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 :

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.
   

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 [5212]0 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 :

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.
   

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 :

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.
   

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.
   

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