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**Exercise 11.1 : ** Solutions of Questions on Page Number : **219**

<< Previous Chapter 10 : Circles
Next Chapter 12 : Areas Related to Circles >>
###
**Exercise 11.2 : ** Solutions of Questions on Page Number : **221**

<< Previous Chapter 10 : Circles
Next Chapter 12 : Areas Related to Circles >>
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Q1 :
**
**

Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give the justification of the construction.

**Answer :**

A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

**Step 1** Draw line segment AB of 7.6 cm and draw a ray AX
making an acute angle with line segment AB.

**Step 2** Locate 13 (= 5 + 8) points, A_{1},
A_{2}, A_{3}, A_{4}
……..
A_{13}, on AX such that AA_{1} =
A_{1}A_{2} = A_{2}A_{3} and so
on.

**Step 3** Join BA_{13}.

**Step 4** Through the point A_{5}, draw a line
parallel to BA_{13} (by making an angle equal to
∠AA_{13}B) at A_{5}
intersecting AB at point C.

C is the point dividing line segment AB of 7.6 cm in the required ratio of 5:8.

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 cm respectively.

**Justification**

The construction can be justified by proving that

By construction, we have A_{5}C || A_{13}B. By
applying Basic proportionality theorem for the triangle
AA_{13}B, we obtain

… (1)

From the figure, it can be observed that AA_{5} and
A_{5}A_{13} contain 5 and 8 equal divisions of
line segments respectively.

… (2)

On comparing equations (1) and (2), we obtain

This justifies the construction.

Answer needs Correction? Click Here

Q2 :
**
**

Construct a triangle of sides 4 cm, 5cm and 6cm and then a triangle similar to it whose sides areof the corresponding sides of the first triangle.

Give the justification of the construction.

**Answer :**

**Step 1**

Draw a line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ΔABC is the required triangle.

**Step 2**

Draw a ray AX making an acute angle with line AB on the opposite side of vertex C.

**Step 3**

Locate 3 points A_{1}, A_{2}, A_{3} (as 3
is greater between 2 and 3) on line AX such that AA_{1} =
A_{1}A_{2} = A_{2}A_{3}.

**Step 4**

Join BA_{3} and draw a line through A_{2}
parallel to BA_{3} to intersect AB at point B'.

**Step 5**

Draw a line through B' parallel to the line BC to intersect AC at C'.

ΔAB'C' is the required triangle.

**Justification**

The construction can be justified by proving that

By construction, we have B'C' || BC

∴ ∠A = ∠ABC (Corresponding angles)

In ΔAB'C' and ΔABC,

∠ = ∠ABC (Proved above)

∠ = ∠BAC (Common)

∴ Δ Ã¢Ë†Â¼ ΔABC (AA similarity criterion)

… (1)

In ΔAA_{2}B' and
ΔAA_{3}B,

∠A_{2}AB' =
∠A_{3}AB (Common)

∠AA_{2}B' =
∠AA_{3}B (Corresponding
angles)

∴
ΔAA_{2}B'
Ã¢Ë†Â¼
ΔAA_{3}B (AA similarity
criterion)

From equations (1) and (2), we obtain

This justifies the construction.

Answer needs Correction? Click Here

Q3 :
**
**

Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are of the corresponding sides of the first triangle.

Give the justification of the construction.

**Answer :**

Q4 :
**
**

Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are times the corresponding sides of the isosceles triangle.

Give the justification of the construction.

**Answer :**

Q5 :
**
**

Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides areof the corresponding sides of the triangle ABC.

Give the justification of the construction.

**Answer :**

Q6 :
**
**

Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are times the corresponding side of ΔABC. Give the justification of the construction.

**Answer :**

Q7 :
**
**

Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. the construct another triangle whose sides are times the corresponding sides of the given triangle. Give the justification of the construction.

**Answer :**

Q1 :
**
**

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

**Answer :**

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. Give the justification of the construction.

**Answer :**

Q3 :
**
**

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. Give the justification of the construction.

**Answer :**

Q4 :
**
**

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

**Answer :**

Q5 :
**
**

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

**Answer :**

Q6 :
**
**

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 circles. Give the justification of the construction.

**Answer :**

Q7 :
**
**

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

**Answer :**

Maths : CBSE ** NCERT ** Exercise Solutions for Class 10th for ** Constructions ** ( Exercise 11.1, 11.2 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

- 10th Maths Paper Solutions Set 2 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 3 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE Abroad Previous Year 2015
- 10th Maths Paper Solutions Set 1 : CBSE All India Previous Year 2015

- Chapter 3 - Pair of Linear Equations in Two Variables Class 10
- Chapter 2 - Polynomials Class 10
- Chapter 1 - Real Numbers Class 10
- Chapter 6 - Triangles Class 10
- Chapter 8 - Introduction to Trigonometry Class 10
- Chapter 13 - Surface Areas and Volumes Class 10
- Chapter 4 - Quadratic Equations Class 10
- Chapter 14 - Statistics Class 10
- Chapter 9 - Some Applications of Trigonometry Class 10
- Chapter 5 - Arithmetic Progressions Class 10

Exercise 11.1 |

Question 1 |

Question 2 |

Question 3 |

Question 4 |

Question 5 |

Question 6 |

Question 7 |

Exercise 11.2 |

Question 1 |

Question 2 |

Question 3 |

Question 4 |

Question 5 |

Question 6 |

Question 7 |