Constructions — Previous Year Questions (Class 10 Mathematics)
Geometric constructions develop spatial reasoning and understanding of mathematical properties. This chapter covers construction of tangents, division of l
TL;DR: Geometric constructions develop spatial reasoning and understanding of mathematical properties. This chapter covers construction of tangents, division…
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Geometric constructions develop spatial reasoning and understanding of mathematical properties. This chapter covers construction of tangents, division of l
Constructions — Previous Year Questions with Solutions
Q (2023, 4 marks): Construct a triangle ABC with AB = 5 cm, BC = 6 cm, CA = 7 cm. Then construct a triangle similar to triangle ABC with scale factor 3/2.
Answer: Steps to construct similar triangle with scale factor 3/2:
1. Construct triangle ABC with sides AB=5 cm, BC=6 cm, CA=7 cm
2. Extend side BC beyond C
3. Divide BC in ratio 2:1 internally using compass (scale factor 3/2 means new:original = 3:2)
4. Mark point C' on extended BC such that BC' = (3/2)*BC = 9 cm
5. Draw line from A parallel to CC' meeting extended BC at C'
6. Construct triangle AB'C' similar to ABC with sides 7.5 cm, 9 cm, 10.5 cm
7. The triangle AB'C' is the required triangle
Note: Scale factor 3/2 means each side of new triangle is 1.5 times the original
Q (2022, 4 marks): Draw a circle with center O and radius 4 cm. Construct tangents to the circle from an external point P at distance 6 cm from the center.
Answer: Steps to draw tangents from external point:
1. Draw circle with center O and radius 4 cm
2. Mark point P at distance 6 cm from O
3. Join OP
4. Find midpoint M of OP using perpendicular bisector
5. With M as center and MO (= 3 cm) as radius, draw a circle
6. This circle intersects the original circle at points T1 and T2
7. Join PT1 and PT2 - these are the required tangents
Verification:
- OT1 perpendicular to PT1 (radius to tangent)
- OT2 perpendicular to PT2 (radius to tangent)
- Both tangents are equal in length: PT1 = PT2 = sqrt(36-16) = sqrt(20) = 2sqrt(5) cm
Q (2023, 3 marks): Divide a line segment AB of length 8 cm internally in the ratio 3:2.
Answer: Steps to divide AB in ratio 3:2:
1. Draw line segment AB = 8 cm
2. At point A, draw any ray AC making acute angle with AB
3. Mark 5 points (3+2=5) A1, A2, A3, A4, A5 on AC at equal intervals using compass
4. Join A5B
5. At point A3, draw line parallel to A5B (using compass to match angles)
6. This parallel line intersects AB at point P
7. Point P divides AB in ratio AP:PB = 3:2
Verification:
AP = (3/5)*8 = 4.8 cm
PB = (2/5)*8 = 3.2 cm
Ratio AP:PB = 4.8:3.2 = 3:2
Q (2021, 4 marks): Construct a right-angled triangle with hypotenuse 10 cm and one side 6 cm. Then construct a triangle similar to it with scale factor 2/3.
Answer: Steps to construct similar right triangle:
1. First construct right triangle:
- Draw line segment BC = 10 cm (hypotenuse)
- Find midpoint O of BC
- Draw semicircle with BC as diameter
- From B, mark point A on semicircle at distance 6 cm (using compass)
- Join AB and AC to complete right triangle ABC
2. Construct similar triangle with scale factor 2/3:
- This means new triangle sides are 2/3 of original
- New hypotenuse = (2/3)*10 = 20/3 cm
- New side = (2/3)*6 = 4 cm
- Repeat construction with new measurements
Alternate method:
- Extend BC beyond C
- Mark C' on BC extended such that BC':BC = 2:3
- Draw line from A parallel to CC' meeting BC at C'
- Triangle AB'C' is the required similar triangle
Q (2022, 3 marks): Construct a tangent to a circle at a given point on the circle.
Answer: Steps to construct tangent at a point on circle:
1. Draw circle with center O
2. Mark point P on the circle
3. Join OP (radius to the point)
4. At point P, construct a line perpendicular to OP using:
- Place compass at P with suitable radius
- Mark arcs on both sides of P on OP
- From these arc endpoints, draw arcs above and below
- Join intersection points - this is the perpendicular
5. The perpendicular line through P is the required tangent
Key Property:
- A tangent is always perpendicular to the radius at the point of contact
Q (2021, 3 marks): Draw a triangle ABC with AB = 4 cm, BC = 5 cm, AC = 6 cm. Construct a triangle similar to ABC with scale factor 4/3.
Answer: Steps to construct similar triangle:
1. Construct triangle ABC with sides AB=4 cm, BC=5 cm, AC=6 cm
2. Extend side BC beyond C
3. Scale factor 4/3 means new:original = 4:3
4. Divide BC in ratio 3:1 to locate point on extended line at distance (4/3)*BC from B
5. Mark point C' on BC extended such that BC' = (4/3)*5 = 20/3 cm
6. Draw line parallel to AC through C'
7. This line meets extended AB at point A'
8. Triangle A'BC' is the required triangle with scale factor 4/3
New sides:
- A'B = (4/3)*4 = 16/3 cm
- B'C' = (4/3)*5 = 20/3 cm
- A'C' = (4/3)*6 = 8 cm
Frequently Asked Questions
What is the difference between construction and drawing?
Construction uses only compass and unmarked straightedge to create exact geometric figures, while drawing may use other tools like protractors and scales.
Can we construct a line parallel to a given line using only compass and straightedge?
Yes, using the properties of alternate angles. We can construct a line through a given point parallel to another line using compass and straightedge.
More Class 10 Mathematics PYQs
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