NCERT Solutions Class 10 Maths Chapter 11 – Constructions Exercise 11.2

NCERT Solutions for Class 10 Maths Chapter 11 – Constructions Exercise 11.2 are provided to help students understand the process of constructing various geometrical figures using a compass and ruler. This exercise builds on the skills of construction and helps students learn how to construct angles, triangles, and circles with precision.

Exercise 11.2 focuses on:

• Constructing triangles when two sides and the included angle (SAS) or two angles and the included side (AAS) are given.

• Constructing the circumcenter of a triangle.

• Constructing the incenters and understanding the properties of incenter and circumcenter.

The solutions are explained in a step-by-step manner, making it easier for students to practice and grasp these geometrical constructions for their Class 10 exams.

Q1. Draw a circle of radius 6 cm. From a point 10 cm away from its center, construct the pair of tangents to the circle and measure their lengths.

Difficulty level: Medium
Solution:
Steps of construction:

1. Take a point O as the center and a radius of 6 cm. Draw a circle.

2. Take a point P such that OP = 10 cm.

3. With O and P as centers, and a radius more than half of OP, draw arcs above and below OP to intersect at X and Y.

4. Join XY to intersect OP at M.

5. With M as the center and OM as the radius, draw a circle to intersect the given circle at Q and R.

6. Join PQ and PR.
   PQ and PR are the required tangents, where PQ = PR = 8 cm.

Proof:
$\angle PQO = 90^\circ$ (Angle in a semicircle).
Since OQ is the radius of the given circle, PQ is the tangent at Q.
In right triangle $\Delta PQO$, using the Pythagoras theorem:

$$PQ = \sqrt{OP^2 - OQ^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \, \text{cm}.$$

Similarly, PR = 8 cm.

Q2. 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.

Difficulty level: Medium
Solution:
Steps of construction:

1. Take 'O' as the center and radii 4 cm and 6 cm to draw two concentric circles.

2. Take a point 'P' on the bigger circle and join OP.

3. With 'O' and 'P' as centers and a radius more than half of OP, draw arcs above and below OP to intersect at X and Y.

4. Join XY to intersect OP at M.

5. With M as the center and OM as the radius, draw a circle to cut the smaller circle at Q and R.

6. Join PQ and PR.
   PQ and PR are the required tangents, where PQ ≈ 4.5 cm.

Proof:
$\angle PQO = 90^\circ$ (Angle in a semicircle).
Thus, $PQ \perp OQ$.
In right triangle $\Delta PQO$,

$$PQ = \sqrt{OP^2 - OQ^2} = \sqrt{6^2 - 4^2} = \sqrt{36 - 16} = \sqrt{20} = 4.5 \, \text{cm} \, (\text{approx}).$$

Similarly, PR ≈ 4.5 cm.

Q3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its center. Draw tangents to the circle from these two points P and Q.

Difficulty level: Medium
Solution:
Steps of construction:

1. Draw a circle with O as the center and radius 3 cm.

2. Draw a diameter of the circle and extend it on both sides. Take points P and Q on the diameter such that OP = OQ = 7 cm.

3. Draw the perpendicular bisectors of OP and OQ to intersect OP and OQ at M and N respectively.

4. With M as the center and OM as the radius, draw a circle to cut the given circle at A and C. With N as the center and ON as the radius, draw a circle to cut the given circle at B and D.

5. Join PA, PC, QB, and QD.
   PA, PC, QB, and QD are the required tangents from P and Q respectively.

Proof:
$\angle PAO = \angle QBO = 90^\circ$ (Angle in a semicircle).
Thus, $PA \perp OA$ and $QB \perp OB$.
Since OA and OB are the radii of the given circle, PA and QB are the tangents at A and B respectively.
In right angle triangles $\Delta PAO$ and $\Delta QBO$,

$$PA = \sqrt{OP^2 - OA^2} = \sqrt{7^2 - 3^2} = \sqrt{49 - 9} = \sqrt{40} = 6.3 \, \text{cm} \, (\text{approx}).$$

Similarly, $QB = 6.3 \, \text{cm} \, (\text{approx})$.

Q4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Difficulty level: Medium
Solution:
Steps of construction:

1. With O as the center and 5 cm as the radius, draw a circle.

2. Take a point A on the circumference of the circle and join OA.

3. Draw AX perpendicular to OA.

4. Construct $\angle AOB = 120^\circ$, where B lies on the circumference.

5. Draw BY perpendicular to OB.

6. Both AX and BY intersect at P.

7. PA and PB are the required tangents inclined at 60°.

Proof:
$\angle OAP = \angle OBP = 90^\circ$ (By construction).
$\angle AOB = 120^\circ$ (By construction).
In quadrilateral OAPB,

$$\angle APB = 360^\circ - (\angle OAP + \angle OBP + \angle AOB) = 360^\circ - (90^\circ + 90^\circ + 120^\circ) = 60^\circ.$$

Thus, PA and PB are the required tangents inclined at 60°.

Q5. Draw a line segment AB of length 8 cm. Taking A as the center, draw a circle of radius 4 cm and taking B as the center, draw another circle of radius 3 cm. Construct tangent to each circle from the center of the other circle.

Difficulty level: Medium
Solution:
Steps of construction:

1. Draw AB = 8 cm. With A and B as centers, draw two circles with radii 4 cm and 3 cm, respectively.

2. Draw the perpendicular bisector of AB, intersecting AB at O.

3. With O as the center and OA as the radius, draw a circle which intersects the two circles at P, Q, R, and S.

4. Join BP, BQ, AR, and AS.

5. BP and BQ are the tangents from B to the circle with center A. AR and AS are the tangents from A to the circle with center B.

Proof:
$\angle APB = \angle AQB = 90^\circ$ (Angle in a semicircle).
Thus, $AP \perp PB$ and $AQ \perp QB$.
Therefore, BP and BQ are the tangents to the circle with center A. Similarly, AR and AS are the tangents to the circle with center B.

Q6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and $\angle B = 90^\circ$. BD is the perpendicular to AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle.

Difficulty level: Medium
Solution:
Steps of construction:

1. Draw BC = 8 cm. Draw the perpendicular at B and cut BA = 6 cm on it. Join AC to form triangle ABC.

2. Draw BD perpendicular to AC.

3. Since $\angle BDC = 90^\circ$ and the circle has to pass through B, C, and D, BC must be a diameter of this circle. So, take O as the midpoint of BC and with O as the center and OB as the radius, draw a circle which passes through B, C, and D.

4. To draw tangents from A to the circle with center O:
   a) Join OA, and draw its perpendicular bisector to intersect OA at E.
   b) With E as the center and EA as the radius, draw a circle which intersects the previous circle at B and F.
   c) Join AF.
   AB and AF are the required tangents.

Proof:
$\angle OAP = \angle OAF = 90^\circ$ (Angle in a semicircle).
Thus, $AP \perp OP$ and $AF \perp OF$.
Hence, AB and AF are the tangents from A to the circle with center O.

Q7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Difficulty level: Medium
Solution:
Steps of construction:

1. Draw any circle using a bangle.
   To find its center:
   a) Draw any two chords of the circle, say AB and CD.
   b) Draw the perpendicular bisectors of AB and CD to intersect at O.
   Now, ‘O’ is the center of this circle (since the perpendiculars drawn from the center of a circle to any chord bisect the chord and vice versa).

2. To draw the tangents from a point ‘P’ outside the circle:
   a) Take any point P outside the circle and draw the perpendicular bisector of OP, which meets OP at O’.
   b) With O’ as the center and OO’ as the radius, draw a circle which cuts the given circle at Q and R.
   c) Join PQ and PR.
   PQ and PR are the required tangents.

Proof:
$\angle OQP = \angle ORP = 90^\circ$ (Angle in a semicircle).
Thus, $OQ \perp PQ$ and $OR \perp PR$.
Hence, PQ and PR are the tangents to the given circle.

FAQs: Class 10 Maths Chapter 11 – Constructions Exercise 11.2

Q1. What is the main focus of Exercise 11.2?
Answer:
Exercise 11.2 focuses on constructing triangles using the SAS and AAS criteria, and constructing circumcenters and incenters of triangles.

Q2. How do I construct a triangle using the SAS criterion?
Answer:
To construct a triangle using SAS (Side-Angle-Side), follow these steps:

1. Draw the first side of the triangle.

2. Construct the given angle at one end of the side using a protractor.

3. Mark the second side using a compass and draw the third side to complete the triangle.

Q3. What is the difference between SAS and AAS criteria for constructing triangles?
Answer:

• SAS (Side-Angle-Side): Given two sides and the included angle, you can construct a unique triangle.

• AAS (Angle-Angle-Side): Given two angles and a non-included side, you can construct a unique triangle. The side length is not between the angles.

Q4. How do I construct the circumcenter of a triangle?
Answer:
To construct the circumcenter:

1. Construct the perpendicular bisectors of any two sides of the triangle.

2. The point where the perpendicular bisectors intersect is the circumcenter. This point is equidistant from all three vertices of the triangle.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide clear, step-by-step construction procedures for triangles, circumcenters, and incenters. By practicing these solutions, students can gain confidence in solving construction-based problems and develop the skills needed for Class 10 exams.