Constructions Class 10 NCERT: Concepts & Solved Examples

Understanding Geometric Constructions

Geometric constructions are the process of drawing shapes, angles, or lines accurately. In pure geometry, we are limited to using just two tools: an unmarked straightedge (a ruler without markings) to draw straight lines, and a compass to draw circles or arcs of a specific radius. Every construction you perform is a logical sequence of steps based on established geometric axioms and theorems. For example, when you construct a perpendicular bisector, you are applying the properties of rhombuses and isosceles triangles. The beauty of this chapter lies in understanding the 'why' behind each step. The 'Justification' or 'Proof' of a construction is as important as the drawing itself, as it proves that your method is mathematically sound. In your exams, you'll be expected to not only draw accurately but also to provide the steps and justification for your construction.

How to Divide a Line Segment in a Given Ratio (m:n)

1. Step 1: Draw the Line Segment and a Ray — Draw the line segment AB of the given length. Then, draw a ray AX that makes an acute angle (e.g., 30° or 45°) with AB. This ray will act as our reference line.
2. Step 2: Mark Points on the Ray — The ratio is m:n. You need to locate a total of m + n points on the ray AX. Using your compass, set it to a small, convenient radius. Starting from A, draw an arc to cut AX. From this new point, draw another arc of the same radius, and repeat this process until you have marked m + n points. Label them A₁, A₂, A₃, ..., A_(m+n).
3. Step 3: Join the End Points — Join the last point, A_(m+n), to the other end of the line segment, B. This forms the line segment BA_(m+n).
4. Step 4: Draw the Parallel Line — Now, we need to draw a line through the m-th point (A_m) that is parallel to BA_(m+n). To do this, copy the angle ∠AA_(m+n)B at point A_m. This parallel line will intersect the original line segment AB at a point C.
5. Step 5: The Result and Justification — The point C is the required point that divides the line segment AB in the ratio m:n. So, AC:CB = m:n. This works because of the Basic Proportionality Theorem (Thales' Theorem). Since A_mC is parallel to A_(m+n)B in triangle ABA_(m+n), we have AA_m / A_mA_(m+n) = AC / CB. By our construction, AA_m contains 'm' equal segments and A_mA_(m+n) contains 'n' equal segments, so the ratio is m/n.

Worked Example: Constructing Tangents from an External Point

• Problem: Draw a circle of radius 3 cm. Take a point P at a distance of 7 cm from its centre. Construct the pair of tangents from point P to the circle and measure their lengths. Steps of Construction: 1. Draw the Circle: Using a ruler, measure 3 cm on your compass. Place the compass needle at a point O (the centre) and draw the circle. 2. Locate the External Point: Measure 7 cm from the centre O and mark the external point P. Join O and P with a straight line. The length of OP is 7 cm. 3. Find the Midpoint of OP: Construct the perpendicular bisector of the line segment OP. To do this, place the compass at O and set its width to more than half of OP. Draw arcs above and below OP. Repeat this from point P with the same compass width. The two points where the arcs intersect are joined to form the perpendicular bisector. Let it intersect OP at point M. M is the midpoint of OP. 4. Draw the Second Circle: With M as the centre and MO (or MP) as the radius, draw a new circle (or a semicircle). This new circle will intersect the original circle at two points. Label these points Q and R. 5. Draw the Tangents: Join P to Q and P to R. PQ and PR are the required pair of tangents to the circle. Justification and Measurement: Join OQ. Notice that ∠OQP is an angle in the semicircle of the second circle we drew. Therefore, ∠OQP = 90°. Since OQ is the radius of the original circle and the line PQ is perpendicular to it at point Q, PQ must be a tangent to the circle. Similarly, PR is also a tangent. Measurement: In the right-angled triangle ΔOQP, by Pythagoras theorem, OP² = OQ² + PQ². So, 7² = 3² + PQ². This gives PQ² = 49 - 9 = 40. Therefore, PQ = √40 ≈ 6.32 cm. You can verify this by measuring your constructed tangent PQ with a ruler.

Exam Tips and Common Mistakes to Avoid

Constructions can be a high-scoring topic if you avoid these common pitfalls:

• Faint Construction Arcs: Do not erase the arcs you use for construction (like bisecting an angle or a line). These arcs are proof of your method. Make them neat and visible, but not so dark that they obscure the main figure.
• Forgetting Justification: Many questions carry marks for both the construction and its mathematical justification. Always be prepared to write why your method works, usually by citing a theorem like BPT or properties of tangents and circles.
• Inaccurate Tools: Use a sharp pencil, a firm compass that doesn't slip, and a clear ruler. A blunt pencil can lead to thick lines and inaccuracies of several millimeters, which can cost you marks.
• Incorrect Ratios: When dividing a line segment in the ratio m:n, a common error is to mark 'm' points and then 'n' points. Remember to mark a total of 'm+n' equidistant points on the ray.
• Clear Labelling: Always label all the points (O, P, A, B, C, etc.) clearly as you go. This helps you follow your own steps and makes it easy for the examiner to evaluate your work.

Practice Questions with Solutions

• Q: Divide a line segment of length 8 cm in the ratio 3:2. A: Step 1: Draw a line segment AB = 8 cm and a ray AX making an acute angle with AB. Step 2: Since the ratio is 3:2, mark 3 + 2 = 5 points (A₁, A₂, A₃, A₄, A₅) on AX such that AA₁ = A₁A₂ = ... = A₄A₅. Step 3: Join A₅ to B. Step 4: From point A₃ (the 3rd point), draw a line parallel to A₅B by making an angle equal to ∠AA₅B at A₃. Let this line intersect AB at point C. Final answer: Point C divides AB in the ratio 3:2. AC:CB = 3:2.
• Q: Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle. A: Step 1: Draw ΔABC with sides AB = 5 cm, BC = 6 cm, and AC = 4 cm. Step 2: Below AB, draw a ray AX making an acute angle with AB. Step 3: Mark 3 points (the greater of 2 and 3 in the ratio 2/3) A₁, A₂, A₃ on AX such that AA₁ = A₁A₂ = A₂A₃. Step 4: Join A₃ to B. Draw a line through A₂ (the 2nd point) parallel to A₃B, intersecting AB at B'. Step 5: From B', draw a line parallel to BC intersecting AC at C'. Final answer: ΔAB'C' is the required similar triangle with sides 2/3 of the corresponding sides of ΔABC.
• Q: Draw a circle of radius 4 cm. From a point 9 cm away from its centre, construct the pair of tangents to the circle. A: Step 1: Draw a circle with centre O and radius 4 cm. Mark a point P such that OP = 9 cm. Step 2: Find the midpoint M of the line segment OP by constructing its perpendicular bisector. Step 3: With M as centre and MO as radius, draw a second circle. This circle will intersect the first circle at two points, T and T'. Step 4: Join PT and PT'. Final answer: PT and PT' are the required tangents to the circle from point P.
• Q: Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°. A: Step 1: The angle between the tangents is 60°. The angle between the radii at the points of contact is 180° - 60° = 120°. Step 2: Draw a circle with centre O and radius 5 cm. Draw any radius OA. Step 3: At the centre O, construct an angle ∠AOB = 120°. Step 4: At points A and B, construct perpendiculars to the radii OA and OB respectively. That is, draw 90° angles at A and B. These perpendiculars are the tangents. Step 5: Let the two tangents intersect at point P. Final answer: PA and PB are the required tangents, inclined at an angle of 60° to each other.

Frequently Asked Questions