# Class 10 NCERT Solutions- Chapter 11 Constructions – Exercise 11.1

**In each of the following, give the justification of the construction also:**

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

**Solution:**

Steps of construction:To divide the line segment of 7.6 cm in the ratio of 5 : 8.

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

Step 2.Draw a ray AC which forms an acute angle with the line segment AB.

Step 3.Mark the points = 13 as (5+8=13) points, such as A_{1}, A_{2}, A_{3}, A_{4}…….. A_{13}, on the ray AC such that it becomes AA_{1}= A_{1}A_{2}= A_{2}A_{3}and such like this.

Step 4.Now join the line segment and the ray, BA_{13}.

Step 5.Hence, the point A_{5}, construct a line parallel to BA_{13}which makes an angle equal to ∠AA_{13}B.

Step 6.Point A_{5}intersects the line AB at point X.

Step 7. X is that point which divides line segment AB into the ratio of 5:8.

Step 8.Thus, measure the lengths of the line AX and XB. Hence, it measures 2.9 cm and 4.7 cm respectively.

Justification:The construction can be justified by proving that

From construction, we have A

_{5}X || A_{13}B. By the Basic proportionality theorem for the triangle AA_{13}B, we will get….. (1)

By the figure we have constructed, it can be seen that AA

_{5}and A_{5}A_{13}contains 5 and 8 equal divisions of line segments respectively.Thus,

… (2)

Comparing the equations (1) and (2), we get

Thus, Justified.

**Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.**

**Solution:**

Steps of Construction:

Step 1.Draw a line segment XY which measures 4 cm, So XY = 4 cm.

Step 2.Taking point X as centre, and construct an arc of radius 5 cm.

Step 3.Similarly, from the point Y as centre, and draw an arc of radius 6 cm.

Step 4.Thus, the following arcs drawn will intersect each other at point Z.

Step 5.Now, we have XZ = 5 cm and YZ = 6 cm and therefore ΔXYZ is the required triangle.

Step 6.Draw a ray XA which will make an acute angle along the line segment XY on the opposite side of vertex Z.

Step 7.Mark the 3 points such as X_{1}, X_{2}, X_{3}(as 3 is greater between 2 and 3) on line XA such that it becomes XX_{1}= X_{1}X_{2}= X_{2}X_{3}.

Step 8.Join the point YX_{3}and construct a line through X_{2 }which is parallel to the line YX_{3}that intersect XY at point Y’.

Step 9.From the point Y’, construct a line parallel to the line YZ that intersect the line XZ at Z’.

Step 10.Hence, ΔXY’Z’ is the required triangle.

Justification:The construction can be justified by proving that

From the construction, we get Y’Z’ || YZ

∴ ∠XY’Z’ = ∠XYZ (Corresponding angles)

In ΔXY’Z’ and ΔXYZ,

∠XYZ = ∠XY’Z (Proved above)

∠YXZ = ∠Y’XZ’ (Common)

∴ ΔXY’Z’ ∼ ΔXYZ (From AA similarity criterion)

Therefore,

…. (1)

In ΔXXY’ and ΔXXY,

∠X

_{2}XY’ =∠X_{3}XY (Common)From the corresponding angles, we get,

∠AA

_{2}B’ =∠AA_{3}BThus, by the AA similarity criterion, we get

ΔXX

_{2}Y’ and XX_{3}YSo,

Therefore, ……. (2)

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

It is written as

XY’ =

Y’Z’ =

XZ’=

Therefore, justified.

**Question 3. Construct a triangle with sides 5 cm, 6 cm**,** and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle**

**Solution:**

Steps of construction:

Step 1.Construct a line segment XY =5 cm.

Step 2.By taking X and Y as centre, and construct the arcs of radius 6 cm and 5 cm respectively.

Step 3.These two arcs will intersect each other at point Z and hence ΔXYZ is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.

Step 4.Construct a ray XA which will make an acute angle with the line segment XY on the opposite side of vertex Z.

Step 5.Pinpoint the 7 points such as X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6}, X_{7}(as 7 is greater between 5 and 7), on the line XA such that it becomes XX_{1}= X_{1}X_{2}= X_{2}X_{3}= X_{3}X_{4}= X_{4}X_{5}= X_{5}X_{6}= X_{6}X_{7}

Step 6.Join the points YX_{5}and construct a line from X_{7}to YX_{5}that is parallel to the line YX_{5}where it intersects the extended line segment XY at point Y’.

Step 7.Now, construct a line from Y’ the extended line segment XZ at Z’ that is parallel to the line YZ, and it intersects to make a triangle.

Step 8.Hence, ΔXY’Z’ is the needed triangle.

Justification:The construction can be justified by proving that

XY’ =

Y’Z’ =

XZ’=

By the construction, we have Y’Z’ || YZ

Therefore,

∠XY’Z’ = ∠XYZ {Corresponding angles}

In ΔXY’Z’ and ΔXYZ,

∠XYZ = ∠XY’Z {As shown above}

∠YXZ = ∠Y’XZ’ {Common}

Therefore,

ΔXY’Z’ ∼ ΔXYZ { By AA similarity criterion}

Therefore,

…. (1)

In ΔXX

_{7}Y’ and ΔXX_{5}Y,∠X

_{7}XY’=∠X_{5}XY (Common)From the corresponding angles, we will get,

∠XX

_{7}Y’=∠XX_{5}YHence, By the AA similarity criterion, we will get

ΔXX

_{2}Y’ and XX_{3}YThus,

Hence, ……. (2)

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

It can be also shown as

XY’ =

Y’Z’ =

XZ’=

Thus, justified.

**Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are **** times the corresponding sides of the isosceles triangle**

**Solution:**

Steps of construction:

Step 1.Construct a line segment YZ of 8 cm.

Step 2.Now construct the perpendicular bisector of the line segment YZ and intersect at the point A.

Step 3.Taking the point A as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point X.

Step 4.Join the lines XY and XZ and the triangle is the required triangle.

Step 5.Construct a ray YB which makes an acute angle with the line YZ on the side opposite to the vertex X.

Step 6.Mark the 3 points Y_{1}, Y_{2}and Y_{3}on the ray YB such that YY_{1}= Y_{1}Y_{2}= Y_{2}Y_{3}

Step 7.Join the points Y_{2}Z and construct a line from Y_{3}which is parallel to the line Y_{2}Z where it intersects the extended line segment YZ at point Z’.

Step 8.Now, draw a line from Z’ the extended line segment XZ at X’, that is parallel to the line XZ, and it intersects to make a triangle.

Step 9.Hence, ΔX’YZ’ is the required triangle.

Justification:The construction can be justified by proving that

X’Y =

YZ’ =

X’Z’=

By the construction, we will obtain X’Z’ || XZ

Therefore,

∠ X’Z’Y = ∠XZY {Corresponding angles}

In ΔX’YZ’ and ΔXYZ,

∠Y = ∠Y (common)

∠X’YZ’ = ∠XZY

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Hence,

Thus, the corresponding sides of the similar triangle are in the same ratio, we get

Thus, justified.

**Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.**

**Solution:**

Steps of construction:

Step 1.Construct a ΔXYZ with base side YZ = 6 cm, and XY = 5 cm and ∠XYZ = 60°.

Step 2.Construct a ray YA that makes an acute angle with YZ on the opposite side of vertex X.

Step 3.Mark 4 points (as 4 is greater in 3 and 4), such as Y_{1}, Y_{2}, Y_{3}, Y_{4}, on line segment YA.

Step 4.Join the points Y_{4}Z and construct a line through Y_{3}, parallel to Y_{4}Z intersecting the line segment YZ at Z’.

Step 5.Construct a line through Z’ parallel to the line XZ which intersects the line XY at X’.

Step 6.Therefore, ΔX’YZ’ is the required triangle.

Justification:The construction can be justified by proving that

Since here the scale factor is ,

We need to prove

X’Y =

YZ’ =

X’Z’=

From the construction, we will obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore,

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

Thus, the corresponding sides of the similar triangle are in the same ratio, we get

Therefore,

Thus, it becomes

Hence, justified.

**Question 6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.**

**Solution:**

To find ∠Z:

Given:

∠Y = 45°, ∠X = 105°

∠X+∠Y +∠Z = 180° {Sum of all interior angles in a triangle is 180°}

105°+45°+∠Z = 180°

∠Z = 180° − 150°

∠Z = 30°

Thus, from the property of triangle, we get ∠Z = 30°

Steps of construction:

Step 1.Construct a ΔXYZ with side measures of base YZ = 7 cm, ∠Y = 45°, and ∠Z = 30°.

Step 2.Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.

Step 3.Mark 4 points (as 4 is greater in 4 and 3), such as Y_{1}, Y_{2}, Y_{3}, Y_{4}, on the ray YA.

Step 4.Join the points Y_{3}Z.

Step 5.Construct a line through Y_{4}parallel to Y_{3}Z which intersects the extended line YZ at Z’.

Step 6.Through Z’, construct a line parallel to the line YZ that intersects the extended line segment at Z’.

Step 7.Hence, ΔX’YZ’ is the required triangle.

Justification:The construction can be justified by proving that

Here the scale factor is , we have to prove

X’Y =

YZ’ =

X’Z’=

From the construction, we obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore.

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

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

Therefore,

We get,

Thus, justified.

**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 5/3 times the corresponding sides of the given triangle.**

**Solution:**

Given:

The sides other than hypotenuse are of lengths 4cm and 3cm. Hence, the sides are perpendicular to each other.

Step of construction:

Step 1.Construct a line segment YZ =3 cm.

Step 2.Now measure and draw ∠= 90°

Step 3.Now taking Y as centre and draw an arc with the radius of 4 cm and intersects the ray at the point Y.

Step 4.Join the lines XZ and the triangle XYZ is the required triangle.

Step 5.Construct a ray YA makes an acute angle with YZ on the opposite side of vertex X.

Step 6.Mark 5 such as Y_{1}, Y_{2}, Y_{3}, Y_{4}, on the ray YA such that YY_{1}= Y_{1}Y_{2}= Y_{2}Y_{3}= Y_{3}Y_{4}= Y_{4}Y_{5}

Step 7.Join the points Y_{3}Z.

Step 8.Construct a line through Y_{5}parallel to Y_{3}Z which intersects the extended line YZ at Z’.

Step 9.Through Z’, draw a line parallel to the line XZ that intersects the extended line XY at X’.

Step 10.Therefore, ΔX’YZ’ is the required triangle.

Justification:The construction can be justified by proving that

Here the scale factor is , we need to prove

X’Y =

YZ’ =

X’Z’=

From the construction, we obtain X’Z’ || XZ

In ΔX’YZ’ and ΔXYZ,

Therefore,

∠X’Z’Y = ∠XZY {Corresponding angles}

∠Y = ∠Y {common}

Therefore,

ΔX’YZ’ ∼ ΔXYZ {By AA similarity criterion}

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

Therefore,

So, it becomes

Therefore, justified.

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