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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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