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NCERT Solutions Class 10 Maths Chapter 11 Constructions – Here are all the NCERT solutions for Class 10 Maths Chapter 11. This solution contains questions, answers, images, explanations of the complete Chapter 11 titled Constructions of Maths taught in Class 10. If you are a student of Class 10 who is using NCERT Textbook to study Maths, then you must come across Chapter 11 Constructions. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 10 Maths Chapter 11 Constructions in one place.

NCERT Solutions Class 10 Maths Chapter 11 Constructions

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Class 10
Subject Maths
Book Mathematics
Chapter Number 11
Chapter Name  

Constructions

NCERT Solutions Class 10 Maths chapter 11 Constructions

Class 10, Maths chapter 11, Constructions solutions are given below in PDF format. You can view them online or download PDF file for future use.

Constructions

Q.1: In the question, give the justification of the construction also Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.

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

Q.2: In the question, give the justification of the construction also , construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

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

Q.3: In the question, give the justification of the construction also ,construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

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

Q.4: \(\begin{array}{l}{\text { In the question, give the justification of the construction also , construct an isosceles triangle whose base is } 8 \mathrm{cm} \text { and altitude } 4 \mathrm{cm} \text { and then another }} \\ {\text { triangle whose sides are } 1 \frac{1}{2} \text { times the corresponding sides of the isosceles triangle. }}\end{array}\)

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

Q.5: In the question, give the justification of the construction also , draw a triangle ABC with side BC = 6 cm, AB=5 cm and \(\angle \mathrm{ABC}\)=\(60^{\circ}\) . Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.

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

Q.6: \(\begin{array}{l}{\text {In the question, give the justification of the construction also , draw a triangle } A B C \text { with side } B C=7 \mathrm{cm}, \angle B=45^{\circ}, \angle A=105^{\circ} . \text { Then, construct a }} \\ {\text { triangle whose sides are } \frac{4}{3} \text { times the corresponding sides of } \Delta \text { ABC. }}\end{array}\)

Ans : \(\begin{array}{l}{\angle B=45^{\circ}, \angle A=105^{\circ}} \\ {\text { Sum of all interior angles in a triangle is } 180^{\circ} .}\end{array}\) \(\begin{array}{l}{\angle A+\angle B+\angle C=180^{\circ}} \\ {105^{\circ}+45^{\circ}+\angle C=180^{\circ}} \\ {\angle C=180^{\circ}-150^{\circ}} \\ {\angle C=30^{\circ}}\end{array}\) The required triangle can be drawn as follows. \(\begin{array}{l}{\text { Step } 1} \\ {\text { Draw a } \triangle A B C \text { with side } B C=7 \mathrm{cm}, \angle B=45^{\circ}, \angle C=30^{\circ}}\end{array}\) \(\begin{array}{l}{\text { Step } 2} \\ {\text { Draw a ray BX making an acute angle with } B C \text { on the opposite side of vertex } A \text { . }}\end{array}\) Step 3 \(Join B_{3} C.\) Draw a line through \(B_{4}\) parallel to \(B_{3}C\) intersecting extended BC at \(C^{\prime}\). Step 4 \(\begin{array}{l}{\text { Through } \mathrm{C}^{\prime}, \text { draw a line parallel to } \mathrm{AC} \text { intersecting extended line segment at } \mathrm{C}^{\prime} .} \\ {\triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime} \text { is the required triangle. }}\end{array}\) Justification The construction can be justified by proving that \(\begin{array}{l}{\mathrm{A}^{\prime} \mathrm{B}=\frac{4}{3} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{4}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{4}{3} \mathrm{AC}} \\ {\text { In } \triangle \mathrm{ABC} \text { and } \triangle \mathrm{A}^{\prime} \mathrm{BC}^{\prime}}\end{array}\) \(\begin{array}{l}{\angle A B C=\angle A^{\prime} B C^{\prime}(\text { Correspon })} \\ {\angle A C B=\angle A^{\prime} C^{\prime} B(\text { Corresponding angles })} \\ {\therefore \triangle A B C-\triangle A^{\prime} B C^{\prime} \text { (AA similarity criterion) }}\end{array}\) \(\begin{array}{l}{\Rightarrow \frac{A B}{A^{\prime} B}=\frac{B C}{B C^{\prime}}=\frac{A C}{A^{\prime} C^{\prime}}} \\ {\text { In } \Delta B B_{3} C \text { and } \Delta B B_{4} C^{\prime}} \\ {\angle B_{3} B C=\angle B_{4} B C^{\prime}(\text { Common })}\end{array}\) \(\begin{array}{l}{\angle B B_{3} C=\angle B B_{4} C^{\prime}(\text { Corresponding angles) }} \\ {\therefore \triangle B B_{3} C-\Delta B B_{4} C^{\prime}(\text { AA similarity criterion) }} \\ {\Rightarrow \frac{B C}{B C^{\prime}}=\frac{B B_{3}}{B B_{4}}} \\ {\Rightarrow \frac{B C}{B C^{\prime}}=\frac{3}{4}}\end{array}\) On comparing equations (1) and (2), we obtain \(\begin{array}{l}{\frac{A B}{A^{\prime} B}=\frac{B C}{B C^{\prime}}=\frac{A C}{A^{\prime} C^{\prime}}=\frac{3}{4}} \\ {\Rightarrow A B=\frac{4}{3} A B, B C^{*}=\frac{4}{3} B C, A C^{\prime}=\frac{4}{3} A C} \\ {\text { This justifies the construction. }}\end{array}\)

Q.7: In the question, give the justification of the construction also , 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.

Ans : It is given that sides other than hypotenuse are of lengths 4 cm and 3 cm . Clearly,these will be perpendicular to each other. The required triangle can be drawn as follows. Step 1 Draw a line segment AB=4 cm . Draw a ray SA making 90\(^ \circ\) with it. Step 2 Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C . Join BC .\(\triangle ABC\) is the required triangle. Step 3 Draw a ray AX making an acute angle with AB , opposite to vertex C. Step 4 Locate 5 points as 5 is greater in 5 and 3, \(A_{1}, A_{2}, A_{3}, \)As, on line segment AX such that \(\mathrm{AA}_{1}=\mathrm{A}_{1} \mathrm{A}_{2}=\mathrm{A}_{2} \mathrm{A}_{3}=\mathrm{A}_{3} \mathrm{A}_{4}=\mathrm{A}_{4} \mathrm{A}_{5}\) Step 5 Join \(A_{3}\) B . Draw a line through \(A_{5}\) parallel to \(A_{3}\) B intersecting extended line segment AB at \(B^{\prime}\) Step 6 \(\begin{array}{l}{\Rightarrow \frac{A B}{A B^{\prime}}=\frac{A A_{3}}{A A_{5}}} \\ {\Rightarrow \frac{A B}{A B^{\prime}}=\frac{3}{5}}\end{array}\) \(\begin{array}{l}{\text { On comparing equations }(1) \text { and }(2), \text { we obtain }} \\ {\frac{A B}{A B}=\frac{B C}{B C^{\prime}}=\frac{A C}{A C^{\prime}}=\frac{3}{5}} \\ {\Rightarrow A B^{\prime}=\frac{5}{3} A B, B C^{\prime}=\frac{5}{3} BC, A C^{\prime}=\frac{5}{3} A C} \\ {\text { This justifies the construction. }}\end{array}\) Through B', draw a line parallel to BC intersecting extended line segment AC at C . \(\triangle A B^{\prime} C^{\prime} \) is the required triangle. Justification The construction can be justified by proving that \(\mathrm{AB}^{\prime}=\frac{5}{3} \mathrm{AB}, \mathrm{BC}^{\prime}=\frac{5}{3} \mathrm{BC}, \mathrm{AC}^{\prime}=\frac{5}{3} \mathrm{ AC } \ In \ \Delta \mathrm{ABC}\ and \ \Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime} \) \(\begin{array}{l}{\angle \mathrm{ABC}=\angle \mathrm{AB}^{\prime} \mathrm{C}^{\prime}(\text { Corresponding angles })} \\ {\angle \mathrm{BAC}=\angle \mathrm{B}^{\prime} \mathrm{AC}^{\prime}(\text { Common })}\end{array}\) \(\begin{array}{l}{\therefore \triangle \mathrm{ABC}-\Delta \mathrm{AB}^{\prime} \mathrm{C}^{\prime} \text { (AA similarity criterion) }} \\ {\Rightarrow \frac{\mathrm{AB}}{\mathrm{AB}^{\prime}}=\frac{\mathrm{BC}}{\mathrm{B}^{\prime} \mathrm{C}^{\prime}}=\frac{\mathrm{AC}}{\mathrm{AC}^{\prime}}}\end{array}\) \(\begin{array}{l}{\text { In } \triangle \mathrm{AA}_{3} \mathrm{B} \text { and } \triangle \mathrm{AA}_{5} \mathrm{B}^{\prime}} \\ {\angle \mathrm{A}_{3} \mathrm{AB}=\angle \mathrm{A}_{5} \mathrm{AB}^{\prime}(\text { Common })}\end{array}\) \(\begin{array}{l}{\angle A A_{3} B=\angle A A_{5} B^{\prime}(\text { Corresponding angles })} \\ {\therefore \triangle A A_{3} B-\triangle A A_{5} B^{\prime} \text { (AA similarity criterion) }}\end{array}\)

Q.1: In question , give also the justification of the construction : 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.

Ans : A pair of tangents to the given circle can be constructed as follows. Step 1 Taking ny point O of the given plane as centre, draw a circle of 6 cm radius Locate a point P, 10 Jn away from O, Join OP. Step 2 Bisect OP. Let M be the mid-point of PO. Step 3 Taking M as centre and MO as radius, draw a circle, Step 4 Let this circle intersect the previous circle at point Q and R. Step 5 Join PQ and PR, PQ and PR the required tangents. The lengths of tangents PQ and PR are 8 cm each. Justification The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 6 cm). For this, join OQ and OR. \(\begin{array}{l}{\angle \mathrm{PQO} \text { is an angle in the semi-circle. We know that angle in a semicircle is a right }} \\ {\text { angle. }} \\ {\therefore \angle \mathrm{PQO}=90^{\circ}} \\ {\Rightarrow \angle \mathrm{PQ} \perp \mathrm{PQ}} \\ {\text { since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR }} \\ {\text { is a tangent of the circle }}\end{array}\)

Q.2: In question , give also the justification of the construction : 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.

Ans : Tangents on the given circle can be drawn as follows. Step 1 Draw a circle of 4 cm radius with centre as O on the given plane. Step 2 Draw a circle of 5 radius taking O as its centre. Locate a point P on this circle and join OP. Step 3 Bisect OP. Let M be the mid-pont of PO. Step 4 Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R. Step 5 Join PQ and PR. PQ and PR are the required tangents. \(\begin{array}{l}{\text { It can be observed that } \mathrm{PQ} \text { and } \mathrm{PR} \text { are of length } 4.47 \mathrm{cm} \text { each. }} \\ {\text { In } \Delta \mathrm{PQO} \text { , }} \\ {\text { since PQ is a tangent, }} \\ {\angle \mathrm{P} Q \mathrm{O}=90^{\circ}} \\ {\mathrm{PO}=6 \mathrm{cm}} \\ {\mathrm{QO}=4 \mathrm{cm}}\end{array}\) \(\begin{array}{l}{\text { Applying Pythagoras theorem in } \Delta P Q O, \text { we obtain }} \\ {\mathrm{PQ}^{2}+\mathrm{QO}^{2}=\mathrm{PQ}^{2}} \\ {\mathrm{PQ}^{2}+(4)^{2}=(6)^{2}} \\ {\mathrm{PQ}^{2}+16=36} \\ {\mathrm{PQ}^{2}+16=36} \\ {\mathrm{PQ}^{2}=36-16} \\ {\mathrm{PQ}=2 \sqrt{5}} \\ {\mathrm{PQ}=4.47 \mathrm{cm}}\end{array}\) Justification The construction can be justified by proving that PQ and PR are the tangents to the circle (whose centre is O and radius is 4 crn). For this, let us join OQ and OR. \(\begin{array}{l}{\angle \mathrm{PQO} \text { is an angle in the semi-circle. We know that angle in a semi-circle is a right }} \\ {\text { angle. }} \\ {\therefore \angle \mathrm{PQO}=90^{\circ}} \\ {\Rightarrow \angle \mathrm{OQ} \perp \mathrm{PQ}} \\ {\text { since oQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR }} \\ {\text { is a tangent of the circle. }}\end{array}\)

Q.3: In question , give also the justification of the construction : Draw a circle of radius 3 crn, 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.

Ans : The tangent can be constructed on the given circle as follows. Step 1 Taking any point O on the given plane as centre, draw a circle of 3 crn radius. Step 2 Take one of its diameters, PQ, and extend it on both sides. Locate two points on this diameter such that OR = OS = 7 cm Step 3 Bisect OR and OS. Let T and U be the m d-points of OR and OS respectively. Step 4 Taking T and U as its centre and Kith TO and UO as radius, draw two circles. These two circles will intersect the circle at point V, W, X, Y respectively. Join RV, RW, SX, and SY. These are the required tangents. Justification The construction can be justified by proving that RV, RW, SY, and SX are the tangents to the circle (whose centre is O and radius is 3 cm). For this, join OV, OW, OX, and OY. \(\begin{array}{l}{\angle \mathrm{RVO} \text { is an angle in the semi-circle. We know that angle in a semi-circle is a right }} \\ {\text { angle. }} \\ {\therefore \angle \mathrm{RVO}=90^{\circ}}\end{array}\) \(\Rightarrow \text { OV } \perp \mathrm{RV}\) Since OV is the radius of the circle, RV has to be a tangent of the circle. Similarly, OW, OX, and OY are the tangents of the circle.

Q.4: In question , give also the justification of the construction: Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60^{\circ} .

Ans : The tangents can be constructed in the following manner: Step 1 Draw a circle of radius 5 cm and with centre as O. Step 2 Take a point A on the circumference of the circle end join OA. Draw a perpendicular to OA at point A. Step 3 Draw a radius OB, making an angle of \(120^{\circ}\left(180^{\circ}-60^{\circ}\right)\) with OA. Step 4 Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P, PA and PB are the required tangents at angle of 60\(^\circ\) Justification The construction can be justified by proving that \(\angle APB=60^{\circ}\) \(\begin{array}{l}{\text { By our construction }} \\ {\angle \mathrm{OAP}=90^{\circ}} \\ {\angle \mathrm{OBP}=90^{\circ}} \\ {\text { And } \angle \mathrm{AOB}=120^{\circ}}\end{array}\) \(\begin{array}{l}{\text { We know that the sum of all interior angles of a quadrilateral = } 360^{\circ}} \\ {\angle \mathrm{OAP}+\angle \mathrm{AOB}+\angle \mathrm{OBP}+\angle \mathrm{APB}=360^{\circ}} \\ {90^{\circ}+120^{\circ}+90^{\circ}+\angle \mathrm{APB}=360^{\circ}} \\ {\angle \mathrm{APB}=60^{\circ}} \\ {\text { This justifies the construction. }}\end{array}\)

Q.5: In question , give also the justification of the construction: 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.

Ans : The tangents can be constructed on the given circles as follows. Step 1 Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm and 3 cm radius. Step 2 Bisect the line AS. Let the mid-point of AB be C. Taking C as centre, draw a circle of AC radius which will intersect the circles at points P, Q, R, and S. Join AP, AQ, AS, and AR, These are the required tangents. Justification The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is A and radius is 3 cm ) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm ). For this , join AP, AQ , BS and BR. \(\angle\)ASB is an angle in the serni-circle. We know that an angle in a semi-circle is a right angle. \(\begin{array}{l}{\therefore \angle A S B=90^{\circ}} \\ {\Rightarrow B S \perp A S}\end{array}\) Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR, BP, and BQ are the tangents.

Q.6: \(\begin{array}{l}{\text {In question , give also the justification of the construction: Let } \mathrm{ABC} \text { be a right triangle in which } \mathrm{AB}=6 \mathrm{cm}, \mathrm{BC}=8 \mathrm{cm} \text { and } \angle \mathrm{B}=90^{\circ} . \mathrm{BD} \text { is the }} \\ {\text { perpendicular from } \mathrm{B} \text { on } \mathrm{AC} \text { . The circle through } \mathrm{B}, \mathrm{C}, \mathrm{D} \text { is drawn. Construct the tangents }} \\ {\text { from A to this circle. }}\end{array}\)

Ans : Consider the following situation. If a circle is drawn through B, D, and C, BC will be its diameter as \(\angle\)BDC is of measure 90\(\circ\). The centre E of this circle will be the mid- point of BC. The required tangents can be constructed on the given circle as follows. Step 1 Join AE and bisect it. Let F be the mid-point of AE. Step 2 Taking F as centre and FE as its radius, draw a circle which will intersect the circle at point B and G. join AG. AB and AG are the required tangents. Justification The construction can be justified by proving that AG and Aa are the tangents to the circle. For this, join EC. \(\angle\)AGE is an angle in the semi-circle. We know that an angle in a semi-circle is a right angle. \(\begin{array}{l}{\therefore \angle A G E=90^{\circ}} \\ {\Rightarrow E G \perp A G}\end{array}\) Since EG is the radius of the circle, AG has to be a tangent of the circle. \(\begin{array}{l}{\text { Already, } \angle B=90^{\circ}} \\ {\Rightarrow A B \perp B E}\end{array}\) Since BE is the radius of the circle, AB has to be a tangent of the circle.

Q.7: In question , give also the justification of the construction: 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.

Ans : The required tangents can be constructed on the given circle as follows. Step 1 Draw a circle with the help of a bangle. Step 2 Take a point P outside this circle and take two chords QR and ST. Step 3 Draw perpendicular bisectors of these chords. Let them intersect each other at point O. Step 4 Join PO and bisect it. Let U be the mid-point of PO. Taking U as centre, draw a circle of radius OU, which will intersect the circle at V and W. Join PV and PW. PV and PW are the required tangents. Justification The construction can be justified by proving that PV and PW are the tangents to the circle. For this, first of all, it has to be proved that O is the centre of the circle. Let us join OV and OW. We know that perpendicular bisector of a chord passes through the centre. Therefore, the perpendicular bisector of chords QR and ST pass through the centre. It is clear that the intersection point of these perpendicular bisectors is the centre of the circle \(\angle\)PVO is an angle in the serni-circle. We know that an angle in a serni-circle is a right angle. \(\begin{array}{l}{\therefore \angle \mathrm{PVO}=90^{\circ}} \\ {\Rightarrow \mathrm{OV} \perp \mathrm{PV}}\end{array}\) Since OV is the radius of the circle, PV has to be e tangent of the circle. Similarly, PW is a tangent of the circle.

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