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

• Last Updated : 07 Nov, 2021

### 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 A1, A2, A3, A4 â€¦â€¦.. A13, on the ray AC such that it becomes AA1 = A1A2 = A2A3 and such like this.

Step 4. Now join the line segment and the ray, BA13.

Step 5. Hence, the point A5, construct a line parallel to BA13 which makes an angle equal to âˆ AA13B.

Step 6. Point A5 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 A5X || A13B. By the Basic proportionality theorem for the triangle AA13B, we will get

â€¦.. (1)

By the figure we have constructed, it can be seen that AA5 and A5A13 contains 5 and 8 equal divisions of line segments respectively.

Thus,

â€¦ (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 Î”XYZ 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 X1, X2, X3 (as 3 is greater between 2 and 3) on line XA such that it becomes XX1 = X1X2 = X2X3.

Step 8. Join the point YX3 and construct a line through X2 which is parallel to the line YX3 that intersect XY at point Yâ€™.

Step 9. From the point Yâ€™, construct a line parallel to the line YZ that intersect the line XZ at Zâ€™.

Step 10. Hence, Î”XYâ€™Zâ€™ is the required triangle.

Justification:

The construction can be justified by proving that

From the construction, we get Yâ€™Zâ€™ || YZ

âˆ´ âˆ XYâ€™Z’ = âˆ XYZ (Corresponding angles)

In Î”XYâ€™Zâ€™ and Î”XYZ,

âˆ XYZ = âˆ XYâ€™Z (Proved above)

âˆ YXZ = âˆ Yâ€™XZâ€™ (Common)

âˆ´ Î”XYâ€™Zâ€™ âˆ¼ Î”XYZ (From AA similarity criterion)

Therefore,

â€¦. (1)

In Î”XXYâ€™ and Î”XXY,

âˆ X2XYâ€™ =âˆ X3XY (Common)

From the corresponding angles, we get,

âˆ AA2Bâ€™ =âˆ AA3B

Thus, by the AA similarity criterion, we get

Î”XX2Yâ€™ and XX3Y

So,

Therefore, â€¦â€¦. (2)

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

It is written as

XYâ€™ =

Yâ€™Zâ€™ =

XZâ€™=

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 Î”XYZ 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 X1, X2, X3, X4, X5, X6, X7 (as 7 is greater between 5 and 7), on the line XA such that it becomes XX1 = X1X2 = X2X3 = X3X4 = X4X5 = X5X6 = X6X7

Step 6. Join the points YX5 and construct a line from X7 to YX5 that is parallel to the line YX5 where it intersects the extended line segment XY at point Yâ€™.

Step 7. Now, construct a line from Yâ€™ the extended line segment XZ at Zâ€™ that is parallel to the line YZ, and it intersects to make a triangle.

Step 8. Hence, Î”XYâ€™Zâ€™ is the needed triangle.

Justification:

The construction can be justified by proving that

XYâ€™ =

Yâ€™Zâ€™ =

XZâ€™=

By the construction, we have Yâ€™Zâ€™ || YZ

Therefore,

âˆ XYâ€™Zâ€™ = âˆ XYZ {Corresponding angles}

In Î”XYâ€™Zâ€™ and Î”XYZ,

âˆ XYZ = âˆ XYâ€™Z   {As shown above}

âˆ YXZ = âˆ Yâ€™XZâ€™   {Common}

Therefore,

Î”XYâ€™Zâ€™ âˆ¼ Î”XYZ   { By AA similarity criterion}

Therefore,

â€¦. (1)

In Î”XX7Yâ€™ and Î”XX5Y,

âˆ X7XYâ€™=âˆ X5XY (Common)

From the corresponding angles, we will get,

âˆ XX7Yâ€™=âˆ XX5Y

Hence, By the AA similarity criterion, we will get

Î”XX2Yâ€™ and XX3Y

Thus,

Hence,  â€¦â€¦. (2)

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

It can be also shown as

XYâ€™ =

Yâ€™Zâ€™ =

XZâ€™=

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 Y1, Y2 and Y3 on the ray YB such that YY1 = Y1Y2 = Y2Y3

Step 7. Join the points Y2Z and construct a line from Y3 which is parallel to the line Y2Z where it intersects the extended line segment YZ at point Zâ€™.

Step 8. Now, draw a line from Zâ€™ the extended line segment XZ at Xâ€™, that is parallel to the line XZ, and it intersects to make a triangle.

Step 9. Hence, Î”Xâ€™YZâ€™ is the required triangle.

Justification:

The construction can be justified by proving that

X’Y =

YZâ€™ =

X’Zâ€™=

By the construction, we will obtain Xâ€™Zâ€™ || XZ

Therefore,

âˆ  Xâ€™Zâ€™Y = âˆ XZY {Corresponding angles}

In Î”Xâ€™YZâ€™ and Î”XYZ,

âˆ Y = âˆ Y (common)

âˆ Xâ€™YZâ€™ = âˆ XZY

Therefore,

Î”Xâ€™YZâ€™ âˆ¼ Î”XYZ {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 âˆ ABC = 60Â°. 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 Î”XYZ with base side YZ = 6 cm, and XY = 5 cm and âˆ XYZ = 60Â°.

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 Y1, Y2, Y3, Y4, on line segment YA.

Step 4. Join the points Y4Z and construct a line through Y3, parallel to Y4Z intersecting the line segment YZ at Zâ€™.

Step 5. Construct a line through Zâ€™ parallel to the line XZ which intersects the line XY at Xâ€™.

Step 6. Therefore, Î”Xâ€™YZâ€™ 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 =

YZâ€™ =

X’Zâ€™=

From the construction, we will obtain Xâ€™Zâ€™ || XZ

In Î”Xâ€™YZâ€™ and Î”XYZ,

Therefore,

âˆ X’Zâ€™Y = âˆ XZY {Corresponding angles}

âˆ Y = âˆ Y {common}

Therefore,

Î”Xâ€™YZâ€™ âˆ¼ Î”XYZ {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, âˆ  B = 45Â°, âˆ  A = 105Â°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of âˆ† ABC.

Solution:

To find âˆ Z:

Given:

âˆ Y = 45Â°, âˆ X = 105Â°

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

105Â°+45Â°+âˆ Z = 180Â°

âˆ Z = 180Â° âˆ’ 150Â°

âˆ Z = 30Â°

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

Steps of construction:

Step 1. Construct a Î”XYZ with side measures of base YZ = 7 cm, âˆ Y = 45Â°, and âˆ Z = 30Â°.

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 Y1, Y2, Y3, Y4, on the ray YA.

Step 4. Join the points Y3Z.

Step 5. Construct a line through Y4 parallel to Y3Z which intersects the extended line YZ at Zâ€™.

Step 6. Through Zâ€™, construct a line parallel to the line YZ that intersects the extended line segment at Zâ€™.

Step 7. Hence, Î”Xâ€™YZâ€™ is the required triangle.

Justification:

The construction can be justified by proving that

Here the scale factor is , we have to prove

X’Y =

YZâ€™ =

X’Zâ€™=

From the construction, we obtain Xâ€™Zâ€™ || XZ

In Î”Xâ€™YZâ€™ and Î”XYZ,

Therefore.

âˆ Xâ€™Zâ€™Y = âˆ XZY {Corresponding angles}

âˆ Y = âˆ Y {common}

Therefore,

Î”Xâ€™YZâ€™ âˆ¼ Î”XYZ {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 âˆ = 90Â°

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 Y1, Y2, Y3, Y4, on the ray YA such that YY1 = Y1Y2 = Y2Y3= Y3Y4 = Y4Y5

Step 7. Join the points Y3Z.

Step 8. Construct a line through Y5 parallel to Y3Z which intersects the extended line YZ at Zâ€™.

Step 9. Through Zâ€™, draw a line parallel to the line XZ that intersects the extended line XY at Xâ€™.

Step 10. Therefore, Î”Xâ€™YZâ€™ is the required triangle.

Justification:

The construction can be justified by proving that

Here the scale factor is , we need to prove

X’Y =

YZâ€™ =

X’Zâ€™=

From the construction, we obtain Xâ€™Zâ€™ || XZ

In Î”Xâ€™YZâ€™ and Î”XYZ,

Therefore,

âˆ Xâ€™Zâ€™Y = âˆ XZY {Corresponding angles}

âˆ Y = âˆ Y {common}

Therefore,

Î”Xâ€™YZâ€™ âˆ¼ Î”XYZ {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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