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#### Class nine Chapter 11 Construction Questions

## Chapter 11: Construction

**Short
Answer Questions Type – I:**

**Can we construct an angle of 52 1/2 °? Justify your answer.**

**Sol. **An angle of 52 1/2 **° **can be constructed using ruler
and compasses.

52 1/2 ° = 12 (90 + 15)°.

** 2. Can we construct an angle of 67 1/2 °? Justify your answer. **

**Sol. **An angle of 67 1/2 **° **can
be constructed using ruler and compasses.

67 1/2 **° = **1/2 (120 + 15)°

**3. **A triangle ABC can be constructed in which ∠B = 60°, ∠C = 45° and AB + BC + AC = 12 cm. Is this statement true? Justify your answer.

**Sol. **Yes. Since the base angles, 60º and 45º can be constructed using ruler and compasses.

**4. **Can
we construct a triangle ABC in which ∠B
= 105°, ∠C = 90° and

AB + BC + AC = 10 cm.

**Sol. **No, because ∠B + ∠C = 105º + 90º = 195º, which is not possible for a triangle.

**Short
Answer Questions Type – II:**

**1. Draw a line segment AB and by ruler and compass, obtain a line segment of length 3/4 AB. **

**Sol. Steps of Construction **

(i) Draw a line segment AB of a given length.

(ii) Taking A and B as centres and radius equal to more than AB/2, draw arcs, which intersect each other at C and D.

(iii) Join CD which intersects AB at E. AE = 1/2 AB.

(iv) Similarly, locate the point H, such that EH = 1/2 EB = 1/2 ( 1/2 AB) = 1/4 AB

So, AH = AE + EH = 1/2 AB + 1/4 AB = 3/4 AB.

** 2. Draw a line segment AB of 5 cm in length.
Draw a line perpendicular to AB through A and B respectively. Are these lines parallel?**

**Sol.**

Yes, the lines CD and EF are parallel.

**3. ****Draw an angle of 80° with the help of a
protractor. Then construct angles of (i) 40° (ii) 160° (iii) 120° **

**Sol.**

Using a protractor, draw ∠BOC = 80º.

Draw the bisector of ∠BOC to get:

∠BOD = ∠COD = 40º.

Draw ∠DOE = 120º

So, ∠BOE = ∠BOD + ∠DOE

= 40º + 120º = 160º.

**4. ****Construct a ΔABC in which BC = 6.4 cm, ****∠****B = 45° and AB – AC = 3.5 cm. **

**Sol.**

**Steps of
Construction **

(i) Draw BC = 6.4 cm.

(ii) At B, draw ∠PBC = 45º.

(iii) From BP, cut off BD = 3.5 cm.

(iv) Join DC.

(v) Draw the perpendicular bisector of DC, which meets BP produced at A.

(vi) Join AC to get the required triangle ABC

**5. Construct ΔABC in which BC = 4.5 cm, ∠B = 45° and AB – AC = 5.6 cm.**

**Sol.**

**Steps of Construction **

(i) Draw BC = 4.5 cm.

(ii) At B, draw ∠CBP = 45º.

(iii) From BP, cut off BD = 5.6 cm.

(iv) Join DC and draw the perpendicular bisector of DC. It cuts BP at A.

(v) Join AC to get the required triangle ABC.

**6. Construct a rhombus whose side is of length 3.4 cm and one of its angles is 45°. **

**Sol.**

**Steps of Construction **

(i) Draw AB = 3.4 cm.

(ii) At A, draw ∠BAX = 45º

From AX, cut off AD = 3.4 cm.

(iii) At B, draw ∠YBZ = 45º.

From BZ cut off BC = 3.4 cm.

(iv) Join CD to get the required rhombus as ABCD.

**7. Construct a triangle ABC in which ∠A = 45°, ∠B = 120° and AB + BC + AC = 10.4 cm. **

**Sol.**

**Steps of Construction **

(i) Draw PQ = 10.4 cm.

(ii) At P, draw ∠QPR = 45º and at Q, draw ∠PQS = 120º.

(iii) Draw the bisector of ∠QPR and ∠PQS. These bisectors meet at A.

(iv) Draw the perpendicular bisectors of AP and AQ. These perpendicular bisectors meet PQ at C and B respectively.

(v) Join AB and AC to get the required ΔABC.

**8. Construct a triangle ABC such that AB = BC = 6 cm and median AD = 4 cm. **

**Sol. Steps of Construction **

(i) Draw BC = 6 cm and mark its mid-point as D.

(ii) With D as centre and radius equal to 4 cm, draw an arc.

(iii) With B as centre and radius equal to 6 cm, draw an arc, which cuts the previous arc at A.

(iv) Join AB, AD and AC to get the required triangle ABC.

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