# NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Ex 11.1) Exercise 11.1 Free PDF download of NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1 (Ex 11.1) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 10 Maths NCERT Solutions Chapter 11 Constructions Exercise 11.1 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise CBSE Solutions in your emails. Subjects like Science, Maths, English will become easy to study if you have access to Class 10 Science NCERT Solutions, Maths solutions, and solutions of other subjects that are available on Vedantu only.

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Exercise 11.1

1. 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.

Ans: To divide a line segment of length $7.6$ cm in the ratio of $5:8$ follow the steps below:

Step 1: Draw line segment $AB$ of $7.6$

Step 2: Draw a ray $AX$ making an acute angle with line segment $AB$.

Step 3: Mark $13$ points ${{A}_{1}},{{A}_{2}},...,{{A}_{13}}$ on $AX$such that the length of each segment is equal.

Step 4: Join $B$ with ${{A}_{13}}$.

Step 5: Draw a line parallel to $B{{A}_{13}}$ passing through ${{A}_{5}}$ and intersecting $AB$ at point $C$.

The point $C$ divides the line segment $AB$ in the ratio of $5:8$. The length of required segments are: $AC=2.9cm$ and $BC=4.7cm$.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle A{{A}_{5}}C,\vartriangle AB{{A}_{13}}$ with ${{A}_{5}}C\parallel B{{A}_{13}}$ and vertex $A$  common. Therefore,

$\dfrac{AC}{BC}=\dfrac{A{{A}_{5}}}{{{A}_{5}}{{A}_{13}}}$

Since the distance between ${{A}_{i}},{{A}_{i+1}}$ are equal. Therefore,

$\dfrac{A{{A}_{5}}}{{{A}_{5}}{{A}_{13}}}=\dfrac{5}{13-5}=\dfrac{5}{8}$

Hence, the point $C$ divides the line segment $AB$ in the ratio of $5:8$.

2. Construct a triangle of sides $4$ cm, $5$ cm and $6$ cm and then a triangle similar to it whose sides are $\dfrac{2}{3}$ of the corresponding sides of the first triangle. Give the justification of the construction.

Ans: To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{AB}$ of length $4$ cm.

Step 2: Now, take point $A$ as a centre and draw an arc of $5$ cm radius.

Step 3: Now, take point $B$ as a centre and draw an arc of $6$ cm radius.

Step 4: The point at which the arcs from step 3 and 2 meet. Name it as $C$. $\vartriangle ABC$ is the required triangle.

Step 5: Draw a ray $AX$ on the opposite side of vertex $C$ making an acute angle with line $AB$.

Step 6: Mark $3$ points ${{A}_{1}},{{A}_{2}},{{A}_{3}}$ on $AX$ such that the length of each segment is equal.

Step 7: Join $B$ with ${{A}_{3}}$.

Step 8: Draw a line parallel to $B{{A}_{3}}$ passing through ${{A}_{2}}$ and intersecting $AB$ at point $B'$.

Step 9: Draw a line parallel to $BC$ passing through $B'$ and intersecting $AC$ at point $C'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $BC\parallel B'C'$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle ABC=\angle AB'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle A{{A}_{2}}B',\vartriangle A{{A}_{3}}B$ with ${{A}_{2}}B'\parallel {{A}_{3}}B$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle A{{A}_{2}}B'=\angle A{{A}_{3}}B$

Hence, by  $AA$ similarity criteria, $\vartriangle A{{A}_{2}}B'\sim \vartriangle A{{A}_{3}}B$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{A{{A}_{2}}}{A{{A}_{3}}}=\dfrac{2}{3}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{2}{3}AB,\text{ }AC'=\dfrac{2}{3}AC\text{ and }B'C'=\dfrac{2}{3}BC$

3. Construct a triangle with sides $5$ cm, $6$ cm and $7$ cm and then another triangle whose sides are $\dfrac{7}{5}$ of the corresponding sides of the first triangle. Give the justification of the construction.

Ans: To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{AB}$ of length $5$ cm and extend it from $B$ end.

Step 2: Now, take point $A$ as a centre and draw an arc of $6$ cm radius.

Step 3: Now, take point $B$ as a centre and draw an arc of $7$ cm radius.

Step 4: The point at which the arcs from step 3 and 2 meet. Name it as $C$. $\vartriangle ABC$ is the required triangle.

Step 5: Draw a ray $AX$ on the opposite side of vertex $C$ making an acute angle with line $AB$.

Step 6: Mark $7$ points ${{A}_{1}},{{A}_{2}},...,{{A}_{7}}$ on $AX$such that the length of each segment is equal.

Step 7: Join $B$ with ${{A}_{5}}$.

Step 8: Draw a line parallel to $B{{A}_{5}}$ passing through ${{A}_{7}}$ and intersecting extended $AB$ at point $B'$.

Step 9: Draw a line parallel to $BC$ passing through $B'$ and intersecting extended $AC$ at point $C'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $BC\parallel B'C'$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle ABC=\angle AB'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle A{{A}_{7}}B',\vartriangle A{{A}_{5}}B$ with ${{A}_{7}}B'\parallel {{A}_{5}}B$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle A{{A}_{7}}B'=\angle A{{A}_{5}}B$

Hence, by  $AA$ similarity criteria, $\vartriangle A{{A}_{7}}B'\sim \vartriangle A{{A}_{5}}B$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{A{{A}_{7}}}{A{{A}_{5}}}=\dfrac{7}{5}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{7}{5}AB,\text{ }AC'=\dfrac{7}{5}AC\text{ and }B'C'=\dfrac{7}{5}BC$

4. Construct an isosceles triangle whose base is $8$ cm and altitude $4$ cm and then another triangle whose side are $1\dfrac{1}{2}$ times the corresponding sides of the isosceles triangle. Give the justification of the construction.

Ans: To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{AB}$ of length $8$ cm and extend it from $B$ end.

Step 2: Now, take point $A$ and $B$ as centres and draw arcs of $6$ cm radius. The point at which the arcs meet is $O,O'$. Join the line $OO'$.

Step 3: Let the line $OO'$ meet the line $AB$ at $D$.  Now, take point $D$ as centre and draw an arc of $4$ cm radius.

Step 4: Let the arc cuts the line $OO'$ at point $C$. $\vartriangle ABC$ is the required isosceles triangle.

Step 5: Draw a ray $AX$ on the opposite side of vertex $C$ making an acute angle with line $AB$.

Step 6: Mark $3$ points ${{A}_{1}},{{A}_{2}},{{A}_{3}}$ on $AX$such that the length of each segment is equal.

Step 7: Join $B$ with ${{A}_{2}}$.

Step 8: Draw a line parallel to $B{{A}_{2}}$ passing through ${{A}_{3}}$ and intersecting extended $AB$ at point $B'$.

Step 9: Draw a line parallel to $BC$ passing through $B'$ and intersecting extended $AC$ at point $C'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $BC\parallel B'C'$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle ABC=\angle AB'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle A{{A}_{3}}B',\vartriangle A{{A}_{2}}B$ with ${{A}_{3}}B'\parallel {{A}_{2}}B$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle A{{A}_{3}}B'=\angle A{{A}_{2}}B$

Hence, by  $AA$ similarity criteria, $\vartriangle A{{A}_{3}}B'\sim \vartriangle A{{A}_{2}}B$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{A{{A}_{3}}}{A{{A}_{2}}}=\dfrac{3}{2}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{3}{2}AB,\text{ }AC'=\dfrac{3}{2}AC\text{ and }B'C'=\dfrac{3}{2}BC$

5. Draw a triangle $\mathbf{ABC}$ with side $BC=6$ cm, $\mathbf{AB}=\mathbf{5}$ cm and $\angle ABC={{60}^{\circ }}$ . Then construct a triangle whose sides are $\dfrac{3}{4}$ of the corresponding sides of the triangle $\mathbf{ABC}$. Give the justification of the construction.

Ans: To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{BC}$ of length $6$ cm.

Step 2: Now, take point $B$ as centre and draw an arc of $5$ cm radius making an angle of ${{60}^{\circ }}$ with $BC$. Let the point be known as $A$

Step 3: Join $AC$. $\vartriangle ABC$ is the required triangle.

Step 4: Draw a ray $BX$ on the opposite side of vertex $A$ making an acute angle with line $BC$.

Step 5: Mark $4$ points ${{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}}$ on $BX$such that the length of each segment is equal.

Step 6: Join $C$ with ${{B}_{4}}$.

Step 7: Draw a line parallel to $C{{B}_{4}}$ passing through ${{B}_{3}}$ and intersecting $BC$ at point $C'$.

Step 8: Draw a line parallel to $AC$ passing through $C'$ and intersecting $AB$ at point $B'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $AC\parallel A'C'$ and vertex $B$  common. Therefore, by corresponding angle property, $\angle BAC=\angle BA'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle B{{B}_{3}}C',\vartriangle B{{B}_{4}}C$ with ${{B}_{4}}C\parallel {{B}_{3}}C'$ and vertex $B$ common. Therefore, by corresponding angle property, $\angle B{{B}_{4}}C=\angle B{{B}_{3}}C'$

Hence, by  $AA$ similarity criteria, $\vartriangle B{{B}_{3}}C'\sim \vartriangle B{{B}_{4}}C$

$\Rightarrow \dfrac{BC'}{BC}=\dfrac{B{{B}_{3}}}{B{{B}_{4}}}=\dfrac{3}{4}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{3}{4}AB,\text{ }AC'=\dfrac{3}{4}AC\text{ and }B'C'=\dfrac{3}{4}BC$

6. Draw a triangle $\mathbf{ABC}$ with side $\mathbf{BC}=\mathbf{7}$ cm,  $\angle B={{45}^{\circ }}$, $\angle A={{105}^{\circ }}$. Then, construct a triangle whose sides are $\dfrac{4}{3}$ times the corresponding side of $\vartriangle ABC$. Give the justification of the construction.

Ans: We know that sum of interior angles of a triangle is ${{180}^{\circ }}$. Therefore,

$\angle A+\angle B+\angle C={{180}^{\circ }}$

$\Rightarrow {{105}^{\circ }}+{{45}^{\circ }}+\angle C={{180}^{\circ }}$

$\therefore \angle C={{30}^{\circ }}$

To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{BC}$ of length $7$ cm and extend it from end $C$.

Step 2: Now, take point $B$ as centre and draw an arc making an angle of ${{45}^{\circ }}$ with $BC$.

Step 3: Now, take point $C$ as centre and draw an arc making an angle of ${{30}^{\circ }}$ with $BC$.

Step 4: Let the point where the arcs from step 2 and 3 meet be known as $A$. $\vartriangle ABC$ is the required triangle.

Step 5: Draw a ray $BX$ on the opposite side of vertex $A$ making an acute angle with line $BC$.

Step 6: Mark $4$ points ${{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}}$ on $BX$such that the length of each segment is equal.

Step 7: Join $C$ with ${{B}_{3}}$.

Step 8: Draw a line parallel to $C{{B}_{3}}$ passing through ${{B}_{4}}$ and intersecting $BC$ at point $C'$.

Step 9: Draw a line parallel to $AC$ passing through $C'$ and intersecting $AB$ at point $B'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $AC\parallel A'C'$ and vertex $B$  common. Therefore, by corresponding angle property, $\angle BAC=\angle BA'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle B{{B}_{3}}C,\vartriangle B{{B}_{4}}C'$ with ${{B}_{4}}C'\parallel {{B}_{3}}C$ and vertex $B$ common. Therefore, by corresponding angle property, $\angle B{{B}_{4}}C'=\angle B{{B}_{3}}C$

Hence, by  $AA$ similarity criteria, $\vartriangle B{{B}_{3}}C\sim \vartriangle B{{B}_{4}}C'$

$\Rightarrow \dfrac{BC'}{BC}=\dfrac{B{{B}_{4}}}{B{{B}_{3}}}=\dfrac{4}{3}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{4}{3}AB,\text{ }AC'=\dfrac{4}{3}AC\text{ and }B'C'=\dfrac{4}{3}BC$

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 $\dfrac{5}{3}$ times the corresponding sides of the given triangle. Give the justification of the construction.

Ans: To construct the required triangle, follow the steps below:

Step 1: Draw a line segment $\text{AB}$ of length $4$ cm and extend it from $B$ end.

Step 2: Now, take point $A$ as centres and draw an arc of $3$ cm radius and making an angle of ${{90}^{\circ }}$ with $AB$. Let the point be known as $C$.

Step 3: Join $BC$. $\vartriangle ABC$ is the required isosceles triangle.

Step 5: Draw a ray $AX$ on the opposite side of vertex $C$ making an acute angle with line $AB$.

Step 6: Mark $5$ points ${{A}_{1}},{{A}_{2}},...,{{A}_{5}}$ on $AX$such that the length of each segment is equal.

Step 7: Join $B$ with ${{A}_{3}}$.

Step 8: Draw a line parallel to $B{{A}_{3}}$ passing through ${{A}_{5}}$ and intersecting extended $AB$ at point $B'$.

Step 9: Draw a line parallel to $BC$ passing through $B'$ and intersecting extended $AC$ at point $C'$.

$\vartriangle AB'C'$ is the required triangle.

(Image Will Be Updated Soon)

Justification

By construction, we have $\vartriangle ABC,\vartriangle AB'C'$ with $BC\parallel B'C'$ and vertex $A$  common. Therefore, by corresponding angle property, $\angle ABC=\angle AB'C'$

Hence, by  $AA$ similarity criteria, $\vartriangle ABC\sim \vartriangle AB'C'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{B'C'}{BC}=\dfrac{AC'}{AC}$  ….. (1)

Consider triangles, $\vartriangle A{{A}_{3}}B,\vartriangle A{{A}_{5}}B'$ with ${{A}_{3}}B\parallel {{A}_{5}}B'$ and vertex $A$ common. Therefore, by corresponding angle property, $\angle A{{A}_{3}}B=\angle A{{A}_{5}}B'$

Hence, by  $AA$ similarity criteria, $\vartriangle A{{A}_{3}}B\sim \vartriangle A{{A}_{5}}B'$

$\Rightarrow \dfrac{AB'}{AB}=\dfrac{A{{A}_{5}}}{A{{A}_{3}}}=\dfrac{5}{3}$  ….. (2)

Hence from (1) and (2),

$\Rightarrow AB'=\dfrac{5}{3}AB,\text{ }AC'=\dfrac{5}{3}AC\text{ and }B'C'=\dfrac{5}{3}BC$.

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions (Ex 11.1) Exercise 11.1

Opting for the NCERT solutions for Ex 11.1 Class 10 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 11.1 Class 10 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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1. What is Class 10 Maths Chapter 11 all about?

Ans: NCERT Class 10 Chapter 11-Constructions is a part of Geometry. This chapter is all about the construction of line segments, division of a Line Segment and Construction of a Circle, Constructions of Tangents to a circle using an analytical approach. The chapter contains four sub-topics as given in the following.

1. Introduction.

2. Division of a Line Segment.

3. Construction of Tangents to a Circle.

4. Summary.

This chapter also contains two exercises at the end which include various types and patterns of questions/ problems.

2. How many questions are there in Class 10 Maths Chapter 11 Exercise 11.1?

Ans: A total of seven questions are there in Class 10 Maths Chapter 11 Exercise 11.1 of the NCERT textbook. Answers to those problems are also provided in the NCERT Solutions of Class 10 Maths Chapter 11 provided by Vedantu. These NCERT solutions are absolutely accurate as these are created by the experienced Maths experts while adhering to the CBSE curriculum strictly.

3. Why are NCERT Solutions for Maths Chapter 11 beneficial for all the students of Class 11?

Ans: NCERT Solutions for Class 10 Maths Chapter 11 Constructions are designed in a detailed manner, where one can find a step-by-step solution to all the problems/questions for quick and easy revisions. Solutions for Chapter 11 of NCERT Class 10 Maths are prepared by the industry’s subject matter experts by following the strict guidelines of CBSE.

You can get a free PDF of NCERT Solutions for Class 10 Maths, Chapter 11 – Constructions on Vedantu site and app. Download the solutions to all the questions of NCERT exercises along with the required diagrams with a step-by-step procedure. NCERT Solutions help students in boosting their concepts and clear doubts. With the help of these NCERT Solutions for Maths Chapter 11, you will be able to understand the pattern of questions that you are going to encounter in the final exam. Not only that but also you will be able to understand how to solve these problems quickly with the help of our solutions PDF.

4. How to download NCERT Solutions for Class 10 Maths Chapter 11 Exercise 11.1?

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