NCERT Solutions for Class 10 Maths Chapter 11 Constructions

Ans : 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 L3=(= 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 \(\angle A A_{13} B\)) 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 
\(\begin{array}{l}{\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{5}{8}} \\ {\text { By construction, we have } \mathrm{A}_{5} \mathrm{C}\left\|\mathrm{A}_{13} \mathrm{B} \text { . By applying Basic proportionality theorem for }\right.} \\ {\frac{\mathrm{AC}}{\mathrm{CB}}=\frac{\mathrm{AA}_{5}}{\mathrm{A}_{5} \mathrm{A}_{13}}}\end{array}\)
\(\begin{array}{l}{\text { From the figure, it can be observed that } A A_{5} \text { and } A_{5} A_{13} \text { contain } 5 \text { and } 8 \text { equal }} \\ {\text { divisions of line segments respectively. }} \\ {\therefore \frac{A A_{f}}{A_{3} A_{13}}=\frac{5}{8}} \\ {\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{A C}{C B}=\frac{5}{8}} \\ {\text { This justifies the construction. }}\end{array}\) 

Ans : Step 1 Draw a line segment AB = 4 cm. Taking point A as centre, draw and 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 \(\triangle ABC\) is the required triangle.
Step 2 Draw a ray AX  making an acute angle with line AB on the opposite site of vertex C.
Step 3 Locate 3 points (as 3 is greater between 2 and 3) online AX such that \(AA_{1}=A_{1}A_{2}=A_{2}A_{3}\).
Step 4 \(Join \mathrm{BA}_{3}\) and draw a line through \(\mathrm{A}_{2}\) parallel to \(\mathrm{B} \mathrm{A}_{3}\) to intersect \(\mathrm{AB}\) at point \(\mathrm{B}^{\prime} .\)
 Step 5  Draw a line through B\(\prime\) parallel to the line BC to intersect AC at \(C^{\prime}\) \(\triangle A B^{\prime} C^{\prime}\)is the required triangle.

Justification
The construction can be justified by proving that 
\(\mathrm{AB}^{\prime}=\frac{2}{3} \mathrm{AB}, \mathrm{B} \mathrm{C}^{\prime}=\frac{2}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{2}{3} \mathrm{AC}\)
\(\begin{array}{l}{\text { By construction, we have } B^{\prime} C^{\prime}\|B C} \\ {\therefore \angle A B C^{\prime}=\angle A B C(\text {Corresponding angles) }}\end{array}\)
\(\triangle A B^{\prime} C^{\prime} \text { and } \triangle A B C\)
\(\begin{aligned} \angle \mathrm{ABC}^{\prime} &=\angle A B C(\text { Proved above }) \\ \angle \mathrm{B}^{\prime} \mathrm{AC}^{\prime} &=\angle B A C(\text { Common }) \end{aligned}\)
\(\therefore \triangle \mathrm{ABC}^{\prime}-\Delta \mathrm{ABC}(\text { AA similarity criterion })\)
\(\Rightarrow \frac{A B^{\prime}}{A B}=\frac{B^{\prime} C^{\prime}}{B C}=\frac{A C^{\prime}}{A C}\)
\(\triangle \mathrm{AA}_{2} \mathrm{B}^{\prime} \text { and } \triangle \mathrm{A} \mathrm{A}_{3} \mathrm{B}\)
\(\begin{array}{l}{\angle A_{2} A B^{\prime}=\angle A_{3} A B(\text {Corresponding} \text { angles })} \\ {\angle A A_{2} B^{\prime}=\angle A A_{3} B(\text {Corresponding angles) }}\end{array}\)
\(\begin{array}{l}{\therefore \triangle A A_{2} B^{\prime}-\Delta A A_{3} B(\text { AA similarity criterion) }} \\ {\Rightarrow \frac{A B^{\prime}}{A B}=\frac{A A_{2}}{A A_{3}}}\end{array}\)
\(\Rightarrow \frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{2}{3}\)
\(\begin{array}{l}{\text { From equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{\mathrm{AB}^{\prime}}{\mathrm{AB}}=\frac{\mathrm{B} \mathrm{C}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{AC}^{\prime}}{\mathrm{AC}}=\frac{2}{3}} \\ {\Rightarrow \mathrm{AB}^{\prime}=\frac{2}{3} \mathrm{AB}, \mathrm{B}^{\prime} \mathrm{C}^{\prime}=\frac{2}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{2}{3} \mathrm{AC}} \\ {\text { This justifies the construction. }}\end{array}\) 

Ans : Step 1 Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 5 cm radius respectively. Let these arcs intersect each other at point C.\(\triangle ABC\) is the required triangle having length of sides as 5 cm , 6 cm , 7 cm respectively.
Step 2 Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
Step 3 
\(\begin{array}{l}{\text { Locate } 7 \text { points, } A_{1}, A_{2}, A_{3}, A_{4} A_{5}, A_{5}, A_{7}(\text { as } 7 \text { is greater between } 5 \text { and } 7), \text { on line } A X} \\ {\text { such that } A A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4}=A_{4} A_{5}=A_{5} A_{6}=A_{6} A_{7}}\end{array}\)
Step 4 Join \(BA_{5}\) and draw a line through \(A_{7}\) parallel to \(BA_{5}\) to intersect extended line segment AB at point \(B^{\prime}\).
Step 5 Draw a line through \(B^{\prime}\) parallel to BC intersecting the extended line segment AC at \(C^{\prime}\).\(\triangle\)A\(B^{\prime}\)\(C^{\prime}\) is the required triangle.



Justification
The construction can be justified by proving that 
\(\begin{array}{l}{\mathrm{AB}^{\prime}=\frac{7}{5} \mathrm{AB}, \mathrm{B} \mathrm{C}^{\prime}=\frac{7}{5} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{7}{5} \mathrm{AC}} \\ {\text { In } \triangle \mathrm{ABC} \text { and } \Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}} \\ {\angle \mathrm{ABC}=\angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}(\text { Corresponding angles })} \\ {\angle \mathrm{BAC}=\angle \mathrm{B}^{\prime} \mathrm{AC}^{\prime}(\text { Common })}\end{array}\)
\(\therefore \triangle \mathrm{ABC}-\triangle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}\)
\(\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AC}}{\mathrm{AC}^{\prime}} \ldots\)
\(\begin{array}{l}{\text { In } \triangle A A_{5} B \text { and } \triangle A A_{7} B^{\prime}} \\ {\angle A_{5} A B=\angle A_{7} A B^{\prime}(\text { Common })}\end{array}\)
\(\angle A A_{\mathrm{s}} B=\angle A A_{7} B^{\prime}(\text { Corresponding angles })\)
\(\begin{array}{l}{\therefore \triangle \mathrm{AA}_{5} \mathrm{B}-\triangle \mathrm{AA}_{7} \mathrm{B}^{\prime} \text { (AA similarity criterion) }} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{AA}_{5}}{\mathrm{AA}_{7}}} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{5}{7}}\end{array}\)
\(\begin{array}{l}{\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{BC}^{\prime}}=\frac{\mathrm{AC}}{\mathrm{AC}^{\prime}}=\frac{5}{7}} \\ {\Rightarrow \mathrm{AB}^{\prime}=\frac{7}{5} \mathrm{AB}, \mathrm{B}^{\prime} \mathrm{C}=\frac{7}{5} \mathrm{BC}, \mathrm{AC}=\frac{7}{5} \mathrm{AC}} \\ {\text { This justifies the construction. }}\end{array}\) 

Ans : Let us assume that ∠ABC is an isosceles triangle having CA and CB of equal lengths, base AB of 8 crn, and AD is the altitude of 4 cm. \(A \triangle A B^{\prime} C^{\prime}\)whose sides are \(\frac{3}{2}\) times of can be drawn as follows.
Step 1 Draw a line segment Aa of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and as its centre. Let these arcs intersect each other at O and O'. Join OO' Let OO' intersect AB at D.
Step 2 Taking D as centre, draw an arc of 4 cm radius which cuts the extended line segment OO' at point C. An isosceles \(AA_{3}C\) is formed, having CD (altitude) as 4 cm and AB (base) as 8 cm.
Step 3 Draw a ray AX making an acute angle with line segment  AB on the opposite side of vertex C.
Step 4 Locate 3 points (as 3 is greater between 3 and 2) \(A_{1}, A_{2}, \text { and } A_{3}\) on AX such that \(AA_{1}= A_{1}A_{2}= A_{2}A_{3}\).
Step 5 Join \(BA_{2}\) and draw a line through \(A_{3}\) parallel to \(BA_{2}\) to intersect extended line segment AB at point \(B^{\prime}\).
Step 6 Draw a line through \(B^{\prime}\) parallel to BC intersecting the extended the segment AC at \(C^{\prime}\). \(\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime}\) Is the required triangle.



Justification
The construction can be justified by proving that
\(\mathrm{AB}^{\prime}=\frac{3}{2} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{3}{2} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{3}{2} \mathrm{AC}\)
\(\begin{array}{l}{\text { In } \triangle A B C \text { and } \Delta A B^{\prime} C^{\prime}} \\ {\angle A B C=\angle A B^{\prime} C^{\prime} \text { (Corresponding angles) }}\end{array}\)
\(\begin{array}{l}{\angle B A C=\angle B^{\prime} A C^{\prime}(\text { Common })} \\ {\therefore \triangle A B C-\Delta A B^{\prime} C^{\prime} \text { (AA similarity criterion) }}\end{array}\)
\(\begin{array}{l}{\Rightarrow \frac{A B}{A B^{\prime}}=\frac{B C}{B C^{\prime}}=\frac{A C}{A C^{\prime}}} \\ {\text { In } \triangle A A_{2} B \text { and } \Delta A A_{3} B^{\prime}}\end{array}\)
\(\begin{array}{l}{\angle A_{2} A B=\angle A_{3} A B^{\prime}(\text { Corresponding angles })} \\ {\angle A A_{2} B=\angle A A_{3} B^{\prime} \text { (Corresponding angles) }}\end{array}\)
\(\begin{array}{l}{\therefore \triangle \mathrm{AA}_{2} \mathrm{B}-\triangle \mathrm{AA}_{3} \mathrm{B}^{\prime} \text { (AA similarity criterion) }} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}}=\frac{\mathrm{AA}_{2}}{\mathrm{AA}_{3}}} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{2}{3}}\end{array}\)
\(\begin{array}{l}{\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{A B}{A B}=\frac{B C}{B C^{\prime}}=\frac{A C}{A C^{\prime}}=\frac{2}{3}} \\ {\Rightarrow A B^{\prime}=\frac{3}{2} A B, B C=\frac{3}{2} B C, A C=\frac{3}{2} A C} \\ {\text { This justifies the construction. }}\end{array}\) 

Ans : A \(\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}\) whose sides \(\frac{3}{4}\) of the corresponding sides of can be drawn as follows.
Step 1 Draw a \(\triangle\)ABC with side BC = 6 crn, AB = 5 cm and \(\angle\)ABC = 60\(^\circ\). 
Step 2 Draw a ray BX making an acute angle with BC on the opposite side of vertex A. Step 3 Locate 4 points (as 4 is greater in 3 and 4), \(B_{1}, B_{2}, B_{3}, B_{4}\), on line segment BX. 
Step 4 Join \(B_{4} C\)and draw a line through \(B_{3}\) , parallel to \(\mathrm{B}_{4} \mathrm{C}\) intersecting BC at \(C^{\prime}\) 
Step 5 Draw a line through C' parallel to AC intersecting AB at A'.\(\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}\) is the required triangle.

Justification
The construction can be justified by proving that, 
\(\mathrm{AB}=\frac{3}{4} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{3}{4} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{3}{4} \mathrm{AC}\)
\(\begin{array}{l}{\text { In } \triangle A^{\prime} B C^{\prime} \text { and } \triangle A B C \text { , }} \\ {\angle A^{\prime} C^{\prime} B=\angle A C B \text { (Corresponding angles) }} \\ {\angle A^{\prime} B C^{\prime}=\angle A B C(\text { Common})}\end{array}\)
\(\therefore \triangle A^{\prime} B C^{\prime} \sim \triangle A B C(\text { AA similarity criterion })\)
\(\Rightarrow \frac{A B}{A B}=\frac{B C}{B C}=\frac{A C}{A C}\)
\(\begin{array}{l}{\text { In } \triangle B B_{3} C^{\prime} \text { and } \triangle B B_{4} C} \\ {\angle B_{3} B C^{\prime}=\angle B_{4} B C(\text {Common})}\end{array}\)
\(\begin{array}{l}{\angle B B_{3} C^{\prime}=\angle B B_{4} C(\text { Corresponding angles) }} \\ {\therefore \triangle B B_{3} C^{\prime} \sim \Delta B B_{4} C(\text { AA similarity criterion) }}\end{array}\)
\(\begin{array}{l}{\Rightarrow \frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{\mathrm{BB}_{3}}{\mathrm{BB}_{4}}} \\ {\Rightarrow \frac{\mathrm{BC}^{\prime}}{\mathrm{BC}}=\frac{3}{4}} \\ {\text { From equations }(1) \text { and }(2), \text { we obtain }}\end{array}\)
\(\begin{array}{l}{\frac{A B}{A B}=\frac{B C}{B C}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{3}{4}} \\ {\Rightarrow A B=\frac{3}{4} A B, B C^{\prime}=\frac{3}{4} B C, A C^{\prime}=\frac{3}{4} A C} \\ {\text { This justifies the construction. }}\end{array}\)