**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.**

**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 :**

- Draw a line segment PQ = 8.4 cm.
- With P and Q as centres, draw arcs of radius little more than half of PQ. Let his line intersects PQ in M.
- With M and Q as centres, draw arcs of radius little more than half of MQ. Let this line intersects PQ in N.
- 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.

**More Resources for CBSE Class 9**

- NCERT Solutions
- NCERT Solutions Class 9 Maths
- NCERT Solutions Class 9 Science
- NCERT Solutions Class 9 Social Science
- NCERT Solutions Class 9 English
- NCERT Solutions Class 9 Hindi
- NCERT Solutions Class 9 Sanskrit
- NCERT Solutions Class 9 IT
- RD Sharma Class 9 Solutions

**3. Draw any reflex angle. Bisect it using compass. Name the angles so obtained. [CBSE-15-NS72LP7]**

** Answer.**

**4. Why we cannot construct a ΔABC, if ****∠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 using compass only. [CBSE-14-17DIG1U]**

** Answer.**

**SHORT ANSWER QUESTIONS TYPE-I**

**6. Using ruler and compass, construct 4∠XYZ, if ∠XYZ= 20° [CBSE-14-ERFKZ8H]**

** Answer.**

**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 :**

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

**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.**

**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 : **

- Draw a line segment AB = 5 cm.
- With A and B as centres, draw two arcs of radius 6 cm and let they intersect each other in C.
- Join AC and BC to get ΔABC.
- 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.

**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:**

- Draw a line segment BC = 8 cm.
- At B, Draw ∠CBX = 90°.
- From ray BX, cut off BE = 16 cm.
- Join CE .
- Draw the perpendicular bisector of EC meeting BE at A.
- Join AC to obtain the required ΔABC.

**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 ΔABC, 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 :**

- Draw any line segment BC = 4 cm.
- Draw ‘p’ the perpendicular bisector of BC and let it intersect BC in R
- Cut off PQ = 6.5 cm.
- Join QB.
- Draw the perpendicular bisector of BQ and let it intersect PQ in A.
- Join AB and AC. Thus, ΔABC is the required triangle.

**12. Construct an equilateral triangle of altitude 6 cm. [CBSE-15-6DWMW5A]**

** Answer. Steps of construction :**

- Draw any line l.
- Take any point M on it and draw a line p perpendicular to l.
- With M as centre, cut off MC = 6 cm
- 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.

**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 :**

- Draw a line segment QR = 5 cm.
- With Q as centre, construct an angle of 90° and let this line through Q is QX.
- 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.

**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.

**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.**

**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 :**

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