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Learn Exercise 11.1 with Free Lessons & Tips

Question 1:

Draw a line segment of length 7.6 cm and divide it in the ratio 5:3. Measure the two parts.

Solution 1:

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 13 (= 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 $\angle AA_{13}B$ ) 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.

Comments

Abhay

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.

Solution 2:

Step 1

Draw a line segment AB = 4 cm. Taking point A as centre, draw an 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 $\Delta ABC$ is the required triangle.

Step 2

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

Step 3

Locate 3 points $A_{1},A_{2},A_{3}$ (as 3 is greater between 2 and 3) on line AX such that $AA_{1}=A_{1}A_{2}=A_{2}A_{3}$ .

Step 4

Join $BA_{3}$ and draw a line through $A_{2}$ parallel to $BA_{3}$ to intersect AB at point B'.

Step 5

Draw a line through B' parallel to the line BC to intersect AC at C'.

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

Comments

Abhay

Question 3:

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.

Solution 3:

Step 1

Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. $\Delta ABC$ is the required triangle having the length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 2

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

Step 3

Locate 7 points, $A_{1},A_{2},A_{3},A_{4},A_{5},A_{6},A_{7}$ (as 7 is greater between 5and 7), on line AX such that $AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}=A_{5}A_{6}=A_{6}A_{7}$ .

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'.

Step 5

Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. $\Delta AB'C'$ is the required triangle.

Comments

Abhay

Question 4:

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

Solution 4:

Let us assume that $\Delta ABC$ is an isosceles triangle having CA and CB of equal lengths, base AB of 8 cm, and AD is the altitude of 4 cm.

A $\Delta AB'C'$ whose sides are times of $\Delta ABC$ can be drawn as follows.

Step 1

Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the line segment while taking point A and B 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 ΔABC 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) $A_{1},A_{2}$ , 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'.

Step 6

Draw a line through B' parallel to BC intersecting the extended line segment AC at C'. $\Delta AB'C'$ is the required triangle.

Comments

Abhay

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 triangle ABC.

Solution 5:

A $\Delta AB'C'$ whose sides are $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$ can be drawn as follows.

Step 1

Draw a $\Delta ABC$ with side BC = 6 cm, 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 $B_{4}C$ intersecting BC at C'.

Step 5

Draw a line through C' parallel to AC intersecting AB at A'. $\Delta A'BC'$ is the required triangle.

Comments

Abhay

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 $\frac{4}{3}$ times the corresponding sides of $\triangle ABC$ .

Solution 6:

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

The required triangle can be drawn as follows.

Step 1

Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.

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 4 and 3), B1, B2, B3, B4, on BX.

Step 4

Join B3C. Draw a line through B4 parallel to B3C intersecting extended BC at C'.

Step 5

Through C', draw a line parallel to AC intersecting extended line segment at C'. ΔA'BC' is the required triangle.

Comments

Abhay

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.

Solution 7:

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° with it.

Step 2

Draw an arc of 3 cm radius while taking A as its centre to intersect SA at C. Join BC. ΔABC 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), A₁, A₂, A₃, A₄, A₅, on line segment AX such that AA₁ = A₁A₂ = A₂A₃ = A₃A₄ = A₄A₅.

Step 5

Join A₃B. Draw a line through A₅ parallel to A₃B intersecting extended line segment AB at B'.

Step 6

Through B', draw a line parallel to BC intersecting extended line segment AC at C'. ΔAB'C' is the required triangle.

Comments

Abhay

All Exercises - Chapter 11 - Constructions

Exercise 11.1

Exercise 11.2

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Learn Exercise 11.1 with Free Lessons & Tips

All Exercises - Chapter 11 - Constructions

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