CBSE 10th Maths - constructions - Exercise 11.1

Solution:

Construction Procedure

A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows.

Step 1: Draw line segment AB with a length measure of 7.6 cm.

Step 2: Draw a ray AX that makes an acute angle with line segment AB.

Step 3: Locate the points, i.e.,13 (= 5+8) points, such as $A_1, A_2, A_3, A_4 …….. A_{13}$, on the ray AX, such that it becomes $AA_1 = A_1A_2 = A_2A_3$ and so on.

Step 4: Join the line segment and the ray, $BA_{13}$.

Step 5: Through the point $A_5$, draw a line parallel to $BA_{13}$ which makes an angle equal to $\angle AA_{13}B$.

Step 6: Point $A_5$, which intersects line AB at point C.

Step 7: C is the point that divides line segment AB of 7.6 cm in the required ratio of 5:8.

Step 8: Now, measure the lengths of the line AC and CB. It becomes the measure of 2.9 cm and 4.7 cm, respectively.

Justification :

The construction of the given problem can be justified by proving that

$\dfrac{AC}{CB} = \dfrac{5}{8}$

By construction, we have A5C || A13B. From the Basic proportionality theorem for the triangle AA13B, we get

$\dfrac{AC}{CB} = \dfrac{AA_5}{A_5A_{13}}$….. (1)

From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments, respectively.

Therefore, it becomes

$\dfrac{AA_5}{A_5A_{13}} = \dfrac{5}{8}$… (2)

Compare the equations (1) and (2), we obtain

$\dfrac{AC}{CB} = \dfrac{5}{8}$

Hence, justified.

Solution:

Construction Procedure

Step 1: Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.

Step 2: Take point A as the centre, and draw an arc of radius 5 cm.

Step 3: Similarly, take point B as its centre, and draw an arc of radius 6 cm.

Step 4: The arcs drawn will intersect each other at point C.

Step 5: Now, we have obtained AC = 5 cm and BC = 6 cm, and therefore, ΔABC is the required triangle.

Step 6: Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

Step 7: Locate 3 points such as $A_1, A_2, and A_3$ (as 3 is greater between 2 and 3) on line AX such that it becomes $AA_1= A_1A_2 = A_2A_3$.

Step 8: Join point $BA_3$ and draw a line through $A_2$, which is parallel to the line $BA_3$ that intersects AB at point B’.

Step 9: Through the point B’, draw a line parallel to line BC that intersects the line AC at C’.

Step 10: Therefore, ΔAB’C’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

AB’ = $(\dfrac{2}{3})$AB

B’C’ = $(\dfrac{2}{3})$BC

AC’= $(\dfrac{2}{3})$AC

From the construction, we get B’C’ || BC

$\angle AB’C’ = \angle ABC$ (Corresponding angles)

In ΔAB’C’ and ΔABC,

$\angle ABC = \angle AB’C$ (Proved above)

$\angle BAC = \angle B’AC$’ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, $\dfrac{AB’}{AB} = \dfrac{B’C’}{BC} = \dfrac{AC’}{AC}$ …. (1)

In ΔAAB’ and ΔAAB,

$\angle A_2AB’ =\angle A_3AB$ (Common)

From the corresponding angles, we get

$\angle AA_2B’ =\angle AA_3B$

Therefore, from the AA similarity criterion, we obtain

$ΔAA_2B’ and AA_3B$

So,$\dfrac{ AB’}{AB} = \dfrac{AA_2}{AA_3}$

Therefore, $\dfrac{AB’}{AB} = \dfrac{2}{3}$ ……. (2)

From equations (1) and (2), we get

$\dfrac{AB’}{AB} = \dfrac{B’C’}{BC} = \dfrac{AC’}{ AC} = \dfrac{2}{3}$

This can be written as

AB’ = $(\dfrac{2}{3})$AB

B’C’ = $(\dfrac{2}{3})$BC

AC’=$ (\dfrac{2}{3})$AC

Hence, justified.

Solution:

Construction Procedure

Step 1: Draw a line segment AB =5 cm.

Step 2: Take A and B as the centre, and draw the arcs of radius 6 cm and 7 cm, respectively.

Step 3: These arcs will intersect each other at point C, and therefore, ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm, respectively.

Step 4: Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.

Step 5: Locate the 7 points, such as $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ (as 7 is greater between 5 and 7), on line AX such that it becomes $AA_1 = A_1A_2 = A_2A_3 = A_3A_4 = A_4A_5 = A_5A_6 = A_6A_7$

Step 6: Join the points $BA_5$ and draw a line from $A_7 to BA_5$, which is parallel to the line $BA_5$ where it intersects the extended line segment AB at point B’.

Step 7: Now, draw a line from B’ to the extended line segment AC at C’, which is parallel to the line BC, and it intersects to make a triangle.

Step 8: Therefore, ΔAB’C’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

AB’ = $(\dfrac{7}{5})AB$

B’C’ = $(\dfrac{7}{5})BC$

AC’= $(\dfrac{7}{5})AC$

From the construction, we get B’C’ || BC

$ \angle AB’C’ = \angle ABC$ (Corresponding angles)

In ΔAB’C’ and ΔABC,

$\angle ABC = \angle AB’C$ (Proved above)

$\angle BAC = \angle B’AC’$ (Common)

∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)

Therefore, $\dfrac{AB’}{AB} = \dfrac{B’C’}{BC} = \dfrac{AC’}{AC}$ …. (1)

In $ΔAA_7B’ and ΔAA_5B$,

$\angle A_7AB’=\angle A_5AB$ (Common)

From the corresponding angles, we get

$\angle A A_7B’=\angle A A_5B$

Therefore, from the AA similarity criterion, we obtain

$ΔA A_2B’ and A A_3B$

So, $\dfrac{AB’}{AB} = \dfrac{AA_5}{AA_7}$

Therefore, $\dfrac{AB}{AB}’ =\dfrac{ 5}{7}$ ……. (2)

From equations (1) and (2), we get

$\dfrac{AB’}{AB} = \dfrac{B’C’}{BC} = \dfrac{AC’}{AC} = \dfrac{7}{5}$

This can be written as

AB’ = $(\dfrac{7}{5})$AB

B’C’ = $(\dfrac{7}{5})$BC

AC’= $(\dfrac{7}{5})$AC

Hence, justified.

Solution:

Construction Procedure:

Step 1: Draw a line segment BC with a measure of 8 cm.

Step 2: Now, draw the perpendicular bisector of the line segment BC and intersect at point D.

Step 3: Take the point D as the centre and draw an arc with a radius of 4 cm, which intersects the perpendicular bisector at the point A.

Step 4: Now, join the lines AB and AC, and the triangle is the required triangle.

Step 5: Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.

Step 6: Locate the 3 points $B_1, B_2$ and $B_3$ on the ray BX such that $BB_1 = B_1B_2 = B_2B_3$

Step 7: Join the points $B_2C$ and draw a line from $B_3$, which is parallel to the line $B_2C$ where it intersects the extended line segment BC at point C’.

Step 8: Now, draw a line from C’ to the extended line segment AC at A’, which is parallel to the line AC, and it intersects to make a triangle.

Step 9: Therefore, ΔA’BC’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

A’B = $(\dfrac{3}{2})AB$

BC’ = $(\dfrac{3}{2})BC$

A’C’= $(\dfrac{3}{2})AC$

From the construction, we get A’C’ || AC

$ \angle A’C’B = \angle ACB$ (Corresponding angles)

In ΔA’BC’ and ΔABC,

$\angle B = \angle B$ (Common)

$\angle A’BC’ = \angle ACB$

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Therefore, $\dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC}$

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

$\dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC} = \dfrac{3}{2}$

Hence, justified.

Solution:

Construction Procedure

Step 1: Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and $\angle ABC$ = 60°.

Step 2: Draw a ray BX which makes 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), such as $B_1, B_2, B_3, B_4$, on line segment BX.

Step 4: Join the points $B_4C$ and also draw a line through $B_3$, parallel to $B_4C$ intersecting the line segment BC at C’.

Step 5: Draw a line through C’ parallel to the line AC, which intersects the line AB at A’.

Step 6: Therefore, ΔA’BC’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

Since the scale factor is $\dfrac{3}{4}$, we need to prove

A’B = $(\dfrac{3}{4})AB$

BC’ = $(\dfrac{3}{4})BC$

A’C’= $(\dfrac{3}{4})AC$

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

$ \angle A’C’B = \angle ACB$ (Corresponding angles)

$\angle B = \angle B$ (Common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, $\dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC}$

So, it becomes $\dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC} = \dfrac{3}{4}$

Hence, justified.

Solution:

To find $\angle C$:

Given:

$\angle B$ = 45°, $\angle A$ = 105°

We know that,

The sum of all interior angles in a triangle is 180°.

$\angle A+\angle B +\angle C = 180°$

105°+45°+$\angle C$ = 180°

$\angle C$ = 180° − 150°

$\angle C$ = 30°

So, from the property of the triangle, we get $\angle C$ = 30°

Construction Procedure

The required triangle can be drawn as follows.

Step 1: Draw a ΔABC with side measures of base BC = 7 cm, $\angle B$ = 45°, and $\angle C$ = 30°.

Step 2: Draw a ray BX that makes 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), such as $B_1, B_2, B_3, B_4$, on the ray BX.

Step 4: Join the points $B_3C$.

Step 5: Draw a line through $B_4$ parallel to $B_3C$, which intersects the extended line BC at C’.

Step 6: Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.

Step 7: Therefore, ΔA’BC’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

Since the scale factor is $\dfrac{4}{3}$, we need to prove

A’B = $(\dfrac{4}{3})AB$

BC’ =$ (\dfrac{4}{3})BC$

A’C’= $(\dfrac{4}{3})AC$

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

$ \angle A’C’B = \angle ACB$ (Corresponding angles)

$\angle B = \angle B$ (Common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore,$ \dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC}$

So, it becomes $\dfrac{A’B}{AB} = \dfrac{BC’}{BC}= \dfrac{A’C’}{AC} = \dfrac{4}{3}$

Hence, justified.

Solution:

Given:

The sides other than the hypotenuse are of lengths 4cm and 3cm. It defines that the sides are perpendicular to each other

Construction Procedure

The required triangle can be drawn as follows.

Step 1: Draw a line segment BC =3 cm.

Step 2: Now, measure and draw an angle 90°

Step 3: Take B as the centre and draw an arc with a radius of 4 cm, and intersects the ray at point B.

Step 4: Now, join the lines AC, and the triangle ABC is the required triangle.

Step 5: Draw a ray BX that makes an acute angle with BC on the opposite side of vertex A.

Step 6: Locate 5 such as $B_1, B_2, B_3, B_4$, on the ray BX, such that $BB_1 = B_1B_2 = B_2B_3= B_3B_4 = B_4B_5$

Step 7: Join the points $B_3C$.

Step 8: Draw a line through $B_5$ parallel to $B_3C$, which intersects the extended line BC at C’.

Step 9: Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.

Step 10: Therefore, ΔA’BC’ is the required triangle.

Justification

The construction of the given problem can be justified by proving that

Since the scale factor is $\dfrac{5}{3}$, we need to prove

A’B = $(\dfrac{5}{3})AB$

BC’ = $(\dfrac{5}{3})BC$

A’C’= $(\dfrac{5}{3})AC$

From the construction, we get A’C’ || AC

In ΔA’BC’ and ΔABC,

$ \angle A’C’B = \angle ACB$ (Corresponding angles)

$\angle B = \angle B$ (Common)

∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the corresponding sides of the similar triangle are in the same ratio, it becomes

Therefore, $\dfrac{A’B}{AB} = \dfrac{BC’}{BC}= \dfrac{A’C’}{AC}$

So, it becomes $\dfrac{A’B}{AB} = \dfrac{BC’}{BC} = \dfrac{A’C’}{AC} = \dfrac{5}{3}$

Hence, justified.