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

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

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**Access NCERT Solutions for Class 10 Maths Chapter 11 – Construction**

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

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

**Ans.**

$\begin{array}{l}\mathrm{Following}\text{are the steps to divide a line segment of length}\\ \text{7.6 cm in the ratio of 5:8.}\\ \text{Step 1: Draw line segment AB of 7.6 cm and draw a ray AX}\\ \text{making an acute angle with line segment AB.}\\ \text{Step 2: Locate 13(}=5+8){\text{points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3},\text{\hspace{0.17em}\hspace{0.17em}}...,\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{13}\text{on AX}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} such that AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}=...={\mathrm{A}}_{12}{\mathrm{A}}_{13}.\\ {\text{Step 3: Join the points B and A}}_{13}.\\ {\text{Step 4: Through the point A}}_{5}{\text{, draw a line parallel to BA}}_{13}\text{at}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} A}}_{5}\text{intersecting AB at point C.}\\ \text{C is the point which divides line segment AB of length 7.6 cm}\\ \text{in the ratio 5}:8.\\ \mathrm{On}\text{measuring lengths of AC and BC, we get AC}=2.9\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}\text{}\mathrm{and}\\ \mathrm{BC}=4.7\text{cm.}\end{array}$

**Q.2 **

$\begin{array}{l}\text{Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are}\frac{\text{2}}{3}\text{of the}\\ \text{corresponding sides of the first triangle.}\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{Step}1:\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{segment}\mathrm{AB}=4\hspace{0.17em}\hspace{0.17em}\mathrm{cm}.\mathrm{Taking}\mathrm{point}\mathrm{A}\mathrm{as}\\ \mathrm{centre},\mathrm{draw}\mathrm{an}\mathrm{arc}\mathrm{of}5\mathrm{cm}\mathrm{radius}.\mathrm{Again},\mathrm{taking}\\ \mathrm{point}\mathrm{B}\mathrm{as}\mathrm{centre},\mathrm{draw}\mathrm{an}\mathrm{arc}\mathrm{of}6\mathrm{cm}.\mathrm{These}\mathrm{arcs}\\ \mathrm{intersect}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{point}\mathrm{C}.\mathrm{So},\mathrm{we}\mathrm{have}\\ \mathrm{AC}=5\mathrm{\hspace{0.17em}}\mathrm{cm}\mathrm{}\mathrm{and}\mathrm{BC}=6\hspace{0.17em}\hspace{0.17em}\mathrm{cm}.\mathrm{}\mathrm{\Delta}\mathrm{\hspace{0.17em}}\mathrm{ABC}\mathrm{is}\mathrm{the}\mathrm{required}\\ \mathrm{triangle}\mathrm{.}\\ \mathrm{Step}2:\mathrm{Draw}\mathrm{a}\mathrm{ray}\mathrm{AX}\mathrm{making}\mathrm{an}\mathrm{acute}\mathrm{angle}\mathrm{with}\mathrm{line}\mathrm{AB}\\ \mathrm{on}\mathrm{the}\mathrm{opposite}\mathrm{side}\mathrm{of}\mathrm{vertex}\mathrm{C}\mathrm{.}\\ \mathrm{Step}3:\mathrm{Locate}3\mathrm{points}{\mathrm{A}}_{1},{\mathrm{A}}_{2},\hspace{0.17em}\hspace{0.17em}{\mathrm{A}}_{3}\mathrm{on}\mathrm{AX}\mathrm{such}\mathrm{that}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}{\mathrm{AA}}_{1}={\mathrm{A}}_{1}{\mathrm{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}.\\ \mathrm{Step}4:\mathrm{Join}\mathrm{the}\mathrm{points}\mathrm{B}\mathrm{and}{\mathrm{A}}_{3}.\\ \mathrm{Step}5:\mathrm{Through}\mathrm{the}\mathrm{point}{\mathrm{A}}_{2},\mathrm{draw}\mathrm{a}\mathrm{line}\mathrm{parallel}\mathrm{to}{\mathrm{BA}}_{3}\mathrm{}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{intersecting}\mathrm{AB}\mathrm{at}\mathrm{point}\mathrm{B}\u2018\mathrm{.}\\ \mathrm{Step}6:\mathrm{Draw}\mathrm{a}\mathrm{line}\mathrm{through}\mathrm{B}\u2018\mathrm{parallel}\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{BC}\mathrm{to}\\ \mathrm{intersect}\mathrm{AC}\mathrm{at}\mathrm{C}\u2018\mathrm{.}\\ \mathrm{The}\mathrm{required}\mathrm{triangle}\mathrm{is}\mathrm{\Delta AB}\u2018\mathrm{C}\u2018\mathrm{.}\end{array}$

**Q.3 **

$\begin{array}{l}\text{Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are}\frac{\text{7}}{5}\text{of the}\\ \text{corresponding sides of the first triangle.}\end{array}$

**Ans.**

$\begin{array}{l}\text{Step 1: Draw a line segment AB}=5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{Taking point A as}\\ \text{centre, draw an arc of 6 cm radius. Again, taking}\\ \text{point B as centre, draw an arc of 7 cm. These arcs}\\ \text{ intersect each other at point C. So, we have}\\ \text{AC}=6\text{\hspace{0.17em}}\mathrm{cm}\text{}\mathrm{and}\text{BC}=7\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{}\mathrm{\Delta}\text{\hspace{0.17em}ABC is the required}\\ \text{triangle.}\\ \text{Step 2: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 3: Locate 7 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{,}...\text{,}{\mathrm{A}}_{7}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}=...={\mathrm{A}}_{6}{\mathrm{A}}_{7}.\\ {\text{Step 4: Join the points B and A}}_{5}.\\ {\text{Step 5: Through the point A}}_{7}{\text{, draw a line parallel to BA}}_{5}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B\u2019.}\\ \text{Step 6: Draw a line through B\u2019 parallel to the line BC to}\\ \text{intersect extended line segment AC at C\u2019.}\\ \text{The required triangle is}\mathrm{\Delta}\text{AB\hspace{0.33em}\u2019C\u2019.}\end{array}$

**Q.4 **

$\begin{array}{l}\text{Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose}\\ \text{sides are}\mathrm{}\text{\hspace{0.17em}}1\frac{1}{2}\text{times the corresponding sides of the isosceles triangle}.\mathrm{}\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a line segment AB}=8\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm}.\text{Draw arcs of same}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}radius on both sides of the line segment while taking}\\ \text{point A and B as its centre. Let these arcs intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} each other at O and O\u2019. Join OO\u2019. Let OO\u2019\hspace{0.33em}intersect}\\ \text{\hspace{0.17em}AB at D.}\\ \text{Step 2: Taking D as centre, draw an arc of 4 cm radius which}\\ \text{cuts the extended line segment OO\u2019 at point C. An}\\ \text{isosceles}\mathrm{\Delta}\text{\hspace{0.17em}}\mathrm{ABC}\text{is formed havind CD as 4 cm and AB}\\ \text{as 8 cm.}\\ \text{Step 3: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 4: Locate 3 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\text{A}}_{1}{\text{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}.\\ {\text{Step 5: Join the points B and A}}_{2}.\\ {\text{Step 6: Through the point A}}_{3}{\text{, draw a line parallel to BA}}_{2}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B\u2019.}\\ \text{Step 7: Draw a line through B\u2019 parallel to the line BC to}\\ \text{intersect extended line segment AC at C\u2019.}\\ \text{The required triangle is}\mathrm{\Delta}\text{AB\hspace{0.33em}\u2019C\u2019.}\end{array}$

**Q.5 **

$\begin{array}{l}\mathrm{}\text{Draw a triangle ABC with side BC}=\text{6 cm, AB}=\text{5 cm and}\angle \text{ABC}=\text{60\xb0. Then construct a triangle whose sides are}\\ \frac{\text{3}}{\text{4}}\text{\hspace{0.17em}\hspace{0.17em}of the corresponding sides of the triangle ABC.}\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a}\mathrm{\Delta}\text{\hspace{0.17em}ABC with sides AB}=5\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{cm},\text{BC}=6\text{cm and}\\ \text{}\angle \text{ABC}=60\xb0.\text{}\\ \text{Step 2: Draw a ray BX making an acute angle with line BC}\\ \text{on the opposite side of vertex A.}\\ {\text{Step 3: Locate 4 points B}}_{1},{\text{B}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{B}}_{3}\text{, \hspace{0.17em}}{\mathrm{B}}_{4}\text{on BX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} BB}}_{1}={\text{B}}_{1}{\text{B}}_{2}={\mathrm{B}}_{2}{\mathrm{B}}_{3}={\mathrm{B}}_{3}{\mathrm{B}}_{4}.\\ {\text{Step 4: Join the points C and B}}_{4}.\\ {\text{Step 5: Through the point B}}_{3}{\text{, draw a line parallel to CB}}_{4}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting line segment BC at point C\u2019.}\\ \text{Step 6: Draw a line through C\u2019\hspace{0.33em}parallel to the line AC to}\\ \text{intersect line segment AB at A\u2019.}\\ \text{The required triangle is}\mathrm{\Delta}\text{A\hspace{0.33em}\u2019BC\u2019.}\end{array}$

**Q.6 **

$\begin{array}{l}\text{Draw a triangle ABC with sides BC=7 cm,}\angle =\xb0,\angle =\xb0,\text{}\\ \text{then construct a triangle whose sides are}\frac{4}{3}\text{times the}\\ \text{corresponding sides of}\mathrm{\Delta}\text{\hspace{0.17em}}\mathrm{ABC}.\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{It}\text{}\mathrm{is}\text{}\mathrm{given}\text{}\mathrm{that}\text{}\angle \text{B}=45\xb0,\text{}\angle \mathrm{A}=105\xb0.\\ \therefore \text{In}\mathrm{\Delta}\text{\hspace{0.17em}ABC, we have}\angle \mathrm{C}=180\xb0-105\xb0-45\xb0=30\xb0\\ \mathrm{StepsofConstruction}:\\ \text{Step 1: Draw a}\mathrm{\Delta}\text{\hspace{0.17em}ABC with side BC}=7\text{cm and}\\ \text{}\angle \text{B}=45\xb0,\text{}\angle \mathrm{C}=30\xb0.\text{}\\ \text{Step 2: Draw a ray BX making an acute angle with line BC}\\ \text{on the opposite side of vertex A.}\\ {\text{Step 3: Locate 4 points B}}_{1},{\text{B}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{B}}_{3}\text{, \hspace{0.17em}}{\mathrm{B}}_{4}\text{on BX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} BB}}_{1}={\text{B}}_{1}{\text{B}}_{2}={\mathrm{B}}_{2}{\mathrm{B}}_{3}={\mathrm{B}}_{3}{\mathrm{B}}_{4}.\\ {\text{Step 4: Join the points C and B}}_{3}.\\ {\text{Step 5: Through the point B}}_{4}{\text{, draw a line parallel to CB}}_{3}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment BC at point C\u2019.}\\ \text{Step 6: Draw a line through C\u2019 parallel to the line AC to}\\ \text{intersect extended line segment BA at A\u2019.}\\ \text{The required triangle is}\mathrm{\Delta}\text{A\hspace{0.33em}\u2019BC\u2019.}\end{array}$

**Q.7 **

$\begin{array}{l}\text{Draw a right triangle in which the sides (other than hypotenuse})\text{are of length 4 cm and 3 cm}.\text{Then}\\ \text{construct another triangle whose sides are}\frac{\text{5}}{3}\text{times the corresponding sides of the given triangle}.\mathrm{}\end{array}$

**Ans.**

$\begin{array}{l}\mathrm{Steps}\mathrm{of}\mathrm{Construction}:\\ \text{Step 1: Draw a right angle}\mathrm{\Delta}\text{\hspace{0.17em}ABC with base AB}=4\text{cm,}\\ \text{AC}=3\text{cm and}\angle \mathrm{A}=90\xb0.\text{}\\ \text{Step 2: Draw a ray AX making an acute angle with line AB}\\ \text{on the opposite side of vertex C.}\\ {\text{Step 3: Locate 5 points A}}_{1},{\text{A}}_{2},\text{\hspace{0.17em}\hspace{0.17em}}{\mathrm{A}}_{3}\text{, \hspace{0.17em}}{\mathrm{A}}_{4}{\text{, A}}_{5}\text{on AX such that}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} AA}}_{1}={\mathrm{A}}_{1}{\mathrm{A}}_{2}={\mathrm{A}}_{2}{\mathrm{A}}_{3}={\mathrm{A}}_{3}{\mathrm{A}}_{4}={\mathrm{A}}_{4}{\mathrm{A}}_{5}.\\ {\text{Step 4: Join the points B and A}}_{3}.\\ {\text{Step 5: Through the point A}}_{5}{\text{, draw a line parallel to BA}}_{3}\text{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} intersecting extended line segment AB at point B\u2019.}\\ \text{Step 6: Draw a line through B\u2019 parallel to the line BC to}\\ \text{intersect extended line segment AC at C\u2019.}\\ \text{The required triangle is}\mathrm{\Delta}\text{AB\u2019\hspace{0.33em}}\mathrm{C}\u2018\text{.}\end{array}$

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