GSEB Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

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

Steps to Construct:

1. Sketch a line segment AB of length 7.6 cm.
2. From point A, create an acute angle \( \angle BAX \) underneath BA.
3. Along AX, mark 13 (which is 5 + 8) arc segments \( A_1, A_2, ..., A_{13} \) such that \( AA_1 = A_1A_2 = ... = A_{12}A_{13} \).
4. Connect B with \( A_{13} \).
5. Starting from \( A_5 \), draw a line \( A_5C \) parallel to \( A_{13}B \), where C is a point on AB.

The lengths of the two sections will be AC = 2.9 cm and CB = 4.7 cm.

Justification:

In \( \triangle ABA_{13} \), the line \( A_5C \) is parallel to \( A_{13}B \).
\( \implies \frac{AC}{CB} = \frac{AA_5}{A_5A_{13}} \) (By Basic Proportionality Theorem)
However, from our construction, \( \frac{AA_5}{A_5A_{13}} = \frac{5}{8} \).
Therefore, \( \frac{AC}{CB} = \frac{5}{8} \).
By measurement, AC = 2.9 cm and CB = 4.7 cm.
In simple words: First, draw a line 7.6 cm long. Then, from one end, draw a diagonal line and mark 13 equal spaces. Connect the 13th mark to the other end of the first line. Finally, from the 5th mark, draw a line parallel to the connecting line. This splits the original line into a 5:8 ratio.

Exam Tip: Remember to clearly label all points and lines. A neat diagram is crucial for scoring well in construction questions.

 

Question 2. 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.
Answer:

Construction Steps:

1. Sketch \( \triangle ABC \) with AC = 6 cm, AB = 5 cm, and BC = 4 cm.
2. From point A, create an acute angle \( \angle CAX \) underneath the base AC.
3. Place 3 arc points along AX, labeled \( A_1, A_2, A_3 \), so that \( AA_1 = A_1A_2 = A_2A_3 \).
4. Connect \( A_3 \) and C.
5. From \( A_2 \), draw \( A_2C' \parallel A_3C \), where \( C' \) is a point on AC.
6. From \( C' \), construct \( B'C' \parallel BC \), where \( B' \) is a point on AB.

Therefore, \( \triangle AB'C' \) is the desired triangle.

Justification:

We have \( \triangle AB'C' \sim \triangle ABC \) (by AA similarity Rule).
\( \implies \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC} \) ..........(1) (Sides are proportional)
However, by construction, \( A_2C' \parallel A_3C \).
Using the Basic Proportionality Theorem, \( \frac{AC'}{AC}=\frac{AA_2}{AA_3}=\frac{2}{3} \) ...........(2)
From (1) and (2), we get \( \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC}=\frac{2}{3} \).
In simple words: First, draw a triangle with sides 4 cm, 5 cm, and 6 cm. Then, from one corner, draw another line below the triangle. Mark three equal points on this new line. Connect the third point to the opposite side of the original triangle. Now, draw a parallel line from the second point. This creates a smaller, similar triangle where the sides are two-thirds the size of the first triangle.

Exam Tip: When constructing similar triangles, always ensure that the parallel lines are drawn accurately. This maintains the proportionality of sides and the similarity of angles.

 

Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \( \frac{5}{8} \) of the corresponding sides of the first triangle.
Answer:

Construction Steps:

1. Make \( \triangle ABC \) with AB = 7 cm, AC = 5 cm, and BC = 6 cm.
2. From A, draw a sharp angle \( \angle BAX \) underneath AB.
3. Along AX, place 8 arc segments \( A_1, A_2, ..., A_8 \) so that \( AA_1 = A_1A_2 = ... = A_7A_8 \).
4. Connect \( A_8 \) with B.
5. Starting from \( A_5 \), draw a line \( A_5B' \parallel A_8B \), which meets AB at point B'.
6. From B', create \( B'C' \parallel BC \), which meets AC at point C'.

Therefore, \( \triangle AB'C' \) is the desired triangle whose sides are \( \frac{5}{8} \) of the corresponding sides of \( \triangle ABC \).

Justification:

We have \( \triangle AB'C' \sim \triangle ABC \) (by AA similarity Rule).
\( \implies \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC} \) ..........(1) (Sides are proportional)
However, by construction, \( A_5B' \parallel A_8B \).
Using the Basic Proportionality Theorem, \( \frac{AB'}{AB}=\frac{AA_5}{AA_8}=\frac{5}{8} \) ...........(2)
From (1) and (2), we get \( \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC}=\frac{5}{8} \).
In simple words: Draw a triangle with sides 5 cm, 6 cm, and 7 cm. Then, from one corner, draw another line and mark 8 equal points. Connect the 8th point to a corner of the main triangle. Now, draw a parallel line from the 5th point. This creates a new, smaller triangle where each side is five-eighths the length of the original.

Exam Tip: When the scale factor is less than 1, the constructed triangle will be smaller and inside the original triangle. Pay attention to which point to connect to the base and from which point to draw the parallel line, based on the numerator and denominator of the ratio.

 

Question 4. Construct an isosceles triangle whose base is 8 cm and altitude is 4 cm and then another triangle whose sides are \( 1\frac{1}{2} \) times the corresponding sides of the isosceles triangle. (CBSE 2017)
Answer:

Steps to Construct:

1. Sketch a line segment BC = 8 cm.
2. Make its perpendicular bisector AD, where AD = 4 cm.
3. By connecting AB and AC, we obtain the isosceles triangle ABC.
4. Create a sharp angle \( \angle CBX \) pointing downwards from B.
5. On BX, mark 3 arc segments \( B_1, B_2, B_3 \) such that \( BB_1 = B_1B_2 = B_2B_3 \).
6. Connect C to \( B_2 \) and create a line \( B_3C' \parallel B_2C \) (extend BC to C' if needed).
7. From C', construct \( C'A' \parallel CA \) (extend BA to A' if needed).

Therefore, \( \triangle A'BC' \) is the desired triangle.

Justification:

We have \( \triangle ABC \sim \triangle A'BC' \) (by AA similarity criteria).
\( \implies \frac{A'B}{AB}=\frac{A'C'}{AC}=\frac{BC'}{BC} \) ..........(1) (Sides are proportional)
Now, \( B_3C' \parallel B_2C \) (By construction).
Therefore, \( \triangle BB_3C' \sim \triangle BB_2C \) (by AA similarity criteria).
\( \implies \frac{BC'}{BC}=\frac{BB_3}{BB_2} \) (by Thales theorem)
However, by construction, \( \frac{BB_3}{BB_2}=\frac{3}{2} \) ...........(2)
From (1) and (2), we get \( \frac{A'B}{AB}=\frac{A'C'}{AC}=\frac{BC'}{BC}=\frac{3}{2} \).
In simple words: First, draw an isosceles triangle with an 8 cm base and a 4 cm height. Then, from one base corner, draw a line downwards and mark 3 equal points. Connect the second point to the top corner of the original triangle. Now, draw a parallel line from the third point. This will create a new, larger isosceles triangle, with sides one and a half times bigger than the first one.

Exam Tip: For constructions where the scale factor is greater than 1, the new triangle will be an enlargement, meaning it will be outside the original triangle. Remember to extend the sides of the original triangle if required.

 

Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and \( \angle ABC = 60^\circ \). Then construct a triangle whose sides are \( \frac{3}{4} \) of the corresponding sides of the ABC. (CBSE 2015)
Answer:

Construction Method:

1. Make a line segment BC = 6 cm.
2. From B, create angle \( \angle CBY = 60^\circ \) and measure AB = 5 cm along this line.
3. Connect AC. \( \triangle ABC \) is the initial triangle we need.
4. From B, make a sharp angle \( \angle CBX \) pointing downwards.
5. Along BX, place 4 arc marks \( B_1, B_2, B_3, B_4 \) so that \( BB_1 = B_1B_2 = B_2B_3 = B_3B_4 \).
6. Connect \( B_4 \) to C, and from \( B_3 \), draw line \( B_3C' \parallel B_4C \), meeting BC at C'.
7. From C', construct \( A'C' \parallel AC \), meeting AB at A'.

Then \( \triangle A'BC' \) is the required triangle whose sides are \( \frac{3}{4} \) of the corresponding sides of \( \triangle ABC \).

Justification:

\( C'A' \parallel AC \) (By construction)
Therefore, \( \triangle A'BC' \sim \triangle ABC \) (AA similarity Rule).
\( \implies \frac{A'B}{AB}=\frac{A'C'}{AC}=\frac{BC'}{BC} \) ..........(1) (Sides are proportional)
However, \( B_3C' \parallel B_4C \) (By construction).
Therefore, \( \frac{BC'}{BC}=\frac{BB_3}{BB_4} \) ...........(2) (Thales theorem)
By construction, \( \frac{BB_3}{BB_4}=\frac{3}{4} \).
From (1) and (2), we get \( \frac{A'B}{AB}=\frac{BC'}{BC}=\frac{A'C'}{AC}=\frac{3}{4} \).
In simple words: Draw a triangle ABC with specific side lengths and an angle. Then, from point B, draw a line downwards and mark 4 equal sections. Connect the 4th mark to C. From the 3rd mark, draw a line parallel to the one you just drew. This will create a smaller triangle inside the first one, where each side is three-fourths the length of the original.

Exam Tip: When given two sides and an included angle (SAS criteria), always construct the base first, then the angle, and then mark the second side before joining the third. Make sure the parallel lines are drawn accurately.

 

Question 6. Draw a triangle ABC with side BC = 7 cm, \( \angle B = 45^\circ \), \( \angle A = 105^\circ \). Then construct a triangle whose sides are \( \frac{3}{4} \) times the corresponding sides of the ∆ΑΒC. (CBSE 2012, 2017)
Answer:

Method of Construction:

1. Create \( \triangle ABC \) with side BC = 7 cm, \( \angle B = 45^\circ \), \( \angle C = 30^\circ \) (since \( \angle C = 180^\circ - 45^\circ - 105^\circ = 180^\circ - 150^\circ = 30^\circ \)).
2. From B, make a sharp angle \( \angle CBX \) below line BC.
3. On BX, place 4 arc segments \( B_1, B_2, B_3, B_4 \) such that \( BB_1 = B_1B_2 = B_2B_3 = B_3B_4 \).
4. Connect \( B_4 \) and C.
5. Starting from \( B_3 \), draw a line \( B_3C' \parallel B_4C \), which meets BC at point C'.
6. Construct C'A' parallel to CA (extend BA to A' if necessary).

Therefore, \( \triangle A'BC' \) is the desired triangle.

Justification:

\( C'A' \parallel CA \) (By construction)
Therefore, \( \triangle ABC \sim \triangle A'BC' \) (AA similarity Criteria).
\( \implies \frac{A'B}{AB}=\frac{A'C'}{AC}=\frac{BC'}{BC} \) ..........(1) (Corresponding sides are proportional)
Now, \( B_3C' \parallel B_4C \) (By construction).
Therefore, \( \frac{BC'}{BC}=\frac{BB_3}{BB_4} \) ..........(2) (Thales theorem)
However, by construction, \( \frac{BB_3}{BB_4}=\frac{3}{4} \) ..........(3)
From (1) and (2), we get \( \frac{A'B}{AB}=\frac{A'C'}{AC}=\frac{BC'}{BC}=\frac{3}{4} \).
In simple words: First, find the third angle of triangle ABC. Then draw triangle ABC with the given information. Draw another line from B downwards and mark 4 equal parts. Join the 4th mark to C. Now, draw a parallel line from the 3rd mark to BC. This creates a smaller, similar triangle inside the first one, with sides three-fourths the original size.

Exam Tip: When two angles are given, first calculate the third angle using the angle sum property of a triangle. This helps in drawing the triangle more accurately.

 

Question 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. (CBSE 2010)
Answer:

Construction Steps:

1. Make a line segment AB = 4 cm.
2. By forming a right angle at B, draw a line BC that is 3 cm long.
3. Join AC. \( \triangle ABC \) is the required right-angled triangle.
4. From A, draw a sharp angle \( \angle BAX \) pointing downwards.
5. On AX, mark 5 arc segments \( A_1, A_2, A_3, A_4, A_5 \) such that \( AA_1 = A_1A_2 = ... = A_4A_5 \).
6. Join \( A_3 \) to B.
7. From \( A_5 \), draw \( A_5B' \parallel A_3B \) (Extend AB to B').
8. From B', draw \( B'C' \parallel BC \) (Extend AC to C').

Hence, \( \triangle AB'C' \) is the required triangle.

Justification:

We have \( \triangle A'BC' \sim \triangle ABC \) (by AA similarity Rule, as \( A'C' \parallel AC \) and \( B'C' \parallel BC \)).
\( \implies \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC} \) ..........(1) (Sides are proportional)
However, by construction, \( A_5B' \parallel A_3B \).
Using the Basic Proportionality Theorem, \( \frac{AB'}{AB}=\frac{AA_5}{AA_3}=\frac{5}{3} \) ..........(2)
From (1) and (2), we get \( \frac{AB'}{AB}=\frac{AC'}{AC}=\frac{B'C'}{BC}=\frac{5}{3} \).
In simple words: First, draw a right triangle with two sides measuring 4 cm and 3 cm. Then, from one of its sharp corners, draw a line downwards and mark 5 equal sections. Connect the 3rd mark to the opposite corner of the main triangle. Now, from the 5th mark, draw a line parallel to the one you just drew, and extend the side of the main triangle. This creates a larger, similar triangle with sides five-thirds the size of the original.

Exam Tip: For right-angled triangle constructions, drawing one of the non-hypotenuse sides as the base helps simplify the process. Always double-check the scale factor to determine if the new triangle will be larger or smaller and if extensions are needed.