ML Aggarwal Class 10 Maths Solutions Chapter 16 Constructions

Question 1. Use ruler and compass only for answering this question. Draw a circle of radius 4 cm. Mark the centre as O. Mark a point P outside the circle at a distance of 7 cm from the centre. Construct two tangents to the circle from the external point P. Measure and write down the length of any one tangent.
Answer: Steps of construction:
(1) Construct a circle with centre O and radius = 4 cm.
(2) Mark a point P at a distance of 7 cm from the centre. Join OP. Draw its perpendicular bisector to meet OP at M.
(3) With M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the given circle with centre O at points A and B.
(4) Join PA and PB.

Thus, PA and PB are tangents to the circle with centre O. On measuring, the length of tangent = 5.7 cm.
In simple words: First, make a circle with centre O and radius 4 cm. Then place a point P outside it, 7 cm away from O. Connect O and P, and find the middle of this line segment. Use this midpoint as a new centre and draw another circle. Where this new circle cuts the original circle, mark those two spots as A and B. Join P to both A and B - these are your tangents. When you measure, you should get about 5.7 cm.

Exam Tip: The perpendicular bisector is essential - it ensures both tangents from P have equal length. Always verify your tangents by checking that PA and PB are equal in your construction.

Question 2. Draw a line AB = 6 cm. Construct a circle with AB as diameter. Mark a point P at a distance of 5 cm from the mid-point of AB. Construct two tangents from P to the circle with AB as a diameter. Measure the length of each tangent.
Answer: Steps of construction:
(1) Draw a line segment AB = 6 cm.
(2) Mark the mid-point of AB as O. Now as radius OA = 3 cm or OB = 3 cm construct a circle.
(3) Mark a point P at a distance of 5 cm from O. Join OP. Draw its perpendicular bisector to meet OP at M.
(4) With M as centre and OM (or MP) as radius, draw a circle. Let this circle intersect the given circle at points C and D.
(5) Join PC and PD.

Thus, PC and PD are tangents to the circle with diameter AB. On measuring, the length of tangent = 4 cm.
In simple words: Draw a line 6 cm long and find its middle point. Use this middle point as the centre and draw a circle passing through both ends of the line. Place point P about 5 cm away from the middle point. Then, like in the previous question, find the midpoint between O and P, draw another circle, and find where it cuts the first circle. Connect P to those two cut points to get your tangents of about 4 cm each.

Exam Tip: Since AB is the diameter, the radius is half the length (3 cm). Verify that both tangents are equal - this confirms your construction is correct.

Question 3. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.
Answer: Steps of Construction:
(1) Draw two concentric circles of radii 4 cm and 6 cm with point O as their centre.
(2) Let P be a point on the outer circle. Join OP and draw its perpendicular bisector to meet OP at M.
(3) Taking M as centre and OM (or MP) as radius, draw a circle. Let the circle intersect the smaller circle (i.e. circle of radius 4 cm) at points A and B.
(4) Join PA and PB. Then PA and PB are the required tangents. On measuring, PA (or PB), we find that PA = 4.5 cm.

Calculation of length of PA:
Join OA
In triangle OAP, angle OAP = 90° (As angle in semicircle = 90°.)
By Pythagoras theorem we get:
\( OA^2 + PA^2 = OP^2 \)
\( \Rightarrow PA^2 = OP^2 - OA^2 \)
\( \Rightarrow PA^2 = 6^2 - 4^2 = 36 - 16 = 20 \) cm
\( \Rightarrow PA = \sqrt{20} = 4.5 \) cm

Hence, the length of tangent = 4.5 cm.
In simple words: Make two circles at the same centre - one with radius 4 cm and another with radius 6 cm. Pick a point P on the outer circle. Follow the same method as before to find the tangent. When you measure, you get 4.5 cm, which you can check using the Pythagorean theorem: the tangent squared plus the inner radius squared equals the outer radius squared (4.5² + 4² = 6²).

Exam Tip: This question combines construction with calculation - always verify both methods give the same answer. The angle in a semicircle is 90 degrees - this is the key property that makes the Pythagorean theorem work here.

Question 4. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.
Answer: Steps of Construction:
(1) Draw a circle of radius = 3 cm with centre as O.
(2) Extend the diameter (AB) of the circle on both the sides and mark points P and Q on opposite sides such that OP = OQ = 7 cm.
(3) Draw perpendicular bisector of OP and OQ such that they meet OP and OQ at M and N respectively.
(4) Construct circle with M as centre and OM or MP as radius. Let this circle touch the circle with centre O at points C and D.
(5) Construct circle with N as centre and ON or NQ as radius. Let this circle touch the circle with centre O at points E and F.
(6) Join PC, PD, QE and QF.

Hence, PC, PD, QE and QF are the tangents from point P and Q to the circle with centre as O.
In simple words: Draw a circle and extend its diameter on both sides. Mark two points P and Q on this extended line, each 7 cm away from the centre but on opposite sides. For each point, follow the tangent construction method. You will get two tangents from P and two tangents from Q, making four tangents total.

Exam Tip: Since P and Q are on opposite sides of the centre on the same line, this construction shows symmetry - PC = PD and QE = QF. Always extend the diameter correctly on both sides to avoid mistakes.

Question 1. Draw an equilateral triangle of side 4 cm. Draw its circumcircle.
Answer: Steps of construction:
(1) Draw a line segment BC = 4 cm.
(2) With centres B and C, draw two arcs of radius 4 cm which intersect each other at A.
(3) Join AB and AC. Hence, equilateral triangle ABC is formed.
(4) Draw the perpendicular bisectors of AB and BC. Let these bisectors meet at the point O.
(5) With O as center and radius equal to OA, draw a circle. The circle so drawn passes through the points A, B and C, and is the required circumcircle of triangle ABC.
In simple words: Make an equilateral triangle by drawing arcs from each endpoint of a 4 cm base line. Then find the midpoints of two sides and draw lines that cross at 90 degrees to each side. Where these lines meet is the centre of your circumcircle. Draw the circle from this centre through all three corners.

Exam Tip: For any triangle, the circumcircle centre is where the perpendicular bisectors of any two sides meet. Using two sides is sufficient - the third bisector will pass through the same point.

Question 2. Using a ruler and a pair of compasses only, construct:
(i) a triangle ABC, given AB = 4 cm, BC = 6 cm and ∠ABC = 90°.
(ii) a circle which passes through the points A, B and C and mark its centre as O.
Answer:
(i) Steps of construction:
(1) Draw a line segment BC = 6 cm.
(2) Draw the perpendicular from point B and cut AB from that perpendicular such that AB = 4 cm.
(3) Join points A, B and C.
Hence, the triangle ABC is formed.

(ii) Steps of construction:
(1) In the above triangle, draw the perpendicular bisector of AB and BC. Let these bisectors meet at the point O.
(2) With O as center and radius equal to OA, draw a circle. The circle so drawn passes through the points A, B and C, and is required circumcircle of triangle ABC.
In simple words: First, build a right-angled triangle by drawing one side, then a perpendicular side at 90 degrees, and connecting the endpoints. Then draw perpendicular bisectors of two sides - where they cross is your circle's centre. The radius from this centre to any corner stays the same for all three corners.

Exam Tip: For a right-angled triangle, the circumcircle's centre always sits at the midpoint of the hypotenuse (the side opposite the right angle). This is a quick way to verify your construction.

Question 3. Use ruler and compass, construct a triangle ABC where AB = 3 cm, BC = 4 cm and ∠ABC = 90°. Hence, construct a circumcircle circumscribing the triangle ABC. Measure and write down the radius of the circle.
Answer: Steps of construction:
(1) Draw a line segment AB = 3 cm
(2) From B draw a ray BX such that ∠XBA = 90°.
(3) From B draw an arc of 4 cm cutting XB at C.
(4) Join AC. ABC is the required triangle.
(5) Construct perpendicular bisectors of AB and BC, such that they intersect at O.
(6) With O as center and OA as radius draw a circle passing through A, B and C.
(7) Measure OA.

Hence, above is the required circumcircle of triangle ABC.
On measuring we get OA = 2.5 cm
Hence, the length of the radius of the circle = 2.5 cm
In simple words: Build a right-angled triangle with sides 3 cm and 4 cm meeting at a right angle at B. Find where the perpendicular bisectors of the two sides meet - this is your circle's centre O. Measure from O to any corner, and you should get 2.5 cm, which is half of the hypotenuse (5 cm).

Exam Tip: For a right triangle with sides 3, 4, and 5 (a Pythagorean triple), the circumradius is always 2.5 (half the hypotenuse). Check this relationship in your answer.

Question 4. Using ruler and compasses only:
(i) Construct a triangle ABC with the following data:
Base AB = 6 cm, AC = 5.2 cm and ∠CAB = 60°.
(ii) In the same diagram, draw a circle which passes through the points A, B and C, and mark its centre O.
Answer:
(i) Steps of construction:
(1) Draw a line segment AB = 6 cm.
(2) Cut an arc of 5.2 cm from A.
(3) Draw a line segment from B such that angle between the line and AB = 60°.
(4) Mark the point as C where the arc from A and line segment from B meets.
(5) Join points A, B and C. Hence, the triangle ABC is formed.

(ii) Steps of construction:
(1) In the above triangle, draw the perpendicular bisector of AB and BC. Let these bisectors meet at the point O.
(2) With O as center and radius equal to OA, draw a circle. The circle so drawn passes through the points A, B and C, and is required circumcircle of triangle ABC.
In simple words: Draw a 6 cm base line AB. Mark a point 5.2 cm away from A using an arc. Draw another line from B at a 60-degree angle. Where the two lines or arcs meet is point C. Then find the perpendicular bisectors of two sides and draw your circumcircle from their meeting point.

Exam Tip: When constructing an angle with a compass, mark the angle carefully using two arcs and their intersections. Accuracy in the angle affects the entire triangle's shape.

Question 5. Using ruler and compasses only, draw an equilateral triangle of side 5 cm and draw its inscribed circle. Measure the radius of the circle.
Answer: Steps of construction:
(1) Draw a line segment BC = 5 cm.
(2) From B and C cut an arc of 5 cm.
(3) Mark the point as A which is intersection of the two arcs.
(4) Join A, B and C. Hence, the equilateral triangle ABC is formed.
(5) Draw the (internal) bisectors of ∠B and ∠C. Let these bisectors meet at point I.
(6) From I, draw IN perpendicular to the side BC.
(7) With I as centre and radius equal to IN, draw a circle. The circle so drawn touches all the sides of the triangle ABC, and is the required incircle of triangle ABC.

On measuring IN, we get the radius of the incircle.
Hence, the radius of the incircle = 1.5 cm
In simple words: Build an equilateral triangle with sides 5 cm each. Find the angle bisectors of two corners - where they meet is the centre I of your inscribed circle. Drop a perpendicular from I to any side to find the radius. When you measure, you should get about 1.5 cm for this 5 cm triangle.

Exam Tip: The inscribed circle (incircle) touches all three sides of the triangle. For an equilateral triangle, the incircle centre coincides with the centroid. Always drop a perpendicular to a side to find the exact radius.

Question 6(i). Construct a triangle ABC with BC = 6.4 cm, CA = 5.8 cm and ∠ABC = 60°. Draw its incircle. Measure and record the radius of incircle.
Answer: Steps of construction:
(1) Draw a line segment BC = 6.4 cm.
(2) Cut an arc from C of 5.8 cm.
(3) From B construct angle 60° and extend the line and mark the point A where it meets arc from C.
(4) Join A, B and C. Hence, the triangle ABC is formed.
(5) Draw the (internal) bisectors of ∠B and ∠C. Let these bisectors meet at point I.
(6) From I, draw IN perpendicular to the side BC.
(7) With I as centre and radius equal to IN, draw a circle. The circle so drawn touches all the sides of the triangle ABC, and is the required incircle of triangle ABC.

On measuring IN, we get the radius of the incircle.
Hence, the radius of incircle is 1.6 cm
In simple words: Build a triangle with base 6.4 cm, a 60-degree angle at one corner, and a side of 5.8 cm from the other corner. Draw angle bisectors from two corners until they meet at I. Drop a straight line from I down to the base - this length is your incircle radius, about 1.6 cm.

Exam Tip: The incircle centre (called the incentre) is always where the three angle bisectors meet. You only need to draw two to find it. Ensure your perpendicular is truly at 90 degrees for an accurate radius measurement.

Question 6(ii). Construct a triangle ABC with BC = 6.5 cm, AB = 5.5 cm, AC = 5 cm. Construct the incircle of the triangle. Measure and record the radius of the incircle.
Answer: Steps of construction:
(1) Draw a line segment BC = 6.5 cm.
(2) Cut an arc from C of 5 cm and an arc of 5.5 cm from B.
(3) Mark the point as A where the arcs from B and C intersect.
(4) Join A, B and C. Hence, the triangle ABC is formed.
(5) Draw the (internal) bisectors of ∠B and ∠C. Let these bisectors meet at point I.
(6) From I, draw IN perpendicular to the side BC.
(7) With I as centre and radius equal to IN, draw a circle. The circle so drawn touches all the sides of the triangle ABC, and is the required incircle of triangle ABC.

On measuring IN, we get the radius of the incircle.
Hence, the radius of incircle is 1.5 cm
In simple words: Given three side lengths, draw arcs from two corners of your base line to place the third corner. Then draw the angle bisectors of two corners, find where they meet at I, and measure perpendicular from I to the base. This gives your incircle radius of about 1.5 cm.

Exam Tip: When all three sides are given (SSS case), mark the third point where two arcs intersect with precision. The incircle will fit snugly inside, touching all three sides if your bisectors are accurate.

Question 8. Construct a triangle ABC with the following data: AB = 5 cm, BC = 6 cm and ∠ABC = 90°. (i) Find a point P which is equidistant from B and C and is 5 cm from A. How many such points are there?
Answer: (i) Steps of construction:
1. Draw a line segment AB = 5 cm.
2. At B, construct a 90° angle. Mark BC = 6 cm along this ray.
3. Join A and C. Triangle ABC is now complete.
4. Draw the perpendicular bisector of BC. This represents all points equidistant from B and C.
5. With A as centre and radius 5 cm, draw a circle. Points where this circle meets the perpendicular bisector of BC are the required points P.
6. The circle intersects the perpendicular bisector at two locations, giving us two points that satisfy both conditions.

Therefore, there are two such points.
In simple words: The perpendicular bisector of BC holds all points equally far from B and C. The circle centred at A with radius 5 cm passes through all points 5 cm away from A. Where these two curves cross, you find the points P. In this case, they cross at two places.

Exam Tip: Always draw both the perpendicular bisector and the circle with care - the number of intersection points depends on the radius of the circle relative to the perpendicular bisector's position. Mark both points clearly and verify distances to A, B, and C.

Question. (ii) Construct the inscribed circle of △ABC drawn above.
Answer: To build the inscribed circle, follow these steps:
(1) Draw the angle bisectors of ∠B and ∠C, allowing them to meet at point O.
(2) From O, construct a perpendicular line to BC, calling the foot of this perpendicular D.
(3) Using O as the centre and OD as the radius, draw a circle that touches all three sides of the triangle.
This circle is the inscribed circle of △ABC.
In simple words: Find where the angle bisectors meet. Drop a line down to one side. Use that distance as the radius and draw a circle. It will touch all three sides.

Exam Tip: The inscribed circle (incircle) always touches all three sides of the triangle, and its centre is found at the meeting point of the angle bisectors.

Question 9. Use ruler and compasses for the following question taking a scale of 10 m = 1 cm. A park in the city is bounded by straight fences AB, BC, CD and DA. Given that AB = 50 m, BC = 63 m, ∠ABC = 75°. D is a point equidistant from the fences AB and BC. If ∠BAD = 90°, construct the outline of the park ABCD. Also locate a point P on the line BD for the flag post which is equidistant from the corners of the park A and B.
Answer: A point that lies the same distance away from two lines that cross each other can be found on one of the angle bisectors formed where those lines meet. Similarly, a point sitting the same distance from two separate points lies on the perpendicular bisector connecting them.

Given values on the scale used:
\[ \text{Scale: } 10 \text{ m} = 1 \text{ cm} \]
\[ BC = 63 \text{ m} = \frac{63}{10} = 6.3 \text{ cm} \]
\[ AB = 50 \text{ m} = \frac{50}{10} = 5 \text{ cm} \]

Steps of construction:
(1) Draw line segment BC measuring 6.3 cm.
(2) At B, construct ∠ABC = 75° and mark off AB = 5 cm along this ray.
(3) Sketch BE as the angle bisector of ∠ABC.
(4) Create ∠BAF = 90°, which intersects BE at the point D.
(5) Connect the vertices A, B, C, and D to complete the quadrilateral ABCD.
(6) Construct XY, which is the perpendicular bisector of AB. This line intersects BD at point P, which is the required location for the flag post.
In simple words: Draw the base BC. Make the angle and mark AB. Find where the angle bisector meets a line at 90 degrees to get point D. Then draw a line that cuts AB in half at right angles - where it crosses BD is your flag post P.

Exam Tip: Ensure the angle bisector of ∠ABC and the perpendicular bisector of AB are drawn with precision - these two lines meeting at the correct points (D and P) are crucial for full marks. Always verify the scale conversion before starting the construction.