Important Questions for CBSE Class 9 Mathematics Chapter 5 Constructions

IMPORTANT QUESTIONS

VERY SHORT ANSWER TYPE QUESTIONS
1. Construct an angle of 90 ° at the initial point of the given ray. [CBSE-15-6DWMW5A]
Answer.
cbse-class-9-mathematics-constructions-1

2. Draw a line segment PQ = 8.4 cm. Divide PQ into four equal parts using ruler and compass. [CBSE-14-ERFKZ8H], [CBSE – 14-GDQNI3W], [CBSE-14-17DIG1U]
Answer. Steps of construction :

  1. Draw a line segment PQ = 8.4 cm.
  2. With P and Q as centres, draw arcs of radius little more than half of PQ. Let his line intersects PQ in M.
  3. With M and Q as centres, draw arcs of radius little more than half of MQ. Let this line intersects PQ in N.
  4. With P and M as centres, draw arcs of radius little more than half of PM. Let this line intersects PQ in L.  Thus, L, M and N divide the line segment PQ in four equal parts.
    cbse-class-9-mathematics-constructions-2

3. Draw any reflex angle. Bisect it using compass. Name the angles so obtained. [CBSE-15-NS72LP7]
Answer.
cbse-class-9-mathematics-constructions-3

4. Why we cannot construct a ΔABC, if  âˆ A=60 °, AB — 6 cm, AC + BC = 5 cm but construction of A ABC is possible if ∠A=60 °, AB = 6 cm and AC – BC = 5 cm. [CBSE-14-GDQNI3W]
Answer. We know that, by triangle inequality property, construction of triangle is possible if sum of two sides of a triangle is greater than the third side.
Here, AC + BC = 5 cm which is less than AB ( 6 cm)
Thus, ΔABC is not possible.
Also, by triangle inequality property, construction of triangle is possible, if difference of two
sides of a triangle is less than the third side
Here, AC – BC = 5 cm, which is less than AB (6 cm)
Thus, ΔABC is possible.

5. Construct angle of {{\left[ 52\frac{1}{2} \right]}^{0}} using compass only. [CBSE-14-17DIG1U]
Answer.
cbse-class-9-mathematics-constructions-4

SHORT ANSWER QUESTIONS TYPE-I
6. Using ruler and compass, construct 4∠XYZ, if ∠XYZ= 20 ° [CBSE-14-ERFKZ8H]
Answer.
cbse-class-9-mathematics-constructions-5

7. Construct an equilateral triangle LMN, one of whose side is 5 cm. Bisect ∠  M of the triangle. [CBSE March 2012]
Answer. Steps of construction :

  1.  Draw a line segment LM = 5 cm.
  2. Taking L as centre and radius 5 cm draw an arc.
  3. Taking M as centre and radius draw an other arc intersecting previous arc at N.
  4. Join LN and MN. Thus, ΔLMN is the required equilateral triangle.
  5. Taking M as centre and any suitable radius, draw an arc intersecting LM at P and MN at Q.
  6. Taking P and Q as centres and same radii, draw arcs intersecting at S.
  7. Join MS and produce it meet LN at R. Thus, MSR is the required bisector of ∠M.
    cbse-class-9-mathematics-constructions-6

SHORT ANSWER QUESTIONS TYPE-II
8. Construct a A ABC with BC = 8 cm, ∠B= 45 ° and AB – AC = 3.1 cm. [CBSE-15-NS72LP7]
Answer.
cbse-class-9-mathematics-constructions-7

9. Construct an isosceles triangle whose two equal sides measure 6 cm each and whose base is 5 cm. Draw the perpendicular bisector of its base and show that it passes through the opposite vertex [CBSE-15-6DWMW5A]
Answer.  Steps of construction :

  1. Draw a line segment AB = 5 cm.
  2. With A and B as centres, draw two arcs of radius 6 cm and let they intersect each other in C.
  3.  Join AC and BC to get ΔABC.
  4. With A and B as centres, draw two arcs of radius little more than half of AB. Let they intersect each other in P and Q. Join PQ and produce, to pass through C.
    cbse-class-9-mathematics-constructions-8

10. Construct a right triangle whose base is 8 cm and sum of the hypotenuse and other side is 16 cm.
Answer. Given : In ΔABC, BC = 8 cm, ∠B= 90 ° and AB + AC = 16 cm.
Required : To construct ΔABC.
Steps of construction:

  1. Draw a line segment BC = 8 cm.
  2. At B, Draw ∠CBX = 90 °.
  3. From ray BX, cut off BE = 16 cm.
  4.  Join CE .
  5. Draw the perpendicular bisector of EC meeting BE at A.
  6. Join AC to obtain the required ΔABC.
    cbse-class-9-mathematics-constructions-15

11. To construct an isosceles ΔABC in which base BC = 4 cm, sum of the perpendicular from A to BC and side AB = 6.5 cm.
Answer. Given : In  Î”ABC, BC = 4 cm and sum of the perpendicular  from A to BC and side AB = 6.5 cm.
Required : To construct ΔABC.
Steps of construction :

  1. Draw any line segment BC = 4 cm.
  2. Draw ‘p’ the perpendicular bisector of BC and let it intersect BC in R
  3. Cut off PQ = 6.5 cm.
  4. Join QB.
  5. Draw the perpendicular bisector of BQ and let it intersect PQ in A.
  6. Join AB and AC. Thus, ΔABC is the required triangle.
    cbse-class-9-mathematics-constructions-9

12. Construct an  equilateral triangle of altitude 6 cm. [CBSE-15-6DWMW5A]
Answer. Steps of construction :

  1. Draw any line l.
  2. Take any point M on it and draw  a line p perpendicular to l.
  3. With M as centre, cut off MC = 6 cm
  4. At C, with initial line CM construct angles of measures 30 ° on both sides and let these lines intersect line l in A and B. Thus, ΔABC is the required triangle.
    cbse-class-9-mathematics-constructions-10

13. Draw a line segment QR = 5 cm. Construct perpendiculars at point Q and R to it. Name them as QX and RY respectively. Are they both parallel ?  [CBSE-15-NS72LP7] [CBSE-14-ERFKZ8H]
Answer. Steps of construction :

  1. Draw a line segment QR = 5 cm.
  2. With Q as centre, construct an angle of 90 ° and let this line through Q is QX.
  3. With R as centre, construct an angle of 90 ° and let this line through R is RY. Yes, the perpendicular lines QX and- RY are parallel.
    cbse-class-9-mathematics-constructions-11

LONG ANSWER TYPE QUESTIONS
14. Construct a triangle ABC in which BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60 °. [CBSE March 2012]
Answer. Given : In ΔABC, BC= 4.7 cm, AB + AC = 8.2 cm and ∠C= 60 °.
Required : To construct ΔABC.
cbse-class-9-mathematics-constructions-12

15. To construct a triangle, given its perimeter and its two base angles, e.g., construct a triangle with perimeter 10 cm and base angles 60 ° and 45 °. [CBSE March 2012]
Answer.
cbse-class-9-mathematics-constructions-13

16. Construct ΔXYZ, if its perimeter is 14 cm, one side of length 5 cm and ∠X= 45 °. [CBSE-14-ERFKZ8H]
Answer. Here, perimeter of ΔXYZ = 14 cm and one side XY = 5 cm
.-. YZ + XZ = 14 – 5 = 9 cm and ∠X = 45 °.
Steps of construction :

  1. Draw a line segment XY = 5 cm.
  2. Construct an ∠YXA = 45 ° with the help of compass and ruler.
  3. From ray XA, cut off XB – 9 cm.
  4. Join BY.
  5. Draw perpendicular bisector of BY and let it intersect XB in Z.
  6. Join ZY. Thus, ΔXYZ is the required triangle.
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