CBSE Class 9 Mathematics Constructions Worksheet Set 02

Very Short Answer Type Questions:

Question: Can a ΔXYZ be constructed, in which XY = 5 cm, ∠X = 50° and YZ + XZ = 5 cm?
Answer: No, ΔXYZ can’t be constructed.
Since, sum of two sides of triangle must be greater than third side, but here, XY = YZ + XZ.

Question: If we bisect a line segment AB, then each of the equal part we get measures 3.8 cm. Find the length of AB.
Answer: If we bisect line segment AB, then we get each part equal to 3.8 cm.
∴ Length of AB = 2 × 3.8 cm = 7.6 cm

Question: If we draw a perpendicular bisector of a line segment AB = 9 cm which bisects AB at M, then find AM and BM.
Answer: 1. Since, perpendicular bisector of a line segment divides it into two equal parts.
∴ AM = BM = 9/2 cm = 4.5 cm

Question: Draw a straight angle. Using compass bisect it. Name the angles obtained.
Answer: 

Steps of construction :
Step I : Draw any straight angle (say ∠AOC)
Step II : Draw OB , the bisector of ∠AOC.
Then, ∠AOB and ∠BOC are the required angles obtained by bisecting straight ∠AOC.

Short Answer Type Questions:

Question: Construct ΔABC such that AB = 5.8 cm, BC + CA = 7 cm and ∠B = 60°.
Answer: Steps of Construction :

Step I : Draw a line segment AB = 5.8 cm and ∠ABX = 60°.
Step II : Cut line segment BD = 7 cm along BX .
Step III : Join AD and draw PQ, perpendicular bisector of AD.
Step IV : Let PQ intersects BD at C.
Step V : Join AC.
Hence, ΔABC is the required triangle.

Question: Draw a line segment AB = 13.2 cm. Divide it into 4 equal parts using ruler and compass. Also, measure the length of each part. 
Answer: Steps of construction :

Step I : Draw a line segment AB = 13.2 cm.
Step II : Draw a perpendicular bisector of AB, which intersect AB at M.
Step III : Again, draw a perpendicular bisector of AM; which intersects AM at P.
Step IV : Also, draw a perpendicular bisector of BM, which intersects BM at Q.
Thus, AB is divided into four equal parts, where
AP = PM = MQ = QB
= 1/4 AB = 13.2/4 = 3.3 cm

Question: Draw a line segment of length 6 cm. Draw perpendicular bisector of this line segment.
Answer: Steps of construction :

Step I : Draw a line segment AB = 6 cm by using a ruler.
Step II : With A as centre and radius more than half of AB, draw arcs on both sides of AB.
Step III : With B as centre and the same radius (as taken in previous step), draw arcs cutting the
previous arcs drawn in Step II at E and F respectively.
Step IV : Join EF intersecting AB at M.
Thus, EF is perpendicular bisector of the line segment AB.

Question: Draw line segment AB = 8.8 cm and draw its perpendicular bisector and measure the length of each part.
Answer: Steps of construction :

Step I : Draw a line segment AB = 8.8 cm by using 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 AB cutting the previous arcs at E and F.
Step IV : Join EF intersecting AB at M. Then, EF is the required perpendicular bisector of AB. On measuring by graduated ruler, we find that
AM = MB = 4.4 cm.

Question: Construct ΔABC such that BC = 6 cm, ∠B = 45° and AB – AC = 3 cm.
Answer: Steps of Construction :

Step I : Draw BC = 6 cm and ∠CBX = 45°.
Step II : On BX, cut BD = 3 cm.
Step III : Join CD.
Step IV : Draw PQ, perpendicular bisector of CD.
Step V : PQ intersects BX at A and CD at L.
Step VI : Join AC.
Hence, ΔABC is the required triangle.

Question: Using ruler and compass only, draw an angle of measure 135°.
Answer: Steps of construction : 

Step I : Draw a ray OP.
Step II : With centre O and a suitable radius, draw an arc which cuts OP at A.
Step III : With the same radius and starting from A, mark points Q, R and S on the arc drawn in Step II such that AQ = QR = RS
Step IV : Draw OL, the bisector of ROS.
Step V : Draw OM , the bisector of ROL.
Thus, ∠POM = ∠POR + ∠ROM = 120° + 15° = 135°

Question: Construct an equilateral triangle, the sum of its two sides is 8 cm.
Answer: We know that, in equilateral triangle, all the angles are of equal measure and all sides are of equal length. 
Since, sum of two-sides of triangle is 8 cm, therefore each side of equilateral triangle will be 4 cm.
Steps of construction :

Step I : Draw the line segment BC = 4 cm.
Step II : Taking B as centre and radius
= 4 cm draw an arc.
Step III : Taking C as centre and same radius as in Step II, draw an arc cutting the previous arc at A.
Step IV : Join AB and AC. 
Then, ABC is the required equilateral triangle.

Question: Construct a square of side 3 cm.
Answer: Steps of construction :

Step I : Draw a line segment AB = 3 cm.
Step II : Draw angle of 90° at points A and B of the line segment AB. Also draw AX parallel to BY.
Step III : Cut AD and BC of length 3 cm on AX and BY respectively.
Step IV : Join CD.
Then, ABCD is the required square of side 3 cm. 

Question: Using ruler and compass only, draw a right angle.
Answer: Steps of construction :

Step I : Draw a ray OA .
Step II : Taking O as centre and suitable radius, draw a semicircle, which cuts OA at B.
Step III : With B as centre and the same radius, as in Step II, draw an arc cutting the semicircle at C. Again, with C ascentre, draw an arc cutting the semicircle at D.
Step IV : Draw OC and OD .
Step V : Draw OF , the bisector of ∠COD.
Thus, ∠AOF = 90° 

Long Answer Type Questions:

Question: Construct a ΔABC in which BC = 5.6 cm, AC – AB = 1.6 cm and ∠B = 45°. Justify your construction.
Answer: Steps of construction :
Step I : Draw BC = 5.6 cm.
Step II : At B, construct ∠CBX = 45°.
Step III : Produce XB to X′ to form line XBX.
Step IV : Along ray BX′, cut-off a line segment BD = 1.6 cm.
Step V : Join CD.
Step VI : Draw perpendicular bisector of CD which cuts BX at A.
Step VII : Join CA to obtain required triangle BAC. 

Justification:
Since A lies on the perpendicular bisector of CD.
∴ AC = AD = AB + DB = AB + 1.6 ⇒ AC – AB = 1.6 cm which justified the construction.

Question: Give reason:
(i) Construction of an angle of 22.5° is possible with the help of ruler and compass.
(ii) It is not possible to construct a ΔABC given that BC = 7 cm, ∠B = 45° and AB – AC = 10 cm.
(iii) We can construct an angle of 67.5° using ruler and compass.
(iv) Construction of ΔDEF, if EF = 5.5 cm, ∠E = 75° and DE – DF = 3 cm is possible.
Answer: (i) Yes, because 22.5° = 45° ÷ 2 and 45° can be constructed.
(ii) Yes, it is not possible to construct a ΔABC in which BC = 7 cm and AB – AC = 10 cm with ∠B = 45° because the difference between the given two sides is not less than the third side.
(iii) Yes, we can construct an angle of 67.5°, because 67.5° = 135° ÷ 2 and 135° = 90° + 45°, which can be constructed.
(iv) Yes, it is possible to construct a ΔDEF in which EF = 5.5 cm, ∠E = 75° and DE – DF = 3 cm because the difference between the given two sides is less than the third side.