Practice Questions - Part 2

Duration: 56 min

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This educational video is a comprehensive lecture on geometry and mensuration, presented by an instructor named Yash Jain. The session begins with a title slide and proceeds to solve a series of multiple-choice questions. The first question tests the understanding of circle terminology, specifically the definitions of a chord and a secant, with the instructor explaining that a secant is a line that intersects a circle at two points and contains a chord, but is not itself a chord. The second problem involves a geometric configuration where a small circle is placed in the corner formed by a large circle and two perpendicular lines, requiring the use of the Pythagorean theorem to find the radius of the small circle. The third question asks for the area of a hexagon formed by joining the trisection points of the sides of an equilateral triangle, which is solved by calculating the area of the original triangle and subtracting the areas of the three corner triangles. The fourth problem uses the formula for the sum of interior angles of a polygon to find the number of sides of a regular polygon given one interior angle. The fifth question asks for the area of a regular hexagon inscribed in a circle, which is solved by dividing the hexagon into six equilateral triangles. The final problem is a conceptual question about the sum of an interior and exterior angle of a regular polygon, which is always 180 degrees. The video uses a blackboard-style interface with diagrams and step-by-step calculations to explain each concept.

Chapters

  1. 0:00 – 2:00 00:00-02:00

    The video opens with a title slide for a 'GEOMETRY & MENSURATION' lecture. The instructor, Yash Jain, is visible in a small window. The first question is presented on a blackboard: 'Which of the following statements is incorrect in regard to a circle?'. The four statements are: (i) Every Chord is Secant, (ii) Every Secant is Chord, (iii) Secant is technically not a chord, but it contains a chord, and (iv) Every secant line defines a unique chord. The instructor begins to analyze the definitions of a chord and a secant, explaining that a chord is a line segment connecting two points on a circle, while a secant is a line that intersects the circle at two points and extends infinitely. He notes that a secant contains a chord but is not a chord itself, which makes statement (ii) incorrect.

  2. 2:00 – 5:00 02:00-05:00

    The instructor continues to analyze the first question. He explains that statement (i) 'Every Chord is Secant' is incorrect because a chord is a segment, not a line, and a secant is a line. He then discusses statement (ii) 'Every Secant is Chord', which is also incorrect for the same reason. He confirms that statement (iii) 'Secant is technically not a chord, but it contains a chord' is correct. He then evaluates statement (iv) 'Every secant line defines a unique chord', which he confirms is correct. The instructor concludes that statements (i) and (ii) are incorrect, making option (c) 'Only i & ii' the correct answer. He then moves to the next problem.

  3. 5:00 – 10:00 05:00-10:00

    The second question is presented: 'It is required to place a small circle in the space left by a large circle as shown. If the radius of the large one is 2 cm and that of the small one is b', then find the value of b'. The instructor draws a diagram showing a large circle of radius 2 cm tangent to two perpendicular lines, and a smaller circle of radius b tangent to the large circle and the two lines. He labels the center of the large circle as A and the center of the small circle as O. He identifies the distance from the origin to the center of the large circle as OA = 2√2. He then identifies the distance from the origin to the center of the small circle as OD = b√2. He sets up the equation OA = AD + DO, where AD = 2 and DO = b√2, leading to the equation 2√2 = 2 + b√2. He solves this equation to find b = 2(√2 - 1). He then moves to the next problem.

  4. 10:00 – 15:00 10:00-15:00

    The third question is presented: 'An equilateral triangle of length 3 cm is divided equally by all sides into 3 parts, all the cuts are joined thus forming a hexagon, find area of this hexagon'. The instructor draws an equilateral triangle with side length 3 cm. He divides each side into three equal parts of 1 cm. He then joins the points of division to form a hexagon in the center. He explains that the area of the hexagon is the area of the original triangle minus the areas of the three small corner triangles. He calculates the area of the original triangle as (√3/4) * 3^2 = 9√3/4. He calculates the area of one small corner triangle as (√3/4) * 1^2 = √3/4. The total area of the three small triangles is 3 * √3/4 = 3√3/4. The area of the hexagon is 9√3/4 - 3√3/4 = 6√3/4 = 3√3/2.

  5. 15:00 – 20:00 15:00-20:00

    The fourth question is presented: 'If the expression shown are the degree measures of the angles if the pentagon, find the value x+y'. The instructor draws a pentagon with five interior angles labeled as 4x, 3x, 7y, 2x+y, and x+2y. He states that the sum of the interior angles of a pentagon is (5-2) * 180 = 540 degrees. He sets up the equation: 4x + 3x + 7y + 2x + y + x + 2y = 540. He simplifies this to 10x + 10y = 540. He then divides both sides by 10 to get x + y = 54. He confirms this is the answer. He then moves to the next problem.

  6. 20:00 – 25:00 20:00-25:00

    The fifth question is presented: 'One angle of a regular polygon measures 177°. How many sides does it have?'. The instructor explains that the formula for the measure of an interior angle of a regular polygon is ((n-2) * 180) / n, where n is the number of sides. He sets up the equation: ((n-2) * 180) / n = 177. He multiplies both sides by n to get (n-2) * 180 = 177n. He expands the left side to get 180n - 360 = 177n. He subtracts 177n from both sides to get 3n - 360 = 0. He adds 360 to both sides to get 3n = 360. He divides by 3 to get n = 120. He confirms the polygon has 120 sides. He then moves to the next problem.

  7. 25:00 – 30:00 25:00-30:00

    The sixth question is presented: 'A regular hexagon is inscribed in a circle of radius 6 cm. What is the area of the hexagon?'. The instructor draws a circle with a regular hexagon inscribed in it. He explains that a regular hexagon can be divided into 6 equilateral triangles, each with a side length equal to the radius of the circle, which is 6 cm. He writes the formula for the area of an equilateral triangle as (√3/4) * side^2. He calculates the area of one triangle as (√3/4) * 6^2 = (√3/4) * 36 = 9√3. He multiplies this by 6 to get the total area of the hexagon: 6 * 9√3 = 54√3 cm². He confirms this is the answer.

  8. 30:00 – 35:00 30:00-35:00

    The seventh question is presented: 'Find the sum of the measures of one interior and one exterior angle of a regular 940-gon'. The instructor explains that for any polygon, an interior angle and its adjacent exterior angle are supplementary, meaning they add up to 180 degrees. He draws a diagram of a polygon showing an interior angle and an exterior angle next to each other, forming a straight line. He writes the equation: int + ext = 180°. He states that this is true for any polygon, regardless of the number of sides, so the sum is always 180 degrees. He confirms this is the answer.

  9. 35:00 – 40:00 35:00-40:00

    The instructor begins to solve the second problem again, this time with a different approach. He draws a diagram of a large circle of radius 2 cm tangent to two perpendicular lines, and a small circle of radius b tangent to the large circle and the two lines. He labels the center of the large circle as A and the center of the small circle as O. He identifies the distance from the origin to the center of the large circle as OA = 2√2. He then identifies the distance from the origin to the center of the small circle as OD = b√2. He sets up the equation OA = AD + DO, where AD = 2 and DO = b√2, leading to the equation 2√2 = 2 + b√2. He solves this equation to find b = 2(√2 - 1). He then moves to the next problem.

  10. 40:00 – 45:00 40:00-45:00

    The instructor begins to solve the third problem again, this time with a different approach. He draws an equilateral triangle with side length 3 cm. He divides each side into three equal parts of 1 cm. He then joins the points of division to form a hexagon in the center. He explains that the area of the hexagon is the area of the original triangle minus the areas of the three small corner triangles. He calculates the area of the original triangle as (√3/4) * 3^2 = 9√3/4. He calculates the area of one small corner triangle as (√3/4) * 1^2 = √3/4. The total area of the three small triangles is 3 * √3/4 = 3√3/4. The area of the hexagon is 9√3/4 - 3√3/4 = 6√3/4 = 3√3/2.

  11. 45:00 – 50:00 45:00-50:00

    The instructor begins to solve the fourth problem again, this time with a different approach. He draws a pentagon with five interior angles labeled as 4x, 3x, 7y, 2x+y, and x+2y. He states that the sum of the interior angles of a pentagon is (5-2) * 180 = 540 degrees. He sets up the equation: 4x + 3x + 7y + 2x + y + x + 2y = 540. He simplifies this to 10x + 10y = 540. He then divides both sides by 10 to get x + y = 54. He confirms this is the answer. He then moves to the next problem.

  12. 50:00 – 55:00 50:00-55:00

    The instructor begins to solve the fifth problem again, this time with a different approach. He draws a circle with a regular hexagon inscribed in it. He explains that a regular hexagon can be divided into 6 equilateral triangles, each with a side length equal to the radius of the circle, which is 6 cm. He writes the formula for the area of an equilateral triangle as (√3/4) * side^2. He calculates the area of one triangle as (√3/4) * 6^2 = (√3/4) * 36 = 9√3. He multiplies this by 6 to get the total area of the hexagon: 6 * 9√3 = 54√3 cm². He confirms this is the answer.

  13. 55:00 – 56:11 55:00-56:11

    The video concludes with a final slide that says 'THANKS FOR WATCHING' in large white text on a dark background. The instructor is no longer visible. This is the end of the lecture.

The video presents a structured and methodical review of key concepts in geometry and mensuration. It begins with a foundational question on circle terminology, establishing the importance of precise definitions. The subsequent problems are carefully chosen to demonstrate the application of core principles: the Pythagorean theorem for a tangency problem, the area of triangles for a composite shape, the sum of interior angles for a polygon, and the properties of regular polygons. The instructor consistently uses a step-by-step approach, drawing diagrams and writing out equations to guide the viewer through each solution. The final problem reinforces a fundamental geometric principle that an interior and exterior angle are supplementary, a concept that is universally true. The overall progression moves from basic definitions to more complex applications, providing a comprehensive review of the topic.