# Constructions Class 9 Mathematics Exam Questions

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.

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.

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.
Hence, ΔSTU is the required triangle.

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 PQ 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°.