NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Constructions |

Exercise |
Ex 11.2 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.2

**In each of the following, give also the justification of the construction:**

**Ex 11.2 Class 10 Maths Question 1.**

**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.**

**Solution:**

**Steps of Construction:
**1. Draw a circle with center and radices = 6 cm.

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

3. Draw the perpendicular bisector of OP. Let M is the mid-point of OP.

4. With center M and radius PM = MO, draw a circle which cuts the given circle at S and T.

5. Join PS and PT.

Thus, PS and PT are the required tangents.

The length of tangents Ps= PT = 8 cm.

**Justification:**

**Ex 11.2 Class 10 Maths Question 2.**

**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.**

**Solution:**

**Steps of Construction:
**1. Draw two given concentric circles of radii 4 cm and 6 cm with common center O.

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

3. Mark the mid-point M of OP by drawing perpendicular bisector of it.

4. With center Mand radius MP = OM. draw a circle which intersects the smaller circle at S and T.

5. Join PS and PT.

Thus, PS and PT are the required tangents to the smaller circle.

On measuring PS = PT = 4.4 cm (app.)

**Justification:**

Join OS. In ∆PSO, ∠PSO = 90°

**Ex 11.2 Class 10 Maths Question 3.**

**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 center. Draw tangents to the circle from these two points P and Q.**

**Solution:**

**Steps of Construction:
**1. With center and radius 3 cm, draw a circle.

2. Produce the diameter of the circle to both the ends up to P and such that OP = OQ = 7 cm

3. Mark the mid-points M and M’ of the OP and OQ respectively

4. With centers M and M’ and radii MP and MO respectively, draw two circles.

5. Circle with center M intersects the given circle at Rand S. The circle with center M intersects the given circle at T and U.

6. Join PR, PS, QT, and QU.

Thus, we have PR and PS as a pair of tangents from P and OT and QU as another pair of tangents from Q drawn to the given circle.

**Justification:**

Join OR. Now in ∆PRO,

∠PRO = 90° [Angle in a semicircle]

Also, OR is the radius of the circle with center O.

∴ Line PR⊥ OR.

We know that a line drawn through the end of a radius and perpendicular to it is a tangent to the circle.

Hence, PR is the tangent to the point R similarly, PS, QT, and QU are the tangents at the points S, T and U respectively.

**Ex 11.2 Class 10 Maths Question 4.**

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

**Solution:**

**Steps of Construction:**

1. Draw a circle with center and radius 5 cm.

2. Draw two radial OQ and OR, such that ∠QOR = 120°

3. Draw perpendiculars to OQ and OR, which intersect each other at a point P.

Thus, PQ and PR are the required tangents which are inclined at 60°.

**Ex 11.2 Class 10 Maths Question 5.**

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

**Solution:**

**Steps of Construction:
**1. Draw a line segment AB = 8 cm.

2. With centers A and B and radial 4 cm and 3 cm respectively draw two circles.

3. Mark the mid-point M of AB,

4. With center M and radius AM = BM, draw a circle intersecting the two circles at P, Q, and R, S.

5. Join AP, AQ, BR, and BS.

Thus, AR and AS are a pair of tangents drawn from A to the given circle, and BP and BO are a pair of tangents drawn from B to the given circle.

**Ex 11.2 Class 10 Maths Question 6.**

**Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠B = 90°. BD is the perpendicular Burn B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.**

**Solution:**

**Steps of Construction:
**1. Construct a right triangle ABC, such that AB = 6 cm, BC = 8 cm and ∠ABC = 90°.

2. Mark the mid-point of BC as O.

3. Join OA.

4. With O as a center and OB radius, draw a circle which intersects the side AC of ∆ABC at D.

5. Join BD, which cuts OA at M.

6. With center M and radius MA – BM, draw a circle intersecting the first circle at B and E.

7. Join AB and AE.

Thus, AB and AE are the required tangents

dawn from point A

**Ex 11.2 Class 10 Maths Question 7.**

**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.**

**Solution:**

**Steps of Construction:
**1. Draw a circle with the bangle.

2. Take two non-parallel chords AB and CD of the circle

3. Draw perpendicular bisectors of these chords intersecting each other at O, which is the center of the circle.

4. Take a point P outside the circle.

5. Join OP.

6. Mark the mid-point M of OP.

7. With M as center and radius equal to MP = OM, draw a circle intersecting the first circle at and R.

8. Join PQ and PR.

Thus, PQ and PR are the required tangents.

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