SHORT ANSWER TYPE QUESTIONS
Q1. Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle.
Sol. Steps of construction:
I. Draw a circle with centre O and radius = 6.4/2 cm or 3.2 cm.
II. Mark a point P outside the circle such that OP = 6.4 cm.
III. Join OP.
IV. Bisect OP such that its mid point is at M.
V. With centre M and radius OM, draw a circle intersecting the given circle at A and B.
VI. Join PA and PB. Thus, PA and PB are the two tangents to the given circle.
Q2. Draw a circle of radius 3.4 cm. Draw two tangents to it inclined at an angle of 60° to each other:
Sol. Steps of construction:
I. Draw a circle with centre O and radius as 3.4 cm.
II. Draw two radii OA and OB such that ∠AOB = 120°.
III. Draw perpendiculars at A and B such that these perpendiculars meet at P.
Obviously, ∠APB = 60°. [Using Angle sum property of a quadrilateral]
IV. Thus, PA and PB are the required tangents to the given circle.
Q3. Draw Δ ABC in which AB = 3.8 cm, ∠ B = 60° and median AD = 3.6 cm. Draw another triangle AB’C’ similar to the first such that AB = (4/3)AB.
Sol. Steps of construction:
I. Draw AB = 3.8 cm.
II. Construct ∠ABY = 60°.
III. With centre A and radius as 3.6 cm mark a ray to intersect BY at D.
IV. With centre D and radius BD, mark an arc to intersect BY at C.
V. Join CA. Thus, ABC is a triangle.
VI. Draw a ray AX, such that ∠BAX is an acute angle.
VII. Mark 4 points X_{1}, X_{2}, X_{3} and X_{4} such that AX_{1} = X_{1}X_{2} = X_{2} X_{3} = X_{3} X_{4}.
VIII. Join X_{3}B.
IX. Through X4 draw X4B’ y X3B X. Through B’ draw B’ C’ ║ BC where C′ lies on AC (produced).
Thus, ΔC’AB is the required triangle.
Q4. Draw an equilateral triangle of height 3.6 cm. Draw another triangle similar to it such that its side is 2/3 of the side of the first.
Sol. Steps at construction:
I. Draw a line segment RS.
II. Mark a point Y on it.
III. Through Y, draw YZ ⊥ RS
IV. Mark a point A on YZ such that YA = 3.6 cm
V. At A draw ∠YAB = 30° such that the point B is on RS.
VI. With centre A and radius = AB, mark a point C on RS.
VII. Join AC.
VIII. Draw a ray BX such that ∠CBX is an acute angle.
IX. Mark three points X_{1}, X_{2}, X_{3} such that AX_{1} = X_{1}X_{2} = X_{2} X_{3}.
X. Join X_{3} and C.
XI. Through X_{2} draw X_{2}C′ ║ X_{3}C.
XII. Through C′ draw C′ A′ ║ CA.
Thus, ∆A′ BC′ is the required triangle.
Q5. Draw an isosceles ∆ ABC, in which AB = AC = 5.6 cm and ∠ ABC = 60°. Draw another ∆ AB′ C′ similar to ∆ABC such that AB′ = (2/3) AB.
Sol. Steps of Construction:
I. Draw a ray BD.
II. Through B, draw another ray BE such that ∠DBE = 60°.
III. Cut off BA = 5.6 cm.
IV. With A as centre and radius 6 cm, mark an arc intersecting BD at C.
V. Join A and C to get ΔABC.
VI. Draw a ray BX such that ∠CBX is an acute angle.
VII. Mark three point X_{1}, X_{2} and X_{3} such that BX_{1} = X_{1}X_{2} = X_{2}X_{3}.
VIII. Join X_{3} and C.
IX. Through X_{2} draw X_{2}C′ ║ X_{3}C
X. Through C′ , draw C′ A′ ║ CA
Thus, Δ A′ BC′ is the required triangle.
Q6. Construct an isosceles triangle whose base is 9 cm and altitude is 5 cm. Then construct another triangle whose sides are 3/4 of the corresponding sides of the first isosceles triangle.
Sol. Steps of construction:
I. Construct a ΔABC such that AB = AC, BC = 9 cm and altitude AD = 5 cm.
II. Through B, draw a ray BX such that ∠CBX is an acute angle.
III. Mark 4 equal points X_{1}, X_{2}, X_{3} and X_{4} on BX. such that BX_{1} = X_{1} X_{2} = X_{2} X_{3} = X_{3} X_{4}
IV. Join X_{4} and C.
V. Through X_{3}, draw X_{3}C′ ║ X_{4}C, intersecting BC in C′.
VI. Through C′ , draw C′ A′ ║ CA, intersecting AB in A′.
Thus, ΔA′BC′ is the required triangle.
Q7. Draw a line segment AB of length 7 cm. Taking A as centre draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2.5 cm. Construct tangents to each circle from the centre of the other circle.
Sol. Steps of construction:
I. Draw a line segment AB = 7 cm
II. With centre A and radius 3 cm, draw a circle.
III. With centre B and radius 2.5 cm, draw another circle.
IV. Bisect AB and let M be the mid point of AB.
V. With centre M and radius AM, draw a circle intersecting the two circles in P,Q and R,S.
VI. Join AP, AQ, BR and BS.
Thus, AP, AQ, BR and BS are required tangents.
Q8. Construct a Δ ABC in which BC = 6.5 cm, AB = 4.5 cm and ∠ABC = 60°. Construct a triangle similar to this triangle whose sides are 3/4 of the corresponding sides of the ΔABC.
Sol. Steps of construction:
I. Construct the ΔABC such that AB = 4.5 cm, ∠B = 60° and BC = 6.5 cm.
II. Construct an acute angle ∠BAX.
III. Mark 4 points X_{1}, X_{2}, X_{3} and X_{4} on AX such that AX_{1} = X_{1}X_{2} = X_{2}X_{3} =X_{3}X_{4}.
IV. Join X_{4} and B.
V. Draw X_{3} B′ ║ BC, meeting AC at C′ .
Thus, ΔC′AB′ is the required Δ.
Q9. Draw a right triangle in which sides (other than hypotenuse) are of lenghts 8 cm and 6 cm. Then construct another triangle whose sides are 3/4 times the corresponding sides of the first triangle.
Sol. Steps of construction:
I. Draw a ΔABC such that AB = 8 cm, ∠B = 90° and BC = 6 cm.
II. Construct an acute angle ∠BAX.
III. Mark 4 points X_{1}, X_{2}, X_{3} and X_{4} on AX such that AX_{1} = X_{1}X_{2}, = X_{2}X_{3} = X_{3}X_{4}.
IV. Join X_{4} and B.
V. Draw X_{3}B′ ║ X4B.
VI. Draw B′ C′ ║ BC.
Thus, ΔAB′C′ is the required rt Δ.
Q10. Construct a ΔABC in which BC = 5 cm, CA = 6 cm and AB = 7 cm. Construct a ΔA′ BC′ similar to ΔABC, each of whose sides are 7/5 times the corresponding sides of ΔABC.
Sol. Steps of construction:
I. Construct ΔABC such that: BC = 5 cm, CA = 6 cm and AB = 7 cm.
II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark 7 points X_{1}, X_{2}, .............. X_{7} such that: BX_{1} = X_{1}X_{2}, = X_{2}X_{3} = X_{3}X_{4} = X_{4}X_{5} = X_{5}X_{6} = X_{6}X_{7}
IV. Join X_{7} and C.
V. Draw a line through X_{5} parallel to X_{7}C to meet BC extended at C′ .
VI. Through C′ , draw a line parallel to CA to meet BA extended at A′ .
Thus, ΔA′BC′ is the required triangle.
Q11. Construct a triangle with sides 4 cm, 5 cm and 7 cm. Then construct a triangle similar to it whose sides are 2/3 of the corresponding sides of the given triangle.
Sol. Steps of construction:
I. Construct the ΔABC such that BC = 7 cm, CA = 5 cm and BA = 4 cm.
II. Draw a ray BX such that ∠CBX is an acute angle.
III. Mark three points X_{1}, X_{2} and X_{3} on BX such that: BX_{1} = X_{1} X_{2} = X_{2} X_{3}
IV. Join X_{3} and C.
V. Draw X_{2}C′ ║ X_{3}C.
VI. Draw C′ A′ ║ CA
Thus, ΔA′ BC′ is the required triangle.
Q12. Construct a ΔABC in which AB = 6.5 cm, ∠ B = 60° and BC = 5.5 cm. Also construct a triangle AB′C′ similar to ΔABC whose each side is 3/2 times the corresponding side of the ΔABC.
Sol. Steps of construction:
I. Construct a ΔABC such that AB = 6.5 cm, ∠B = 60° and BC = 5.5 cm.
II. Draw a ray AX making an acute angle ∠BAX.
III. Mark three points X_{1}, X_{2}, X_{3} on the ray AX such that AX_{1} = X_{1} X_{2} = X_{2} X_{3}
IV. Join X_{2} and B.
V. Draw X_{3}B′ ║ X_{2}B such that B′ is a point on extended AB.
VI. Join B′ C′ y BC such that C′ is a point on AC (extended).
Thus, ΔC′ AB′ is the required triangle.
Q13. Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Construct ΔAB′ C′ similar to ΔABC such that sides of ΔAB′C′ are 3/4 of the corresponding sides of ΔABC.
Sol. Steps of construction:
I. Construct the given ΔABC.
II. Draw a ray AX such that ∠BAC is an acute angle.
III. Mark 4 points X_{1}, X_{2}, X_{3} and X_{4} on such that AX_{1} = X_{1} X_{2} = X_{2} X_{3} = X_{3} X_{4}.
IV. Join X_{4} B.
V. Draw X_{3}B′ ║ X_{4}B
VI. Through B′ draw B′ C′ ║ BC.
Thus, ΔB′ AC′ is the required triangle.
Q14. Draw a circle of radius 3 cm. From a point P, 6 cm away from its centre, construct a pair of tangents to the circle. Measure the lengths of the tangents.
Sol. Steps of construction:
I. Draw the given circle such that its centre is at O and radius = 3 cm.
II. Mark a point P such that OP = 6 cm.
III. Bisect OP. Let M be the mid point of OP.
IV. Taking M as centre and OM as radius draw a circle intersecting the given circle at A and B.
V. Join PA and PB.
Thus, PA and PB are the required tangents to the given circle.
Q15. Construct a triangle whose perimeter is 13.5 cm and the ratio of the three sides is 2 : 3 : 4.
Sol. Steps of construction:
I. Draw a line PQ = 13.5 cm
II. At P, draw a ray PR making a convenient acute angle –QPR with PQ.
III. On PR mark (2 + 3 + 4), 9 points at equal distances.
IV. Join Q and the mark 9.
V. Through the points 2 and 5 draw lines 2A and 5B parallel to 9Q. Let these lines meet PQ at A and B respectively.
VI. With A as centre and radius = AP, draw an arc.
VII. With B as centre and radius = BQ, draw another arc which intersects the arc of step
VI at C.
VIII. Join CA and CB.
ABC is the required triangle.
1. What are the basic tools required in a construction class? 
2. How do you calculate the area of a rectangular room for construction purposes? 
3. What is the purpose of a foundation in construction? 
4. How do you read a construction drawing? 
5. What are the different types of construction materials used in building construction? 

Explore Courses for Class 10 exam
