3: ZRPQ and ZSQP are bisected. The bisectors of these angles intersect each other at point X.

Step 4: Perpendicular bisectors TU of PX and WV of QX are constructed.

Step 5: Let TU intersect PQ at Y and WV intersect PQ at Z. XY and XZ are joined.

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.

Answer:

Steps of Construction:

Step 1 : A line segment BC = 12 cm is drawn.

Step 2: ZCBY = 90° is constructed.

Step 3: Cut off a line segment BD = 18 cm from BY. CD is joined.

Step 4: Perpendicular bisector of CD is constructed intersecting BD at A. AC is joined.

Thus, ∆ABC is the required triangle.