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NCERT Solutions for Class 9 सामाजिक विज्ञान Chapter 13 Surface Areas And Volumes

Surface Areas And Volumes Here is the CBSE सामाजिक विज्ञान Chapter 13 for Class 9 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 9 सामाजिक विज्ञान Surface Areas And Volumes Chapter 13 NCERT Solutions for Class 9 सामाजिक विज्ञान Surface Areas And Volumes Chapter 13 The following is a summary in Hindi and English for the academic year 2025-26. You can save these solutions to your computer or use the Class 9 सामाजिक विज्ञान.

Question 1
CBSEENMA9003174

Construct an angle of 90° at the initial point of a given ray and justify the construction.

Solution

Given: A ray OA.
Required: To construct an angle of 90° at O and justify the construction.
Steps of Construction:
1. Taking O as centre and some radius, draw an arc of a circle, which intersects OA, say at a point B.
2. Taking B as centre and with the same radius as before, draw an are intersecting the previously drawn are, say at a point C.
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3. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 1, say at D.
4. Draw the ray OE passing through C. Then ∠EOA = 60°.
5. Draw the ray OF passing through D. Then ∠FOE = 60°.
6. Next, taking C and D as centres and with the radius more than 1 half ID, draw arcs to intersect each other, say at G.

7. Draw the ray OG. This ray OG is the bisector of the angle ∠FOE, i.e., ∠FOG
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Justification:
(i) Join BC.
Then. OC = OB = BC (By construction)
∴ ∆COB is an equilateral triangle.
∴ ∠COB = 60°.
∴ ∠EOA = 60°.
(ii) Join CD.
Then, OD = OC = CD (By construction)
∴ ∆DOC is an equilateral triangle.

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