Class 9 Maths | Chapter 9: Similar Figures | Exercise 9.3 | Question 3 | Punjab Boards
Автор: MathPhys
Загружено: 2026-03-02
Просмотров: 16
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The video explains how to solve Question 3 from Exercise 9.3 of Class 9 Maths, focusing on similar figures and ratios of volumes, heights, and base areas of two right cones (0:04).
Here's a breakdown of the key concepts and steps:
Part A: Ratio of Heights (0:55-3:38)
The problem states that the ratio of the volumes (V1:V2) of two right cones is 64:125 (0:15).
The fundamental principle for similar figures is applied: the ratio of their volumes is equal to the cube of the ratio of their corresponding sides (or heights in this case) (1:05). So, V1/V2 = (H1/H2)³.
To find the ratio of heights (H1:H2), the cube root of the given volume ratio is taken (1:50).
Calculations show that the cube root of 64 is 4 and the cube root of 125 is 5 (2:20-3:30), resulting in a height ratio of 4:5 (2:33).
Part B: Ratio of Base Areas (3:42-5:01)
The video then moves to finding the ratio of their base areas (A1:A2) (3:44).
Another principle for similar figures is used: the ratio of their areas is equal to the square of the ratio of their corresponding sides (or heights) (4:16). So, A1/A2 = (H1/H2)².
Using the previously calculated height ratio of 4:5 (3:55-4:08), the square of this ratio is computed.
Squaring 4 gives 16, and squaring 5 gives 25 (4:45-4:51), leading to a base area ratio of 16:25 (4:55).
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