Practice Questions - Part 3
Duration: 1 hr 3 min
This video lesson is available to enrolled students.
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An AI-generated summary of this video lecture.
This video is a comprehensive lecture on geometry and mensuration, presented by an instructor from Knowledge Gate. The session systematically covers a series of problems involving the calculation of areas, perimeters, and other properties of various shapes. The instructor begins with a word problem about a wall, using algebraic substitution to find the sum of its length and height. He then moves to a problem on percentage change in the area of a rectangle when its dimensions are altered. The lecture continues with practical applications, such as calculating the number of tiles needed to pave a floor and the cost of levelling a circular track. The instructor demonstrates the use of formulas for the area of a rhombus, the area of a shaded region in a circle, and the radius of an incircle in a triangle, using Heron's formula and the inradius formula. The video is structured as a problem-solving session, with the instructor writing equations and diagrams on a whiteboard to guide the viewer through each step.
Chapters
0:00 – 2:00 00:00-02:00
The video opens with a title card for 'GEOMETRY & MENSURATION'. The instructor, Yash Jain Sir, introduces the first problem: 'The length of a wall is 5/4 times of its height. If the area of wall be 180 sq. meters, what is the sum of the length and height of the wall?'. He begins to solve it by setting up variables, writing 'l = 5x' and 'h = 4x' on the whiteboard, representing the length and height in terms of a common variable 'x'. He then writes the area formula: 'Area = length x height'.
2:00 – 5:00 02:00-05:00
The instructor continues solving the wall problem. He substitutes the variables into the area formula, writing '180 = 5x x 4x'. He simplifies this to '180 = 20x²' and then divides both sides by 20 to get '9 = x²'. He concludes that 'x = 3'. He then calculates the length as '5x = 15' and the height as '4x = 12'. The final answer is the sum: '15 + 12 = 27'. He circles the answer '27' on the board.
5:00 – 10:00 05:00-10:00
The instructor moves to the next problem: 'If the length of a rectangular field is doubled and its breadth is halved. What is % change in its area?'. He uses a numerical example, setting the original length to '1 m' and breadth to '2 m', so the original area is '1 x 2 = 2'. He then calculates the new dimensions: length becomes '2 m' and breadth becomes '1 m'. The new area is '2 x 1 = 2'. He concludes that the area remains unchanged, so the percentage change is '0%'. He writes '0% -> no change' on the board.
10:00 – 15:00 10:00-15:00
The third problem is introduced: 'How many tiles 20 cm by 40 cm will be required to pave the floor of a prayer hall of a room 16 meters long and 9 meters wide?'. The instructor converts the room dimensions to centimeters: '16 m = 1600 cm' and '9 m = 900 cm'. He calculates the area of the prayer hall as '1600 x 900 = 1,440,000 cm²'. He then calculates the area of one tile as '20 x 40 = 800 cm²'. He sets up the equation 'N x 800 = 1,440,000' and solves for N, finding 'N = 1800' tiles.
15:00 – 20:00 15:00-20:00
The fourth problem is presented: 'The outer circumference of a circular track is 220 meters. The track is 7 meters wide everywhere. Calculate the cost of levelling the track at the rate of 50 paise per square meter?'. The instructor draws a diagram of a circular track (an annulus). He uses the formula for circumference, '2πR = 220', to find the outer radius 'R'. He calculates 'R = 220 / (2π) = 35 m'. He then finds the inner radius 'r' as 'R - 7 = 35 - 7 = 28 m'. He calculates the required area as the difference between the areas of the two circles: 'πR² - πr²'.
20:00 – 25:00 20:00-25:00
The instructor completes the calculation for the circular track. He uses the formula for the area of an annulus, 'Required Area = π(R² - r²)'. He substitutes the values: 'π(35² - 28²)'. He simplifies this using the difference of squares: 'π(35 + 28)(35 - 28) = π(63)(7)'. He calculates the area as '22/7 x 63 x 7 = 1386 m²'. He then calculates the cost as '1386 x 50 paise = 693 rupees'. He writes the final answer 'Rs. 693' on the board.
25:00 – 30:00 25:00-30:00
The fifth problem is introduced: 'In the following figure, find the area of the shaded region.'. The figure shows a rectangle inscribed in a circle. The rectangle has sides 8 and 6. The instructor draws the figure and identifies the diagonal of the rectangle as the diameter of the circle. He uses the Pythagorean theorem: 'h² = 8² + 6² = 64 + 36 = 100', so the diagonal 'h = 10'. He concludes the diameter is 10, so the radius is 5.
30:00 – 35:00 30:00-35:00
The instructor calculates the area of the shaded region. He finds the area of the circle: 'πr² = 3.14 x 5² = 78.5'. He finds the area of the rectangle: '8 x 6 = 48'. He subtracts the area of the rectangle from the area of the circle to get the shaded area: '78.5 - 48 = 30.5'. He writes the final answer '30.5' on the board.
35:00 – 40:00 35:00-40:00
The sixth problem is presented: 'The diagonals of a rhombus are 16 cm and 12 cm. Its perimeter is?'. The instructor draws a rhombus and its diagonals, which bisect each other at right angles. He states that the diagonals are 16 cm and 12 cm, so each half is 8 cm and 6 cm. He uses the Pythagorean theorem to find the side length: 'side² = 8² + 6² = 64 + 36 = 100', so the side length is 10 cm.
40:00 – 45:00 40:00-45:00
The instructor calculates the perimeter of the rhombus. He states that all four sides are equal, so the perimeter is '4 x side = 4 x 10 = 40 cm'. He writes the final answer '40 cm' on the board. He also mentions the formula for the area of a rhombus: '1/2 x d1 x d2'.
45:00 – 50:00 45:00-50:00
The seventh problem is introduced: 'The area of the square garden is 625 sq. meters. What is the area of a path of width 2.5 meters around it if the path is outside the garden?'. The instructor draws a square garden with a path around it. He calculates the side of the square garden as 'a = √625 = 25 m'. He then calculates the side of the larger square (garden + path) as '25 + 2.5 + 2.5 = 30 m'.
50:00 – 55:00 50:00-55:00
The instructor calculates the area of the path. He finds the area of the larger square: '30² = 900 m²'. He finds the area of the garden: '25² = 625 m²'. He subtracts the garden area from the larger square area to get the path area: '900 - 625 = 275 m²'. He writes the final answer '275 m²' on the board.
55:00 – 60:00 55:00-60:00
The eighth problem is presented: 'What is the measure of the radius of the circle inscribed in a triangle whose sides measure 8, 15 and 17 units?'. The instructor draws a triangle with an incircle. He identifies the triangle as a right-angled triangle because '17² = 15² + 8²' (289 = 225 + 64). He states the formula for the inradius: 'r = A / s', where A is the area and s is the semi-perimeter.
60:00 – 63:14 60:00-63:14
The instructor calculates the inradius. He calculates the semi-perimeter 's = (8 + 15 + 17) / 2 = 20'. He calculates the area 'A = 1/2 x 8 x 15 = 60'. He then calculates the inradius 'r = A / s = 60 / 20 = 3'. He writes the final answer '3' on the board. The video ends with a logo for Knowledge Gate.
This video is a structured problem-solving session on geometry and mensuration. The instructor systematically works through a series of eight distinct problems, each focusing on a different shape or concept. The progression moves from basic algebraic word problems (wall) to more complex applications involving circles, rectangles, rhombuses, and triangles. Key concepts demonstrated include the use of variables to represent unknowns, the application of fundamental formulas for area and perimeter (e.g., Area = length x breadth, Area of a circle = πr², Area of a rhombus = 1/2 x d1 x d2), and the use of geometric theorems (e.g., Pythagorean theorem, properties of a rhombus). The instructor consistently uses a step-by-step approach, writing out equations and diagrams on a whiteboard to clearly illustrate the solution process for each problem, making it a comprehensive review for students preparing for competitive exams.