CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1
Exercise 11.1

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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

Answer

Steps of construction:
i. At first, we will draw a line segment

A B = 7.6 c m

$\mathrm{AB}=7.6 \mathrm{~cm}$ .

ii. Draw a ray AX such that it makes an acute angle with AB .

iii. Now, locate 13 points

(5 + 8) A_{1}, A_{2}, A_{3}, \dots \dots . A_{13}

$(5+8) \mathrm{A}_{1}, \mathrm{~A}_{2}, \mathrm{~A}_{3}, \ldots \ldots . \mathrm{A}_{13}$ on AX so that;

{A A}_{1} = A_{1} A_{2} = A_{2} A_{3} = A_{3} A_{4} = A_{4} A_{5} = A_{5} A_{6} = A_{6} A_{7} = A_{7} A_{8} = A_{8} A_{9} =

$\mathrm{AA}_{1}=\mathrm{A}_{1}\mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}=\mathrm{A}_{3} \mathrm{~A}_{4}=\mathrm{A}_{4} \mathrm{~A}_{5}=\mathrm{A}_{5} \mathrm{~A}_{6}=\mathrm{A}_{6} \mathrm{~A}_{7}=\mathrm{A}_{7} \mathrm{~A}_{8}=\mathrm{A}_{8} \mathrm{~A}_{9}=$

A_{9} A_{10} = A_{10} A_{11} = A_{11} A_{12} = A_{12} A_{13}

$\mathrm{A}_{9} \mathrm{~A}_{10}=\mathrm{A}_{10} \mathrm{~A}_{11}=\mathrm{A}_{11} \mathrm{~A}_{12}=\mathrm{A}_{12} \mathrm{~A}_{13}$

iv. Join

A_{13}

$\mathrm{A}_{13}$ to B
v. Draw a line

A_{5} C ∥ A_{13} B

$A_{5} C \| A_{13} B$ ; which intersects

$A B$ at point

$C$ and passes Through the point

$A_{5}$ .
Now we have,

$\mathrm{AC} : \mathrm{CB}=5: 8$
On measuring with the scale.

$\mathrm{AC}=2.92 \mathrm{~cm}$
And

$\mathrm{CB}=4.68 \mathrm{~cm}$
Justification:
Since

$\angle \mathrm{AA}_{13} \mathrm{~B}=\angle \mathrm{AA}_{5} \mathrm{C}$
With

$A X$ is a transversal, corresponding angles are equal, Hence

$\mathrm{A}_{13} \mathrm{~B}$ and

$\mathrm{A}_{5} \mathrm{C}$ are parallel.
So by basic proportionality theorem,

$\frac{A A_{5}}{A_{5} A_{13}}=\frac{A C}{C B}$
By construction

$\frac{A A_{5}}{A_{5} A_{13}}=\frac{5}{8}$
So,

$\frac{A C}{C B}=\frac{5}{8}$
Hence C divides AB in the ratio 5:8.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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2. Construct a triangle of sides

$4 \mathrm{~cm}, 5 \mathrm{~cm}$ and

$6\mathrm{~cm}$ and then a triangle similar to it whose sides are

$\frac{2}{3}$ of the corresponding sides of the first triangle.

Answer

Steps of construction:
i. Now in order to make a triangle, draw a line segment

$\mathrm{AB}=4 \mathrm{~cm}$ .

ii. Then, draw an arc at

$5$ cm from point A. From point B, draw an arc at

$6$ cm in such a way that it intersects the previous arc. Join the point of intersection from points A and B. This gives the required triangle ABC.

iii. Now, divide the base in the ratio

$2: 3$ . Draw a ray

$A X$ which is at an acute angle from

$A B$ . Then, plot three points on

$A X$ such that;

$\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}$ . Join

$\mathrm{A}_{3}$ to B .
iv. Draw a line from point

$A_{2}$ parallel to

$A_{3} B$ and intersects

$A B$ at point

$B^{\prime}$ .
v. Draw a line from point

$B^{\prime}$ that is parallel to

$BC$ intersecting

$AC$ at point

$C^{\prime}$ .

Hence, triangle

$\mathrm{AB}^{\prime} \mathrm{C}^{\prime}$ is the required triangle.
justification:

Since the scalar factor is

$\frac{ 2 }{ 3 }$
We need to prove:

$\frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^\prime C^{\prime}}{B C}=\frac{2}{3}$
By construction,

$\frac{A B^{\prime}}{A B}=\frac{A A_{2}}{A A_{3}}=\frac{2}{3} \ldots(1)$
Also,

$B^{\prime} C^{\prime}$ is parallel to

$BC$ .
So, both will make same angle with

$AB$ .

$\therefore \angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}=\angle \mathrm{ABC}$ (corresponding angles)
In

$\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}$ and

$\Delta \mathrm{ABC}$

$\angle \mathrm{A}=\angle \mathrm{A}(\text { common })$

$\angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}=\angle \mathrm{ABC}$
So, by

$AA$ similarity

$\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime} \sim \Delta \mathrm{ABC}$
As we know if two triangles are similar the ratio of their corresponding sides are also equal.
So,

$\frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^{\prime} C^{\prime}}{B C}$
From (1),

$\frac{A B^{\prime}}{A B}=\frac{A C^{\prime}}{A C}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{2}{3}$
Hence proved.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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3. Construct a triangle with sides

$5 \mathrm{~cm}, 6 \mathrm{~cm}$ and 7 cm and then another triangle whose sides are

$\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer

1. Now in order to make a triangle, draw a line segment

$A B$

$=5 \mathrm{~cm}$ . From point A, draw an arc at 6 cm. Draw an arc at 7 cm from point

$B$ intersecting the previous arc.

2. Join the point of intersection from A and B .

Hence, this gives the required triangle ABC
3. Dividing the base, draw a ray

$AX$ which is at an acute angle from AB

4. Plot seven points on

$A X$ such that:

$\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}=\mathrm{A}_{3} \mathrm{~A}_{4}=\mathrm{A}_{4} \mathrm{~A}_{5}=\mathrm{A}_{5} \mathrm{~A}_{6}=\mathrm{A}_{6} \mathrm{~A}_{7}$

5. Join

$\mathrm{A}_{5}$ to B .
6. Draw a line from point

$\mathrm{A}_{7}$ that is parallel to

$\mathrm{A}_{5} \mathrm{B}$ and joins

$\mathrm{AB}^{\prime}(\mathrm{AB}$ extended to

$\left.\mathrm{AB}^{\prime}\right)$ .
7. Draw a line

$\mathrm{B}^{\prime} \mathrm{C}^{\prime} \| \mathrm{BC}$ .

Hence, triangle

$\mathrm{AB}^{\prime} \mathrm{C}^{\prime}$ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are

$1 \frac{1}{2}$ times the corresponding sides of the isosceles triangle.

Answer

Steps of construction:
i. Now in order to make a triangle, draw a line segment

$\mathrm{AB}=8 \mathrm{~cm}$ .

ii. Draw two arcs intersecting at 4 cm distance from points A and B; on either side of AB.
Join these arcs to get perpendicular bisector CD of AB. (Since, altitude is the perpendicular bisector of base of isosceles triangle).

iii. Join points

$A$ and

$B$ to

$C$ in order to get the triangle

$A B C$ .

iv. Now, draw a ray AX which is at an acute angle from point A .
As

$1 \frac{1}{2}=\frac{3}{2}$
And 3 is greater between 3 and 2, So Plot 3 points on

$AX$ such that:

$\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{~A}_{2}=\mathrm{A}_{2} \mathrm{~A}_{3}$ .

v. As 2 is smaller between 2 and 3 . Join

$\mathrm{A}_{2}$ to point B . Draw a line from

$\mathrm{A}_{3}$ which is parallel to

$\mathrm{A}_{2} \mathrm{~B}$ meeting the extension of

$A B$ at

$B^{\prime}$ .

vi. Draw

$\mathrm{B}^{\prime} \mathrm{C}^{\prime} \| \mathrm{BC}$ . Then, draw

$\mathrm{A}^{\prime} \mathrm{C}^{\prime} \| \mathrm{AC}$ .

Triangle

$\mathrm{AB}^{\prime} \mathrm{C}^{\prime}$ is the required triangle.
Justification:
We need to prove,

$\frac{A C \prime}{A C}=\frac{A B \prime}{A B}=\frac{C \prime B \prime}{C B}=\frac{3}{2}$
By construction

$\frac{A B^{\prime}}{A B}=\frac{A A_{3}}{A A_{2}}=\frac{3}{2}$

$\quad\dots.$ (1)

$s C'B'\| CB$
They will maker equal angles with line

$\mathrm{AB} \angle \mathrm{ACB}=\angle \mathrm{AC}^{\prime} \mathrm{B}^{\prime}$

$\quad \dots.$ (corresponding angles)
In

$\triangle \mathrm{ACB}$ and

$\triangle \mathrm{AC}^{\prime} \mathrm{B}^{\prime}$

$\angle \mathrm{A}=\angle \mathrm{A}$ (common)

$\angle \mathrm{ACB}=\angle \mathrm{AC} \mathrm{B}^{\prime}$ (corresponding angles)
So

$\triangle \mathrm{ACB} \sim \triangle \mathrm{AC}^{\prime} \mathrm{B}^{\prime}$
As corresponding sides of similar triangles are in ratio, Hence,

$\frac{A C \prime}{A C}=$

$\frac{A C^{\prime}}{A C}=\frac{A B \prime}{A B}=\frac{C \prime B \prime}{C B}=\frac{3}{2}$

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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5. Draw a triangle

$A B C$ with side

$B C=6 \mathrm{~cm}, A B=5 \mathrm{~cm}$ and

$\angle A B C=60^{\circ}$ . Then construct a triangle whose sides are

$\frac{3}{4}$ of the corresponding sides of the triangle ABC .

Answer

Given in

$\Delta \mathrm{ABC}$ ,
Length of side

$\mathrm{BC}=6 \mathrm{~cm}$ .
Length of side

$\mathrm{AB}=5 \mathrm{~cm}$ .

$\angle \mathrm{ABC}=60^{\circ}$ .
Steps of Construction:
1. Draw a line segment BC of length 6 cm .

2. With B as center, draw a line which makes an angle of

$60^{\circ}$ with BC.
Construction of

$60^{\circ}$ angle at B:
a. With B as centre and with some convenient radius draw an arc which cuts the line BC at
D.

b. With D as radius and with same radius (in step a), draw another arc which cuts the previous arc at E.

c. Join BE. The line BE makes an angle

$60^{\circ}$ with BC.

3. Again with

$B$ as centre and with radius of 5 cm , draw an arc which intersects the line BE at point A .

4. Join AC. This is the required triangle.

5. Now, from B , draw a ray BX which makes an acute angle on the opposite side of the vertex

$A$ .

6. With

$B$ as center, mark four points

$B_{1}, B_{2}, B_{3}$ and

$B_{4}$ on

$B X$ such that they are equidistant. i.e.

$\mathrm{BB}_{1}=\mathrm{B}_{1} \mathrm{~B}_{2}=\mathrm{B}_{2} \mathrm{~B}_{3}=\mathrm{B}_{3} \mathrm{~B}_{4}$ .

7. Join

$B_{4} C$ and then draw a line from

$B_{3}$ parallel to

$B_{4} C$ which meets the line BC at P .

8. From

$P$ , draw a line parallel to

$A C$ and meets the line

$A B$ at

$Q$ . Thus

$\triangle \mathrm{BPQ}$ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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6. Draw a triangle

$ABC$ with side

$\mathrm{BC}=7 \mathrm{~cm}, \angle \mathrm{B}=45^{\circ}, \angle \mathrm{A}=$

$105^{\circ}$ . Then, construct a triangle whose sides are

$\frac{4}{3}$ times the corresponding sides of

$\triangle \mathrm{ABC}$ .

Answer

Steps of construction:
1. Draw a line segment

$\mathrm{BC}=7 \mathrm{~cm}$ .

2. Draw

$\angle \mathrm{ABC}=45^{\circ}$ and

$\angle \mathrm{ACB}=30^{\circ}$ i.e.

$\angle \mathrm{BAC}=105^{\circ}$ .

We obtain

$\triangle \mathrm{ABC}$ .
3. Draw a ray

$BX$ making an acute angle with BC. Mark four points

$\mathrm{B}_{1}, \mathrm{~B}_{2}, \mathrm{~B}_{3}, \mathrm{~B}_{4}$ at equal distances.

4. Through

$\mathrm{B}_{3}$ draw

$\mathrm{B}_{3} \mathrm{C}$ and through

$\mathrm{B}_{4}$ draw

$\mathrm{B}_{4} \mathrm{C}_{1}$ parallel to

$\mathrm{B}_{3} \mathrm{C}$ . Then draw

$\mathrm{A}_{1} \mathrm{C}_{1}$ parallel to AC.

$\therefore \mathrm{A}_{1} \mathrm{BC}_{1}$ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm . Then construct another triangle whose sides are

$\frac{5}{3}$ times the corresponding sides of the given triangle.

Answer

Steps of construction:
1. Now in order to make a triangle, draw a line segment

$\mathrm{AB}=3$ cm .

2. Make a right angle at point A and draw

$\mathrm{AC}=4 \mathrm{~cm}$ from this point.

3. Join points

$A$ and

$B$ to get the right triangle

$A B C$ .

4. Now, Dividing the base, draw a ray

$AX$ such at it forms an acute angle from

$A B$ .

5. Then, plot 5 points on

$A X$ such that:

$\mathrm{AG}=\mathrm{GH}=\mathrm{HI}=\mathrm{IJ}=\mathrm{JK} .$

6. Join I to point line

$A B$ and Draw a line from

$K$ which is parallel to

$IB$ such that it meets

$AB$ at point

$M$ .
7. Draw

$\mathrm{MN} \| \mathrm{CB}$ .

This is the required construction, thus forming

$AMN$ which have all the sides

$\frac{ 5 }{ 3 }$ times the sides of

$ABC$
Triangle

$AMN$ is the required triangle.

CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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CBSE Solutions for Class 10 Maths Chapter 11: Constructions || CBSE Class 10 Maths Chapter 11 Constructions solutions Ex 11.1

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