Q1. Construct a triangle ABC in which BC = 7.5 cm, ∠ B = 45º and the difference between the other two sides is 4 cm.
Steps of construction:
 Take BC = 7.5 cm.
 Construct ∠CBX = 45º.
 From cut off BD = 4 cm.
 Join D and C.
 Draw the perpendicular bisector of DC which intersects at A.
 Join A and C.
Thus, ABC is the required triangle.
Q2. Construct a ΔABC whose perimeter is 12 cm and sides are in the ratio 2 : 3 : 4.
Steps of construction:
 Draw = 12 cm.
 Draw , making an acute angle with XY in the downward direction.
 From X, set off (2 + 3 + 4 = 9) nine equal distances along XZ.
 Mark points P, Q and R on XZ such that XP = 2 units PQ = 3 units QR = 4 units
 Join RY.
 Draw QC  RY and PB  RY.
 With centre B and radius BX, draw an arc.
 With centre C and radius CY, draw an arc that intersects the previous arc at A.
 Join AB and AC.
Thus, ΔABC is the required triangle.
Q3. Construct ΔABC, in which ∠ B = 60º and ∠ C = 45º and the perpendicular from the vertex A to the base BC is 5.2 cm.
Steps of construction:
 Draw a line PQ.
 Mark a point L on PQ.
 Draw LM ⊥ PQ.
 From cut off LA = 5.2 cm.
 Draw XY  PQ, through A.
 Construct ∠XAB = 60º and ∠YAC = 45º meeting PQ at B and C respectively.
Thus, ΔABC is the required triangle.
Q4. Construct an equilateral triangle if its altitude is 6 cm. Give justification for your construction.
Steps of construction:
 Draw a line XY and mark a point D on it.
 Construct perpendicular PD on XY.
 Cut off AD = 6 cm
 Make ∠CAD = 30° and ∠BAD = 30° at A on both sides of AD, such that B and C lie on XY.
Thus, ABC is the required triangle.
Justification:
∵ ∠CAD = 30° and ∠ADC = 90°
∴ In ΔADC, ∠C = 180 ∠ (90° + 30°) = 60°
Similarly, ∠B = 60° Also, ∠BAD + ∠CAD = 60°
⇒ ∠A = 60°
Thus, ΔABC is an equilateral triangle in which altitude AD = 6 cm.
Q5. Why we cannot construct a ∆ABC, if ∠A = 60°, AB = 6 cm and AC + BC = 5 cm but construction of ∆ABC is possible if ∠A = 60°, AB = 6 cm and AC – BC = 5 cm?
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, ∆ABC 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, ∆ABC is possible.
Q6. Construct an angle of 90° at the initial point of the given ray.
Steps of Construction:
 Draw a ray OA.
 With O as centre and any convenient radius, draw an arc, cutting OA at P.
 With P as centre and same radius, draw an arc cutting the arc drawn in step 2 at Q.
 With Q as centre and the same radius as in steps 2 and 3, draw an arc, cutting the arc drawn in step 2 at R.
 With Q and R as centres and same radius, draw two arcs, cutting each other in S.
 Join OS and produce to B.
 Thus, ∠AOB is the required angle of 90°
Q7. Draw a straight angle. Using compass bisect it. Name the angles obtained.
Steps of Construction:
 Draw any straight angle (say ∠AOC).
 Bisect ∠AOC and join BO.
 ∠AOB is the required bisector of straight angle AOC.
Q8. Construct a ∆ABC such that BC = 3.2 cm, ∠B = 45° and AC – AB = 2.1 cm.
Steps of Construction:
 Draw a line segment BC = 3.2 cm.
 At B, construct an angle ∠CBX = 45° and produce it to point X’.
 Cutoff BD = 2.1 cm and join CD.
 Draw the perpendicular bisector of CD and let it intersect X’BX in A. Join AC.
 Thus, ∆ABC is the required triangle.
Q9. Construct a triangle ABC in which BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.
Given: In ∆ABC, BC = 4.7 cm, AB + AC = 8.2 cm and ∠C = 60°.
Required: To construct ∆ABC.
Steps of Construction:
 Draw BC = 4.7 cm.
 Draw From ray CX, cut off CD = 8.2 cm.
 Join BD. Draw the perpendicular bisector of BD meeting CD at A.
 Join AB to obtain the required triangle ABC.
Justification:
∵ A lies on the perpendicular bisector of BD, therefore, AB = AD
Now, CD = 8.2 cm
⇒ AC + AD = 8.2 cm
⇒ AC + AB = 8.2 cm.
Q10. To construct a triangle, with perimeter 10 cm and base angles 60° and 45°.
Given: In ∆ABC,
AB + BC + CA = 10 cm,
∠B = 60° and ∠C = 45°.
Required: To construct ∆ABC.
Steps of Construction:
 Draw DE = 10 cm.
 At D, construct ∠EDP= 5 of 60°= 30° and at E, construct DEQ = 1 of 45° = 22°
 Let DP and EQ meet at A.
 Draw perpendicular bisector of AD to meet DE at B.
 Draw perpendicular bisector of AE to meet DE at C.
 Join AB and AC.
 Thus, ABC is the required triangle.
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