Constructions Exercise 17.3
Question: 1
Draw △ABC in which AB = 3 cm, BC = 5 cm and ∠Q = 70°.
Solution:

Steps of construction:
-
Draw a line segment AB of length 3 cm.
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Draw ∠XBA=70°.
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Cut an arc on BX at a distance of 5 cm at C.
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Join AC to get the required triangle.
Question: 2
Draw △ABC in which ∠A=70°., AB = 4 cm and AC= 6 cm. Measure BC.
Solution:

Steps of construction:
-
Draw a line segment AC of length 6 cm.
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Draw ∠XAC=70°.
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Cut an arc on AX at a distance of 4 cm at B.
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Join BC to get the desired triangle.
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We see that BC = 6 cm.
Question: 3
Draw an isosceles triangle in which each of the equal sides is of length 3 cm and the angle between them is 45°.
Solution:

Steps of construction:
Draw a line segment PQ of length 3 cm.
Draw ∠QPX=45°.
Cut an arc on PX at a distance of 3 cm at R.
Join QR to get the required triangle.
Question: 4
Draw △ABC in which ∠A = 120°, AB = AC = 3 cm. Measure ∠B and ∠C.
Solution:

Steps of construction:
-
Draw a line segment AC of length 3 cm.
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Draw ∠XAC = 120°.
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Cut an arc on AX at a distance of 3 cm at B.
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Join BC to get the required triangle.
By measuring, we get ∠B = ∠C = 30°.
Question: 5
Draw △ABC in which ∠C = 90° and AC = BC = 4 cm.
Solution:

Steps of construction:
-
Draw a line segment BC of length 4 cm.
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At C, draw ∠BCY=90°.
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Cut an arc on CY at a distance of 4 cm at A.
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Join AB. ABC is the required triangle.
Question: 6
Draw a triangle ABC in which BC = 4 cm, AB = 3 cm and ∠B = 45°. Also, draw a perpendicular from A on BC.
Solution:

Steps of construction:
-
Draw a line segment AB of length 3 cm.
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Draw an angle of 45° and cut an arc at this angle at a radius of 4 cm at C.
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Join AC to get the required triangle.
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With A as centre, draw intersecting arcs at M and N.
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With centre M and radius more than half of MN, cut an arc on the opposite side of A.
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With N as centre and radius the same as in the previous step, cut an arc intersecting the previous arc at E.
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Join AE, it meets BC at D, then AE is the required perpendicular.
Question: 7
Draw a triangle ABC with AB = 3 cm, BC = 4 cm and ∠B = 60°. Also, draw the bisector of angles C and A of the triangle, meeting in a point O. Measure ∠COA.
Solution:

Steps of construction:
Draw a line segment BC = 4 cm.
Draw ∠CBX = 60°.
Draw an arc on BX at a radius of 3 cm cutting BX at A.
Join AC to get the required triangle.
Angle bisector for angle A:
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With A as centre, cut arcs of the same radius cutting AB and AC at P and Q, respectively.
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From P and Q cut arcs of same radius intersecting at R.
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Join AR to get the angle bisector of angle A.
Angle bisector for angle C:
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With A as centre, cut arcs of the same radius cutting CB and CA at M and N, respectively.
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From M and N, cut arcs of the same radius intersecting at T
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Join CT to get the angle bisector of angle C.
Mark the point of intersection of CT and AR as 0.
Angle ∠COA = 120°.
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