• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

April 22, 2022 by Safia Leave a Comment

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 11 Constructions Exercise 11.1

In each of the following constructions, 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 :
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 1
1. First of all draw a line segment AB = 7.6 cm.
2. Now draw a ray AC which makes an acute angle with AB.
3. Locate 5 + 8 = 13 equal line segments on ray AC starting from A : AA1, A1A2, A2A3, A3A4, A4A5, A5A6, A6A7, A7A8, A8A9, A9A10, A10A11, A11A12 and A12A13
4. Now join A13B
5. Through A5 draw A5P || A13B which meets AB at R
6. Thus, P divides AB in the ratio 5 : 8.
On measuring the two parts AP = 2.9 cm. and PB = 4.7 cm.
Justification—
In △ABA13
PA5 || BA13
∴ △APA5 ~ △ABA13
⇒ \(\frac{AP}{PB}\) = \(\frac{\mathrm{AA}_{5}}{\mathbf{A}_{5} \mathbf{A}_{13}}\) = \(\frac{5}{8}\)
⇒ AP : PB = 5 : 8

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:
Steps of Construction :
1. First of all draw a line segment BC = 7 cm.
2. Now taking B as centre and radius of 5 cm, draw an arc.
3. With C as centre and radius 4 cm, draw an arc to intersect the former arc at A.
4. Join AB and AC.
Then DABC is the required triangle.
5. Make an acute angle below BC.
6. On BX, mark three points B1, B2 and B3 such that
BB1 = B1B2 = B2B3
7. Now join B3C
8. Through B2 draw B2D || B3C which meets BC at D.
9. Through D draw ED || AC which meets BA at E.
Thus △EBD is the required triangle whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 2
Justification—Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{2}{3}\)
Thus, we get a new triangle which is similar to the given triangle, whose sides are \(\frac{2}{3}\) times the corresponding sides of △ABC.

RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Question 3.
Construct a triangle with sides 5 cm., 6 cm. and 7 cm. and then another triangle where sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.
Solution:
Steps of Construction :
1. First of all construct a triangle ABC with given values in which BC = 7 cm., CA = 6 cm. and AB = 5 cm.
2. Now below BC make an acute angle CBX,
3. Where 7 points on BX :
B1,B2, B3, B4, B5, B6 and B7 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7
4. Now join B5C.
5. Through B7 draw B7D || B5C. Join BC producing upto D.
6. Through D draw DE || CA, which meets BA produced at E.
Thus △EBD is the required triangle whose sides are \(\frac{7}{5}\) times the corresponding sides of △ABC.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 3
Justification—
Since DE || CA
∴ △ABC ~ △EBD
and \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{7}{5}\)
Thus we get a new triangle similar to the given triangle, where sides are \(\frac{7}{5}\) times the
corresponding side of △ABC.

Question 4.
Construct an isosceles triangle where 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:
Steps of Construction :
1. First of all draw BC = 8 cm.
2. Draw the perpendicular bisector PQ of line segment BC which meets BC at M.
3. From MP, cut MA = 4 cm.
4. Join BA and CA.
Now the DABC obtained is the required triangle.
5. Produce BC to D such that
BD = 12 cm.
6. Draw DE || CA which meets BA produced at E.
Then, △EBD will be the required triangle.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 4
Justification—Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{DE}{CA}\) = \(\frac{BD}{BC}\) = \(\frac{12}{8}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{3}{2}\) times, i.e. 1\(\frac{1}{2}\) the corresponding sides of △ABC.

RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

Question 5.
Draw a triangle ABC with side BC = 6 cm., AB = 5 cm. and ∠ABC = 60°. Then construct a triangle whose sides are \(\frac{3}{4}\) of the corresponding sides of the triangle ABC.
Solution:
Steps of Construction :
1. First of all construct DABC with given values in which BC = 6 cm., DABC = 60° and AB = 5 cm.
2. Now make an acute angle DCBX below
3. Locate four points B1, B2, B3 and B4 on BX. Thus BB1 = B1B2 = B2B3 = B3B4
4. Then join B4C and draw B3D || B4C.
5. Now through D, draw ED || AC which meets BA at E.
Then △EBD will be the required triangle whose sides will be \(\frac{3}{4}\) times the corresponding sides of △ABC.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 5
Justification—
Since DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{3}{2}\)
Thus, we get a new triangle similar to the given triangle, where sides are \(\frac{3}{2}\) times the sides of △ABC.

Question 6.
Draw a triangle ABC with side BC = 7 cm., ∠B = 45°, ∠A = 105. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of △ABC.
Solution:
Steps of Construction :
1. First of all construct △ABC with the given values in which BC = 7 cm.,
∠B = 45°, ∠C = 180° – (∠A + ∠B)
⇒ ∠C = 180° – (105° + 45°)
= 180° – 150° = 30°
2. Now make aft acute angle ∠CBX below BC.
3. Locate four points on BX :
B1, B2, B3 and B4
Thus BB1 = B1B2 = B2B3 = B3B4
4. Then join B3C.
5. Through B4, draw B4D || B3C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E.
Then △EBD will be the required triangle where sides will be \(\frac{4}{3}\) times the corresponding sides of △ABC.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 6
Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{4}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{4}{3}\) times the corresponding sides of △ABC.

RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

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 where sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.
Solution:
Steps of Construction :
1. First of all construct △ABC with given values in which BC = 4 cm., ∠B = 90° and BA = 3 cm.
2. Now make an acute angle CBX below BC.
3. Locate five points on BX :
B1, B2, B3, B4 and B5 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5
4. Join B3C.
5. Through B5, draw B5D || B3C which meets BC produced at D.
6. Through D, draw ED || AC which meets BA produced at E. Thus, △EBD is the required triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.
RBSE Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 7
Justification—
Since, DE || CA
∴ △ABC ~ △EBD
and, \(\frac{EB}{AB}\) = \(\frac{BD}{BC}\) = \(\frac{DE}{CA}\) = \(\frac{5}{3}\)
Thus, we get a new triangle similar to the given triangle, whose sides are \(\frac{5}{3}\) times the corresponding sides of △ABC.

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Recent Posts

  • RBSE Solutions for Class 6 Maths Chapter 6 Decimal Numbers Additional Questions
  • RBSE Solutions for Class 11 Psychology in Hindi Medium & English Medium
  • RBSE Solutions for Class 11 Geography in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Hindi
  • RBSE Solutions for Class 3 English Let’s Learn English
  • RBSE Solutions for Class 3 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 Maths in Hindi Medium & English Medium
  • RBSE Solutions for Class 3 in Hindi Medium & English Medium
  • RBSE Solutions for Class 4 Hindi
  • RBSE Solutions for Class 4 English Let’s Learn English
  • RBSE Solutions for Class 4 EVS पर्यावरण अध्ययन अपना परिवेश in Hindi Medium & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2023 RBSE Solutions

 

Loading Comments...