Please refer to Constructions Class 9 Mathematics Exam Questions provided below. These questions and answers for Class 9 Mathematics have been designed based on the past trend of questions and important topics in your class 9 Mathematics books. You should go through all Class 9 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.

**Class 9 Mathematics Exam Questions Constructions**

Class 9 Mathematics students should read and understand the important questions and answers provided below for Constructions which will help them to understand all important and difficult topics.

**Very Short Answer Type Questions:**

**Question. What is the length of bisected part of a line segment 7.8 cm?****Ans. **We know that bisector of the line, divides it into two equal parts.

∴ Length of bisected part of a line segment measuring

7.8 cm = 1/2 (7.8) cm = 3.9 cm.

**Question. Find the measure of each of the two angles formed by bisecting an angle of measure 135°.****Ans.** The measure of each of the two angles formed by bisecting an angle of measure 135° = 1/2/× 135° = 67.5°.

**Question. In order to construct a triangle uniquely, how many minimum parts of triangle are required to be given?****Ans.** To construct a triangle uniquely, we are required at least three values like, 2 sides and 1 included angle or 2 angles and 1 included side or all three sides.

**Short Answer Type Questions:**

**Question. Draw a perpendicular bisector of line segment PQ of length 8.4 cm.****Ans.** Steps of construction :

Step I : Draw a line segment PQ = 8.4 cm.

Step II : With P as centre and radius more than half of PQ, draw two arcs, one on each side of PQ.

Step III : With Q as centre and the same radius as in Step II, draw arcs cutting the arcs drawn in the previous step at L and M respectively.

Step IV : Join LM.

Thus, the line segment LM is required perpendicular bisector of PQ.

**Question. Construct a triangle with base length 5 cm, the sum of other two sides is 7 cm and one base angle is 60°.****Ans. **Steps of construction :

Step I : Draw PQ = 5 cm.

Step II : At P, construct ∠P = 60°.

Step III : From P, cut line segment PT

= 7 cm (= PR + RQ)

Step IV : Join TQ.

Step V : Draw the perpendicular bisector of TQ which meets PT at R.

Step VI : Join RQ.

Thus, ΔPQR is the required triangle.

**Question. Draw a line segment AB = 16 cm. Divide it into (3/4) ^{th} part. Measure the length of (3/4)^{th }part of AB.**

**Ans.**Since, 3/4 = 2 + 1/4 = 2/4 + 1/4 = 1/2 + 1/4

So, to get (3/4)

^{th }of AB, we should bisect AB and then again bisect one of the bisected part of AB.

Steps of construction :

Step I : Draw a line segment AB = 16 cm.

Step II : Draw the perpendicular bisector PQ of AB such that PQ intersects AB at point M.

Step III : Now, draw the perpendicular bisector CD of MB.

Thus, AM + MN i.e., AN is the required line segment.

∴ AN = 3/4 AB = 3/4 × 16

= 3 × 4 = 12 cm

Hence, the measure of AN is 12 cm.

**Question. Construct a right angled triangle whose base is 6 cm and sum of its hypotenuse and the other side is 10 cm.****Ans. **Steps of construction :

Step I : Draw a base BC equal to 6 cm.

Step II : Construct ∠CBX = 90°.

Step III : Cut line segment BD = 10 cm along BX.

Step IV : Join CD and draw PQ, perpendicular bisector of CD.

Step V : PQ intersects BD at A and CD at L.

Step VI : Join AC.

Hence, ΔABC is the required right triangle.

**Question. Draw a line segment of length 6.4 cm. Bisect it and measure the length of each part.****Ans.** Steps of construction :

Step I : Draw a line segment AB = 6.4 cm by using a graduated ruler.

Step II : Taking A as centre and radius equal to more than half of AB, draw arcs on both sides of line segment AB.

Step III : Taking B as centre and same radius as in Step II,draw arcs on both sides of the line segment AB, cutting the previous arcs at E and F.

Step IV : Join EF, intersecting AB at M.

Then, M bisects the line segment AB.

On measuring with graduated ruler, we find that AM

= MB = 3.2 cm

**Question. By using protractor, draw an angle of 108° and taking this angle as given, construct an angle of 54°.****Ans.** Steps of construction :

Step I : Draw a ray OA.

Step II : By using protractor, draw ∠AOB = 108° = 2 × 54°.

Step III : With O as centre and any convenient radius, draw an arc cutting OA and OB at P and Q respectively.

Step IV : With P as centre and radius more than half of PQ, draw an arc.

Step V : With Q as centre and the same radius as taken in Step IV, draw another arc intersecting the previous arc at C. Join OC.

Thus, OC is the bisector of ∠AOB, such that

∠AOC = 1/2 ∠AOB = 108°/2 = 54°

**Question. Draw lines PQ and RS intersecting at point O. Measure a pair of vertically opposite angles. Bisect them. Are the bisecting rays forming a straight line?****Ans. **Steps of construction :

Step I : Draw a line PQ.

Step II : Draw another line RS intersecting PQ at point O.

Step III : Measure pair of vertically opposite angles.

Step IV : Construct OX bisector of ∠QOS and OY bisector of ∠POR.

Yes, from the construction it is clear that the bisecting rays are forming a straight line.

**Question. Can a ΔABC be constructed in which ∠B = 110°, ∠C = 95° and AB = 10 cm? Justify your answer.****Ans.** No, as we know that sum of all three angles of a triangle is 180°.

But, here ∠B + ∠C = 110° + 95° = 205° > 180°

∴ ΔABC cannot be constructed with given conditions.

**Question. Construct a DSTU, in which ∠T = 100°, TU = 5 cm and ST + US = 8 cm.****Ans. **Steps of construction :

Step I : Draw TU = 5 cm.

Step II : Draw ∠UTX = 100°

Step III : Along TX, cut off a line segment TR = ST + US = 8 cm.

Step IV : Join UR.

Step V : Draw the perpendicular bisector of UR, meeting TR at S.

Step VI : Join US.

Hence, ΔSTU is the required triangle.

**Long Answer Type Questions:**

**Question. Construct a triangle having sides of length 6.2 cm, 7.3 cm and 6 cm. Measure all the three angles. Bisect the smallest and the largest angles. Measure any acute angle formed by the bisecting rays at the point of intersection. Also, verify your answer.****Ans.** Steps of construction :

Step I : Draw a line segment AB = 6.2 cm.

Step II : Draw an arc with A as centre and 7.3 cm as radius and draw another arc with B as centre and 6 cm as radius to intersect each other at C.

Step III : Join AC and BC. Thus, we get the required triangle ABC. On measuring all the three angles, we get ∠A = 52°, ∠B = 73° and ∠C = 55°.

Step IV : Since, ∠A is the smallest angle and ∠B is the largest angle in ΔABC. Draw the angle bisectors of ∠A and ∠B, which intersect each other at O.

Step V : On measuring the acute ∠AOY formed by the bisecting rays AX and BY at the point of intersection O, we get ∠AOY = 62.5°.

Verification : In ΔAOB,

∠AOY = ∠OAB + ∠OBA (Exterior angle property)

⇒ ∠AOY = + 1/2 ∠A + 1/2 ∠B ( AO and BO are the bisectors of ∠A and ∠B respectively)

⇒ ∠AOY = 52°/2 + 73°/2 = 26 + 36.5° = 62.5°

**Question. Construct a ΔPQR, in which QR = 6.5 cm,****∠Q = 60° and PR – PQ = 1.5 cm. Also, justify the construction.****Ans.** Here, PR – PQ = 1.5 cm ∴ PR > PQ

i.e., The side containing the base angle Q is less than third side, so it is the case II.

Steps of construction :

Step I : Draw the base QR = 6.5 cm.

Step II : Construct a ray QX making an angle 60° with QR and extent XQ on opposite side of line segment QR, to form a line XQX′.

Step III : From QX , cut-off the line segment QS = 1.5 cm (= PR – PQ).

Step IV : Join SR.

Step V : Draw the perpendicular bisector of SR intersecting QX at point P.

Step VI : Join PR.

Then, PQR is the required triangle.

Justification : Since, point P lies on the perpendicular bisector of SR.

∴ PS = PR ⇒ PQ + QS = PR ⇒ PR – PQ = QS = 1.5 cm, which justified the construction.

**Question. Using a protractor, draw an angle of measure 128°. With this angle construct an angle of measure 96°.****Ans. **In order to construct an angle of measure 96° from an angle of measure 128°, we use the following steps :

Steps of construction :

Step I : Draw an angle ∠AOB of measure 128° by using a protractor.

Step II : With centre O and a convenient radius, draw an arc cutting OA and OB at P and Q respectively.

Step III : With centre P and radius more than 1/2 (PQ), draw an arc.

Step IV : With centre Q and the same radius, as in Step III, draw another arc intersecting the previously drawn arc at R.

Step V : Join OR and produce it to form ray OX . OX cuts arc PQ at S. Then ∠AOX so obtained is equal to (128°/ 2) = 64°.

Step VI : With S as a centre and radius more than half of QS, draw an arc.

Step VII : With centre Q and the same radius, as in Step VI, draw another arc intersecting the arc drawn in Step VI at T.

Step VIII : Join OT and produce it to form a ray OY.

Clearly, ∠XOY = 1/2 ∠XOB = 1/2 (64°) = 32°

∴ ∠AOY = ∠AOX + ∠XOY = 64° + 32° = 96°

Thus, ∠AOY is the desired angle of measure 96°.