Q: In the given figure, $ABC$ is an equilateral triangle which is inscribed inside a circle and whose radius is $r$. Which of the following is the area of the triangle? <img src="/api/images/abc-equilateral-triangle-inscribed-inside-circle-radius.p…

Concept & Area of a Triangle: The area of a triangle is given by the standard formula: Area = (1/2) × Base × Height Here, we must express the base BC and the height AE (where E is the foot of the perpendicular from A to BC passing through center D) in terms of the circle's radius r and the given segment DE. Step-by-Step Solution: 1. Determine the Height: The line segment passing from vertex A through center D to the base BC at point E is the height of the equilateral triangle. AD is the radius r. Thus, the total height is AE = AD + DE = r + DE. 2. Determine the Base: In the right-angled triangle \triangle BDE, the hypotenuse BD is also a radius r. Using the Pythagorean theorem: BE^2 + DE^2 = BD^2 BE^2 = r^2 - DE^2 \implies BE = \sqrt{r^2 - DE^2} The full base BC is twice BE, so BC = 2\sqrt{r^2 - DE^2}. 3. Calculate the Area: Area = (1/2) × BC × AE Area = (1/2) × (2\sqrt{r^2 - DE^2}) × (r + DE) Area = \sqrt{(r - DE)(r + DE)} × (r + DE) Area = (r - DE)^{(1/2)} (r + DE)^{(1/2)} (r + DE)^1 = (r - DE)^{(1/2)} (r + DE)^{(3/2)} Exam Strategy & Shortcut: Recognize that \sqrt{r^2 - DE^2} factors into \sqrt{r-DE}\sqrt{r+DE}. When you multiply this by the height (r+DE), the (r+DE) terms combine their exponents (1/2 + 1 = 3/2), immediately leading to option (d). Common Pitfall: Failing to recognize that AD and BD are both radii (r) of the circumcircle. Students often get stuck looking for the side length of the equilateral triangle using trigonometry, which makes the algebra much messier than simply using the Pythagorean theorem on \triangle BDE. Final Answer: Therefore, the correct answer is (r - DE)^{(1/2)} (r + DE)^{(3/2)}.

Q: Three boys are standing on a circular boundary of a fountain. They are at equal distance from each other. If the radius of the boundary is 5 m, the shortest distance between any two boys is

Concept & Equilateral Triangle Properties: Since the three boys are at equal distances from each other on a circular boundary, they form the vertices of an equilateral triangle inscribed within the circle. The relationship between the side a (distance between the boys) and the circumradius R of an equilateral triangle is: R = (a/\sqrt{3)} \implies a = R\sqrt{3} Step-by-Step Solution: 1. Identify the Given Information: The radius of the circular boundary (circumradius) is R = 5 m. 2. Apply the Formula: We need to find the side length a of the equilateral triangle. a = R × \sqrt{3} 3. Calculate: a = 5 × \sqrt{3} = 5\sqrt{3} m. Exam Strategy & Shortcut: Memorizing the a = R\sqrt{3} relationship for inscribed equilateral triangles turns this into a 2-second read-and-mark question. No calculation is needed beyond attaching \sqrt{3} to the given radius. Common Pitfall: Some students might attempt to calculate the arc length between the boys instead of the straight-line distance, incorrectly choosing a distance involving \pi. The phrase "shortest distance" always implies a straight line. Final Answer: Therefore, the correct answer is 5\sqrt{3} m.

Q: What is the area of an equilateral triangle inscribed in a circle of unit radius?

Concept & Inscribed Triangle Formula: For an equilateral triangle inscribed in a circle of radius R, the side length a is R\sqrt{3}. The area of an equilateral triangle is: Area = \frac{\sqrt{3}}{4} a^2 Step-by-Step Solution: 1. Find the side length (a): The radius R = 1. a = 1 × \sqrt{3} = \sqrt{3} units. 2. Calculate the Area: Area = \frac{\sqrt{3}}{4} (\sqrt{3})^2 Area = \frac{\sqrt{3}}{4} × 3 Area = \frac{3\sqrt{3}}{4} sq. units. Exam Strategy & Shortcut: You can also compute the area of the inscribed triangle directly using the circumradius formula: Area = \frac{3\sqrt{3}}{4} R^2. Since R=1, the area is simply the coefficient \frac{3\sqrt{3}}{4}. Common Pitfall: Squaring \sqrt{3} incorrectly, or forgetting the 1/4 factor in the equilateral triangle area formula, which leads to choosing incorrect options like 3\sqrt{3}. Final Answer: Therefore, the correct answer is \frac{3\sqrt{3}}{4} sq. units.

Q: The sides of a triangle are 6 cm, 11 cm and 15 cm. The radius of its incircle is

Concept & Inradius Formula: The radius of an incircle (r) of any triangle can be found by dividing the area of the triangle (A) by its semi-perimeter (s). r = (A/s) The area A is found using Heron's formula: A = \sqrt{s(s-a)(s-b)(s-c)} Step-by-Step Solution: 1. Calculate the Semi-perimeter (s): s = (a + b + c/2) = (6 + 11 + 15/2) = (32/2) = 16 cm. 2. Calculate the Area (A) using Heron's Formula: A = \sqrt{16(16 - 6)(16 - 11)(16 - 15)} A = \sqrt{16 × 10 × 5 × 1} A = \sqrt{800} = \sqrt{400 × 2} = 20\sqrt{2} cm^2. 3. Calculate the Inradius (r): r = (A/s) = \frac{20\sqrt{2}}{16} Simplify the fraction by dividing the numerator and denominator by 4: r = \frac{5\sqrt{2}}{4} cm. Exam Strategy & Shortcut: To compute \sqrt{800} quickly, group the factors logically: 16 × (10 × 5) = 16 × 50 = 800. You know \sqrt{16} = 4 and \sqrt{50} = 5\sqrt{2}. Multiply them directly: 4 × 5\sqrt{2} = 20\sqrt{2}. This saves time factoring large numbers. Common Pitfall: A frequent error is forgetting to divide the area by the semi-perimeter s and instead dividing by the full perimeter, which would yield exactly half of the correct answer. Always remember r = A/s, not A/P. Final Answer: Therefore, the correct answer is \frac{5\sqrt{2}}{4} cm.

Q: The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is

Concept & Inradius and Area Relationship: The area (A) of any triangle can be expressed as the product of its inradius (r) and its semi-perimeter (s): A = r × s The circumference of the incircle is C = 2\pi r, which allows us to find the inradius r. Step-by-Step Solution: 1. Find the Inradius (r): Given the circumference is 88 cm: 2\pi r = 88 2 × (22/7) × r = 88 (44/7) × r = 88 r = (88 × 7/44) = 2 × 7 = 14 cm. 2. Find the Semi-perimeter (s): Given the perimeter is 30 cm: s = \frac{Perimeter}{2} = (30/2) = 15 cm. 3. Calculate the Area (A): A = r × s A = 14 × 15 = 210 cm^2. Exam Strategy & Shortcut: When you see a circumference of 88, recognize instantly that it's double 44, which corresponds to a radius of 14 (since C=44 means r=7 for \pi=22/7). Then immediately multiply this radius (14) by half the given perimeter (15) to get 14 × 15 = 210. The entire process can be done mentally. Common Pitfall: Students often multiply the inradius by the full perimeter instead of the semi-perimeter, which leads to the incorrect answer of 420 cm^2. Always remember A = rs, not rp. Final Answer: Therefore, the correct answer is 210 cm^2.

Q: If in a triangle, the area is numerically equal to the perimeter, then the radius of the inscribed circle of the triangle is

Concept & Inradius-Area Formula: The fundamental relationship connecting the area of a triangle (A), its perimeter (P), and the radius of its inscribed circle (inradius r) is: A = r × s where s is the semi-perimeter (s = P / 2). Step-by-Step Solution: 1. Express the Formula in Terms of Perimeter: Substitute s = (P/2) into the area formula: A = r × (P/2) 2. Apply the Given Condition: We are given that the area is numerically equal to the perimeter, so A = P. Substitute A for P in our equation: A = r × (A/2) 3. Solve for r: Divide both sides by A (since area cannot be zero): 1 = (r/2) r = 2 Exam Strategy & Shortcut: This is a theoretical property that can be memorized. Whenever a 2D shape's area is calculated as "radius times semi-perimeter" (which is true for all triangles and tangential polygons), if Area = Perimeter, the inradius is always exactly 2. Common Pitfall: A common mistake is attempting to assume a specific type of triangle (like equilateral) and working through complex side-length formulas, which wastes time and often introduces calculation errors, when the relationship holds universally for all triangles. Final Answer: Therefore, the correct answer is 2.

Q: A circle, a square and an equilateral triangle have the same area. The correct increasing order of the perimeters will be

Concept & Isoperimetric Property (Inverse): The Isoperimetric principle establishes that a circle is the most efficient shape for enclosing area. This means for a fixed area, a circle will require the least amount of perimeter. Conversely, shapes with fewer sides are less efficient and require a larger perimeter to enclose that same area. Step-by-Step Solution: 1. Understand Efficiency: Efficiency here means getting the most area for the least perimeter. - Most efficient: Circle - Moderately efficient: Square (4 sides) - Least efficient: Equilateral Triangle (3 sides) 2. Order by Perimeter (Least to Greatest): Since they all have the same area, the most efficient shape will have the smallest perimeter. - Smallest perimeter: Circle - Next largest: Square - Largest perimeter: Triangle 3. Format the Output: The increasing order (smallest to largest) is circle, square, triangle. Exam Strategy & Shortcut: This is the inverse of the standard Isoperimetric rule. If equal perimeters mean Area_{circle} > Area_{square} > Area_{triangle}, then for equal areas, the perimeters must compensate in reverse: Perimeter_{circle} < Perimeter_{square} < Perimeter_{triangle}. Just remember: the circle always "wins" by doing the most with the least. Common Pitfall: Confusing this question with the "equal perimeter" variation. Students often skim the question, assume it's asking for area based on equal perimeters, and pick the exact opposite order. Always underline whether "Area" or "Perimeter" is being held constant. Final Answer: Therefore, the correct answer is circle, square, triangle.

Q: The area of the largest triangle that can be inscribed in a semi-circle of radius $r$, is

Concept & Inscribed Triangle Geometry: Any triangle inscribed in a semi-circle with its base on the diameter will be a right-angled triangle (Thales's Theorem). The area of a triangle is (1/2) × base × height. To maximize the area for a fixed base, the height must be maximized. Step-by-Step Solution: 1. Determine the Base: The largest triangle inscribed in a semi-circle must span the entire flat edge of the semi-circle. Therefore, its base is the diameter of the semi-circle. Base = 2r 2. Determine the Maximum Height: The third vertex must lie on the semi-circle arc. To maximize the height of the triangle, this vertex must be at the highest point of the semi-circle, which is directly above the center. The distance from the center to this highest point is simply the radius. Maximum Height = r 3. Calculate the Maximum Area: Area = (1/2) × Base × Height Area = (1/2) × (2r) × (r) Area = r^2 Exam Strategy & Shortcut: Visualize the semi-circle. The base is locked at 2r. The highest you can drag the peak of the triangle is the top center, which is height r. Half of 2r × r is just r^2. The visualization is instant. Common Pitfall: A common mistake is assuming the formula for the area of the semi-circle ((\pi r^2/2)) is needed, or mistakenly picking 2r^2 by forgetting to multiply by the 1/2 in the triangle area formula. Final Answer: Therefore, the correct answer is r^2.

Q: $ABC$ is a right-angled triangle with right angle at $B$. If the semi-circle on $AB$ with $AB$ as diameter encloses an area of 81 sq. cm and the semi-circle on $BC$ with $BC$ as diameter encloses an area of 36 sq. cm, then the area of the semi-circl…

Concept & Generalized Pythagorean Theorem: The Pythagorean theorem states that in a right triangle, a^2 + b^2 = c^2. This principle generalizes to any similar geometric figures constructed on the sides of a right triangle. If you draw similar shapes (like squares, equilateral triangles, or semi-circles) on the three sides, the area of the shape on the hypotenuse equals the sum of the areas of the shapes on the two legs. Area_{AC} = Area_{AB} + Area_{BC} Step-by-Step Solution: 1. Identify the Components: - Area of semi-circle on leg AB = 81 sq. cm. - Area of semi-circle on leg BC = 36 sq. cm. - AC is the hypotenuse (since the right angle is at B). 2. Apply the Generalized Pythagorean Theorem: The area of the semi-circle on the hypotenuse (AC) is the sum of the areas of the semi-circles on the legs (AB and BC). Area_{AC} = 81 + 36 3. Calculate the Final Area: Area_{AC} = 117 sq. cm. Exam Strategy & Shortcut: Recognize that area scales with the square of the dimension. Since AB^2 + BC^2 = AC^2, multiplying the entire equation by (\pi/8) gives the semi-circle areas: (\pi/8)AB^2 + (\pi/8)BC^2 = (\pi/8)AC^2. Thus, you just add the two given areas directly: 81 + 36 = 117. No need to find radii. Common Pitfall: A massive time-waster is working backward to find the actual radii of the semi-circles, computing the side lengths of the triangle, using the Pythagorean theorem to find AC, and then computing the area of the semi-circle on AC. While correct, it takes ten times longer. Final Answer: Therefore, the correct answer is 117 cm^2.

Q: When the circumference of a toy balloon is increased from $20\text{ cm}$ to $25\text{ cm}$, its radius is increased by

Concept & Circumference of a Circle: The circumference of a circle is directly proportional to its radius. The formula relating them is: C = 2\pi r Therefore, any change in the circumference (\Delta C) can be directly translated to a change in the radius (\Delta r) using \Delta r = (\Delta C/2\pi). Step-by-Step Solution: • Given: Initial circumference C_1 = 20 cm, Final circumference C_2 = 25 cm. • Deduction: The increase in circumference is \Delta C = 25 - 20 = 5 cm. • Calculation: Using the relationship C = 2\pi r, the increase in radius is \Delta r = (\Delta C/2\pi). • Calculation: Substitute the value of \Delta C: \Delta r = (5/2\pi). Exam Strategy & Shortcut: Instead of finding the individual radii r_1 and r_2 and then subtracting them, directly use the difference. Difference in radius = \frac{Difference in circumference}{2\pi}. This saves significant calculation time. Common Pitfall: A common mistake is forgetting the 2 in the denominator and selecting (5/\pi) as the answer, which corresponds to the change in diameter, not radius. Final Answer: Therefore, the correct answer is (5/2\pi).

Question 1

In the given figure, $ABC$ is an equilateral triangle which is inscribed inside a circle and whose radius is $r$. Which of the following is the area of the triangle?  <img    src="/api/images/abc-equilateral-triangle-inscribed-inside-circle-radius.p…

Accepted Answer

Concept & Area of a Triangle: The area of a triangle is given by the standard formula: Area = (1/2) × Base × Height Here, we must express the base BC and the height AE (where E is the foot of the perpendicular from A to BC passing through center D) in terms of the circle's radius r and the given segment DE. Step-by-Step Solution: 1. Determine the Height: The line segment passing from vertex A through center D to the base BC at point E is the height of the equilateral triangle. AD is the radius r. Thus, the total height is AE = AD + DE = r + DE. 2. Determine the Base: In the right-angled triangle 	riangle BDE, the hypotenuse BD is also a radius r. Using the Pythagorean theorem: BE^2 + DE^2 = BD^2 BE^2 = r^2 - DE^2 \implies BE = \sqrt{r^2 - DE^2} The full base BC is twice BE, so BC = 2\sqrt{r^2 - DE^2}. 3. Calculate the Area: Area = (1/2) × BC × AE Area = (1/2) × (2\sqrt{r^2 - DE^2}) × (r + DE) Area = \sqrt{(r - DE)(r + DE)} × (r + DE) Area = (r - DE)^{(1/2)} (r + DE)^{(1/2)} (r + DE)^1 = (r - DE)^{(1/2)} (r + DE)^{(3/2)} Exam Strategy & Shortcut: Recognize that \sqrt{r^2 - DE^2} factors into \sqrt{r-DE}\sqrt{r+DE}. When you multiply this by the height (r+DE), the (r+DE) terms combine their exponents (1/2 + 1 = 3/2), immediately leading to option (d). Common Pitfall: Failing to recognize that AD and BD are both radii (r) of the circumcircle. Students often get stuck looking for the side length of the equilateral triangle using trigonometry, which makes the algebra much messier than simply using the Pythagorean theorem on 	riangle BDE. Final Answer: Therefore, the correct answer is (r - DE)^{(1/2)} (r + DE)^{(3/2)}.

Question 2

Three boys are standing on a circular boundary of a fountain. They are at equal distance from each other. If the radius of the boundary is 5 m, the shortest distance between any two boys is

Accepted Answer

Concept & Equilateral Triangle Properties: Since the three boys are at equal distances from each other on a circular boundary, they form the vertices of an equilateral triangle inscribed within the circle. The relationship between the side a (distance between the boys) and the circumradius R of an equilateral triangle is: R = (a/\sqrt{3)} \implies a = R\sqrt{3} Step-by-Step Solution: 1. Identify the Given Information: The radius of the circular boundary (circumradius) is R = 5 m. 2. Apply the Formula: We need to find the side length a of the equilateral triangle. a = R × \sqrt{3} 3. Calculate: a = 5 × \sqrt{3} = 5\sqrt{3} m. Exam Strategy & Shortcut: Memorizing the a = R\sqrt{3} relationship for inscribed equilateral triangles turns this into a 2-second read-and-mark question. No calculation is needed beyond attaching \sqrt{3} to the given radius. Common Pitfall: Some students might attempt to calculate the arc length between the boys instead of the straight-line distance, incorrectly choosing a distance involving \pi. The phrase "shortest distance" always implies a straight line. Final Answer: Therefore, the correct answer is 5\sqrt{3} m.

Question 3

What is the area of an equilateral triangle inscribed in a circle of unit radius?

Accepted Answer

Concept & Inscribed Triangle Formula: For an equilateral triangle inscribed in a circle of radius R, the side length a is R\sqrt{3}. The area of an equilateral triangle is: Area = \frac{\sqrt{3}}{4} a^2 Step-by-Step Solution: 1. Find the side length (a): The radius R = 1. a = 1 × \sqrt{3} = \sqrt{3} units. 2. Calculate the Area: Area = \frac{\sqrt{3}}{4} (\sqrt{3})^2 Area = \frac{\sqrt{3}}{4} × 3 Area = \frac{3\sqrt{3}}{4} sq. units. Exam Strategy & Shortcut: You can also compute the area of the inscribed triangle directly using the circumradius formula: Area = \frac{3\sqrt{3}}{4} R^2. Since R=1, the area is simply the coefficient \frac{3\sqrt{3}}{4}. Common Pitfall: Squaring \sqrt{3} incorrectly, or forgetting the 1/4 factor in the equilateral triangle area formula, which leads to choosing incorrect options like 3\sqrt{3}. Final Answer: Therefore, the correct answer is \frac{3\sqrt{3}}{4} sq. units.

Question 4

The sides of a triangle are 6 cm, 11 cm and 15 cm. The radius of its incircle is

Accepted Answer

Concept & Inradius Formula: The radius of an incircle (r) of any triangle can be found by dividing the area of the triangle (A) by its semi-perimeter (s). r = (A/s) The area A is found using Heron's formula: A = \sqrt{s(s-a)(s-b)(s-c)} Step-by-Step Solution: 1. Calculate the Semi-perimeter (s): s = (a + b + c/2) = (6 + 11 + 15/2) = (32/2) = 16 cm. 2. Calculate the Area (A) using Heron's Formula: A = \sqrt{16(16 - 6)(16 - 11)(16 - 15)} A = \sqrt{16 × 10 × 5 × 1} A = \sqrt{800} = \sqrt{400 × 2} = 20\sqrt{2} cm^2. 3. Calculate the Inradius (r): r = (A/s) = \frac{20\sqrt{2}}{16} Simplify the fraction by dividing the numerator and denominator by 4: r = \frac{5\sqrt{2}}{4} cm. Exam Strategy & Shortcut: To compute \sqrt{800} quickly, group the factors logically: 16 × (10 × 5) = 16 × 50 = 800. You know \sqrt{16} = 4 and \sqrt{50} = 5\sqrt{2}. Multiply them directly: 4 × 5\sqrt{2} = 20\sqrt{2}. This saves time factoring large numbers. Common Pitfall: A frequent error is forgetting to divide the area by the semi-perimeter s and instead dividing by the full perimeter, which would yield exactly half of the correct answer. Always remember r = A/s, not A/P. Final Answer: Therefore, the correct answer is \frac{5\sqrt{2}}{4} cm.

Question 5

The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is

Accepted Answer

Concept & Inradius and Area Relationship: The area (A) of any triangle can be expressed as the product of its inradius (r) and its semi-perimeter (s): A = r × s The circumference of the incircle is C = 2\pi r, which allows us to find the inradius r. Step-by-Step Solution: 1. Find the Inradius (r): Given the circumference is 88 cm: 2\pi r = 88 2 × (22/7) × r = 88 (44/7) × r = 88 r = (88 × 7/44) = 2 × 7 = 14 cm. 2. Find the Semi-perimeter (s): Given the perimeter is 30 cm: s = \frac{Perimeter}{2} = (30/2) = 15 cm. 3. Calculate the Area (A): A = r × s A = 14 × 15 = 210 cm^2. Exam Strategy & Shortcut: When you see a circumference of 88, recognize instantly that it's double 44, which corresponds to a radius of 14 (since C=44 means r=7 for \pi=22/7). Then immediately multiply this radius (14) by half the given perimeter (15) to get 14 × 15 = 210. The entire process can be done mentally. Common Pitfall: Students often multiply the inradius by the full perimeter instead of the semi-perimeter, which leads to the incorrect answer of 420 cm^2. Always remember A = rs, not rp. Final Answer: Therefore, the correct answer is 210 cm^2.

Question 6

If in a triangle, the area is numerically equal to the perimeter, then the radius of the inscribed circle of the triangle is

Accepted Answer

Concept & Inradius-Area Formula: The fundamental relationship connecting the area of a triangle (A), its perimeter (P), and the radius of its inscribed circle (inradius r) is: A = r × s where s is the semi-perimeter (s = P / 2). Step-by-Step Solution: 1. Express the Formula in Terms of Perimeter: Substitute s = (P/2) into the area formula: A = r × (P/2) 2. Apply the Given Condition: We are given that the area is numerically equal to the perimeter, so A = P. Substitute A for P in our equation: A = r × (A/2) 3. Solve for r: Divide both sides by A (since area cannot be zero): 1 = (r/2) r = 2 Exam Strategy & Shortcut: This is a theoretical property that can be memorized. Whenever a 2D shape's area is calculated as "radius times semi-perimeter" (which is true for all triangles and tangential polygons), if Area = Perimeter, the inradius is always exactly 2. Common Pitfall: A common mistake is attempting to assume a specific type of triangle (like equilateral) and working through complex side-length formulas, which wastes time and often introduces calculation errors, when the relationship holds universally for all triangles. Final Answer: Therefore, the correct answer is 2.

Question 7

A circle, a square and an equilateral triangle have the same area. The correct increasing order of the perimeters will be

Accepted Answer

Concept & Isoperimetric Property (Inverse): The Isoperimetric principle establishes that a circle is the most efficient shape for enclosing area. This means for a fixed area, a circle will require the least amount of perimeter. Conversely, shapes with fewer sides are less efficient and require a larger perimeter to enclose that same area. Step-by-Step Solution: 1. Understand Efficiency: Efficiency here means getting the most area for the least perimeter. - Most efficient: Circle - Moderately efficient: Square (4 sides) - Least efficient: Equilateral Triangle (3 sides) 2. Order by Perimeter (Least to Greatest): Since they all have the same area, the most efficient shape will have the smallest perimeter. - Smallest perimeter: Circle - Next largest: Square - Largest perimeter: Triangle 3. Format the Output: The increasing order (smallest to largest) is circle, square, triangle. Exam Strategy & Shortcut: This is the inverse of the standard Isoperimetric rule. If equal perimeters mean Area_{circle} > Area_{square} > Area_{triangle}, then for equal areas, the perimeters must compensate in reverse: Perimeter_{circle} < Perimeter_{square} < Perimeter_{triangle}. Just remember: the circle always "wins" by doing the most with the least. Common Pitfall: Confusing this question with the "equal perimeter" variation. Students often skim the question, assume it's asking for area based on equal perimeters, and pick the exact opposite order. Always underline whether "Area" or "Perimeter" is being held constant. Final Answer: Therefore, the correct answer is circle, square, triangle.

Question 8

The area of the largest triangle that can be inscribed in a semi-circle of radius $r$, is

Accepted Answer

Concept & Inscribed Triangle Geometry: Any triangle inscribed in a semi-circle with its base on the diameter will be a right-angled triangle (Thales's Theorem). The area of a triangle is (1/2) × base × height. To maximize the area for a fixed base, the height must be maximized. Step-by-Step Solution: 1. Determine the Base: The largest triangle inscribed in a semi-circle must span the entire flat edge of the semi-circle. Therefore, its base is the diameter of the semi-circle. Base = 2r 2. Determine the Maximum Height: The third vertex must lie on the semi-circle arc. To maximize the height of the triangle, this vertex must be at the highest point of the semi-circle, which is directly above the center. The distance from the center to this highest point is simply the radius. Maximum Height = r 3. Calculate the Maximum Area: Area = (1/2) × Base × Height Area = (1/2) × (2r) × (r) Area = r^2 Exam Strategy & Shortcut: Visualize the semi-circle. The base is locked at 2r. The highest you can drag the peak of the triangle is the top center, which is height r. Half of 2r × r is just r^2. The visualization is instant. Common Pitfall: A common mistake is assuming the formula for the area of the semi-circle ((\pi r^2/2)) is needed, or mistakenly picking 2r^2 by forgetting to multiply by the 1/2 in the triangle area formula. Final Answer: Therefore, the correct answer is r^2.

Question 9

$ABC$ is a right-angled triangle with right angle at $B$. If the semi-circle on $AB$ with $AB$ as diameter encloses an area of 81 sq. cm and the semi-circle on $BC$ with $BC$ as diameter encloses an area of 36 sq. cm, then the area of the semi-circl…

Accepted Answer

Concept & Generalized Pythagorean Theorem: The Pythagorean theorem states that in a right triangle, a^2 + b^2 = c^2. This principle generalizes to any similar geometric figures constructed on the sides of a right triangle. If you draw similar shapes (like squares, equilateral triangles, or semi-circles) on the three sides, the area of the shape on the hypotenuse equals the sum of the areas of the shapes on the two legs. Area_{AC} = Area_{AB} + Area_{BC} Step-by-Step Solution: 1. Identify the Components: - Area of semi-circle on leg AB = 81 sq. cm. - Area of semi-circle on leg BC = 36 sq. cm. - AC is the hypotenuse (since the right angle is at B). 2. Apply the Generalized Pythagorean Theorem: The area of the semi-circle on the hypotenuse (AC) is the sum of the areas of the semi-circles on the legs (AB and BC). Area_{AC} = 81 + 36 3. Calculate the Final Area: Area_{AC} = 117 sq. cm. Exam Strategy & Shortcut: Recognize that area scales with the square of the dimension. Since AB^2 + BC^2 = AC^2, multiplying the entire equation by (\pi/8) gives the semi-circle areas: (\pi/8)AB^2 + (\pi/8)BC^2 = (\pi/8)AC^2. Thus, you just add the two given areas directly: 81 + 36 = 117. No need to find radii. Common Pitfall: A massive time-waster is working backward to find the actual radii of the semi-circles, computing the side lengths of the triangle, using the Pythagorean theorem to find AC, and then computing the area of the semi-circle on AC. While correct, it takes ten times longer. Final Answer: Therefore, the correct answer is 117 cm^2.

Question 10

When the circumference of a toy balloon is increased from $20	ext{ cm}$ to $25	ext{ cm}$, its radius is increased by

Accepted Answer

Concept & Circumference of a Circle: The circumference of a circle is directly proportional to its radius. The formula relating them is: C = 2\pi r Therefore, any change in the circumference (\Delta C) can be directly translated to a change in the radius (\Delta r) using \Delta r = (\Delta C/2\pi). Step-by-Step Solution: • Given: Initial circumference C_1 = 20 cm, Final circumference C_2 = 25 cm. • Deduction: The increase in circumference is \Delta C = 25 - 20 = 5 cm. • Calculation: Using the relationship C = 2\pi r, the increase in radius is \Delta r = (\Delta C/2\pi). • Calculation: Substitute the value of \Delta C: \Delta r = (5/2\pi). Exam Strategy & Shortcut: Instead of finding the individual radii r_1 and r_2 and then subtracting them, directly use the difference. Difference in radius = \frac{Difference in circumference}{2\pi}. This saves significant calculation time. Common Pitfall: A common mistake is forgetting the 2 in the denominator and selecting (5/\pi) as the answer, which corresponds to the change in diameter, not radius. Final Answer: Therefore, the correct answer is (5/2\pi).

Question 11

A can go round a circular path 8 times in 40 minutes. If the diameter of the circle is increased to 10 times the original diameter, then the time required by A to go round the new path once, travelling at the same speed as before, is

Accepted Answer

Concept & Distance, Speed, and Time: The distance around a circular path is its circumference, which is given by \pi × diameter (\pi d). Time required is proportional to the distance traveled if the speed remains constant: T = (D/S) Step-by-Step Solution: • Given: 8 rounds take 40 minutes. • Deduction: 1 round of the original path takes (40/8) = 5 minutes. • Calculation: The original distance for 1 round is \pi d. • Calculation: The new diameter is 10d, so the new distance for 1 round is \pi(10d) = 10(\pi d). • Deduction: The new path is 10 times as long as the original path. • Calculation: Since speed is constant, the time required will also be 10 times the original time. 10 × 5 minutes = 50 minutes. Exam Strategy & Shortcut: Realize that distance is directly proportional to diameter. If diameter becomes 10×, distance becomes 10×. Time for 1 old round is 5 min. Therefore, time for 1 new round is 10 × 5 = 50 min. Common Pitfall: A common error is confusing the number of rounds (8) and applying the multiplier to the total 40 minutes directly without finding the time for a single round first. Final Answer: Therefore, the correct answer is 50 min.

Question 12

If the radius of a circle is doubled, its area is increased by

Accepted Answer

Concept & Scaling Factors in Geometry: When the linear dimension (radius) of a 2D shape is scaled by a factor of k, its area is scaled by a factor of k^2. New Area = k^2 × Old Area Step-by-Step Solution: • Given: The radius is doubled. This means the scaling factor k = 2. • Calculation: Let the original radius be r. Original area A_1 = \pi r^2. • Calculation: New radius is 2r. New area A_2 = \pi (2r)^2 = 4\pi r^2. • Deduction: The new area is 4 times the original area. • Calculation: The *increase* in area is 4\pi r^2 - \pi r^2 = 3\pi r^2. • Calculation: Percentage increase = ((3\pi r^2/\pi r^2)) × 100 = 300%. Exam Strategy & Shortcut: "Doubled" means a multiplier of 2. Area multiplier is 2^2 = 4. If you have 4 times the amount you started with, you have gained 3 times your original amount. 3 × 100% = 300% increase. Common Pitfall: A classic trap is selecting 400%. The area *becomes* 400% of its original size, but the question specifically asks what it is *increased by*, which requires subtracting the original 100%. Final Answer: Therefore, the correct answer is 300%.

Question 13

If the circumference of a circle increases form $4\pi$ to $8\pi$, what change occurs in its area?

Accepted Answer

Concept & Circumference and Area Relationship: The scale factor for the circumference of a circle is identical to the scale factor for its radius. Once the scale factor for the radius (k) is known, the area scales by a factor of k^2. k = (C_2/C_1) = (r_2/r_1) Step-by-Step Solution: • Given: Initial circumference C_1 = 4\pi. Final circumference C_2 = 8\pi. • Calculation: Determine the scale factor for circumference: k = (8\pi/4\pi) = 2. • Deduction: Since circumference doubles, the radius also doubles. • Calculation: Let original area be A_1 = \pi r^2. • Calculation: If radius doubles, new area A_2 = \pi(2r)^2 = 4\pi r^2. • Deduction: The new area is 4 times the original area, which means it quadruples. Exam Strategy & Shortcut: Skip finding the actual radius. Notice that 8\pi is twice 4\pi. The linear dimensions (circumference, radius, diameter) doubled (2×). Area is 2-dimensional, so it scales by 2^2 = 4×. Thus, it quadruples. Common Pitfall: A common mistake is assuming that area increases linearly with circumference and choosing "It doubles." Remember area is proportional to the square of linear dimensions. Final Answer: Therefore, the correct answer is It quadruples.

Question 14

If the circumference of a circle is increased by 20%, what will be the effect on the circle?

Accepted Answer

Concept & Circumference and Area Proportionality: An increase in the circumference by a certain percentage implies the exact same percentage increase in the radius, since C = 2\pi r. The effect on the area (which is implied by "effect on the circle" given the options) can be calculated using the successive percentage change formula: x + y + (xy/100) Step-by-Step Solution: • Given: Circumference increases by 20%. • Deduction: Because C \propto r, the radius also increases by 20%. So, x = 20. • Calculation: We need to find the change in area (A = \pi r^2). Apply successive change for two identical 20% increases: 20 + 20 + (20 × 20/100). • Calculation: 40 + (400/100). • Calculation: 40 + 4 = 44%. Exam Strategy & Shortcut: Any 20% linear increase always leads to a 44% area increase. Memorize standard values: 10% \rightarrow 21%, 20% \rightarrow 44%, 30% \rightarrow 69%. This saves calculation time entirely. Common Pitfall: Students often select "Cannot be determined" because the question vaguely asks for "effect on the circle" rather than specifically "effect on the area". However, analyzing the numerical options makes it clear area is being requested. Final Answer: Therefore, the correct answer is 44% increase.

Question 15

Three equal circles are described with vertices of the triangles as centres. If the radius of each circle is $r$, the sum of areas of the portions of the circles intercepted in a triangle is (Campus Recuritment, 2010)

Accepted Answer

Concept & Area of Sectors in a Triangle: The sum of the interior angles of any triangle is always 180^\circ (or \pi radians). The intercepted portions of the circles at each vertex represent three sectors of a circle with a common radius r. Area of a sector = (	heta/360^\circ) × \pi r^2 Step-by-Step Solution: • Given: Three circles with radius r centered at the vertices of a triangle. • Deduction: Let the interior angles of the triangle be 	heta_1, 	heta_2, and 	heta_3. We know that 	heta_1 + 	heta_2 + 	heta_3 = 180^\circ. • Calculation: The area of the sector at the first vertex is (	heta_1/360^\circ) × \pi r^2. • Calculation: The total area of the three sectors is (	heta_1 + 	heta_2 + 	heta_3/360^\circ) × \pi r^2. • Calculation: Substitute the sum of angles: (180^\circ/360^\circ) × \pi r^2. • Calculation: Simplify the fraction: (1/2)\pi r^2. Exam Strategy & Shortcut: Visualize bringing the three corners of any triangle together. Because their angles sum to 180^\circ, joining the three sectors perfectly forms a semi-circle. The area of a semi-circle with radius r is simply (1/2)\pi r^2. Common Pitfall: Students often try to assume the triangle must be equilateral and calculate for three 60^\circ sectors individually, which wastes time (though it yields the same correct answer). The principle works universally for any triangle. Final Answer: Therefore, the correct answer is (1/2)\pi r^2.

Question 16

Three circles of radius $3.5	ext{ cm}$ are placed in such a way that each circle touches the other two. The area of the portion enclosed by the circles is

Accepted Answer

Concept & Area Enclosed by Touching Circles: When three equal circles touch each other, connecting their centers forms an equilateral triangle. The area enclosed between the circles is the area of this triangle minus the area of the three circular sectors within it. Enclosed Area = Area of Equilateral Triangle - Area of 3 Sectors Step-by-Step Solution: • Given: Radius r = 3.5 cm = (7/2) cm. • Deduction: The side of the equilateral triangle formed by the centers is a = 2r = 7 cm. • Calculation: Area of the equilateral triangle = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} × 7^2 = \frac{49\sqrt{3}}{4}. • Calculation: Using \sqrt{3} \approx 1.732, Area = (49 × 1.732/4) \approx 21.217 cm^2. • Calculation: Each interior angle is 60^\circ. The three sectors form a semicircle (180^\circ). Area of sectors = (1/2)\pi r^2 = (1/2) × (22/7) × ((7/2))^2 = (11/7) × (49/4) = (77/4) = 19.25 cm^2. • Calculation: Enclosed Area = 21.217 - 19.25 = 1.967 cm^2. Exam Strategy & Shortcut: For three mutually touching circles of radius r, the standard formula for the enclosed area is r^2(\sqrt{3} - (\pi/2)). Plugging in r = 3.5 directly yields (3.5)^2(1.732 - 1.571) = 12.25 × 0.161 = 1.972. Due to approximation differences in \pi and \sqrt{3}, 1.967 is the intended precise answer using (22/7). Common Pitfall: A common mistake is failing to convert the three 60^\circ sectors into a single semi-circle for calculation, leading to overly complex and error-prone fraction math. Final Answer: Therefore, the correct answer is 1.967 cm^2.

Question 17

Four horses are tethered at four corners of a square plot of side $63	ext{ metres}$ so that they just cannot reach one another. The area left ungrazed is

Accepted Answer

Concept & Area of Grazing Sectors: This is practically identical to the "four touching circles" problem. If the horses "just cannot reach one another", their grazing regions represent four quarter-circles that meet exactly at the midpoints of the square's sides. Ungrazed Area = Total Plot Area - Total Grazed Area Step-by-Step Solution: • Given: Side of the square plot s = 63 m. • Deduction: The radius r of each horse's grazing sector is exactly half the side length: r = (63/2) = 31.5 m. • Calculation: Area of the square plot = 63^2 = 3969 m^2. • Calculation: The four 90^\circ grazing sectors form one full circle of radius 31.5 m. Area grazed = \pi r^2. • Calculation: Grazed Area = (22/7) × (63/2) × (63/2) = 11 × 9 × (63/2) = 99 × 31.5 = 3118.5 m^2. • Calculation: Ungrazed Area = 3969 - 3118.5 = 850.5 m^2. Exam Strategy & Shortcut: Using the derived formula from the four touching circles concept, the ungrazed area is directly (6/7)r^2. (6/7) × ((63/2))^2 = (6/7) × (3969/4) = (3/7) × (3969/2) = 3 × (567/2) = (1701/2) = 850.5. This significantly reduces the arithmetic steps. Common Pitfall: A common misstep is calculating the area of four separate circles instead of realizing the grazing areas inside the plot only amount to four quarter-sectors (one single full circle equivalent). Final Answer: Therefore, the correct answer is 850.5 m^2.

Question 18

The difference between the areas of the circumcircle and the incircle of a regular polygon of $n$ sides with each side of length $2a$, is

Accepted Answer

Concept & Incircle and Circumcircle Geometry: For any regular polygon, the circumradius (R), the inradius (r), and half of the side length form a right-angled triangle. By using the Pythagorean theorem on this fundamental triangle, we can find the difference in their areas without needing the number of sides n. R^2 = r^2 + (Half-side)^2 Step-by-Step Solution: • Let the regular polygon have n sides, each of length 2a. • Let the center of the polygon be O. • The distance from O to a vertex is the circumradius, R. • The perpendicular distance from O to the midpoint of a side is the inradius, r. • This forms a right-angled triangle with the hypotenuse R, height r, and base equal to half the side length, which is (2a/2) = a. • By the Pythagorean theorem: R^2 = r^2 + a^2. • Rearranging gives: R^2 - r^2 = a^2. • The area of the circumcircle is \pi R^2. • The area of the incircle is \pi r^2. • The difference in their areas = \pi R^2 - \pi r^2 = \pi (R^2 - r^2). • Substitute R^2 - r^2 = a^2 into the equation: Area Difference = \pi a^2. Exam Strategy & Shortcut: Notice that the options are mostly independent of n (Option A). In geometry, if a relationship holds for an n-sided polygon, it must hold for the simplest one. Assume a square (n=4) with side 2a. Inradius r = a, circumradius R = a\sqrt{2}. Area difference = \pi(a\sqrt{2})^2 - \pi(a)^2 = 2\pi a^2 - \pi a^2 = \pi a^2. Common Pitfall: Students often assume the area difference depends on the number of sides n and mistakenly guess a complex option like (2n + 1)\pi a^2. Final Answer: Therefore, the correct answer is \pi a^2.

Question 19

Two equal circles are drawn in square in such a way that a side of the square forms diameter of each circle. If the remaining area of the square is 42 cm$^2$. How much will the diameter of the circle measures? [IBPS—RRB Officers Gr. 'B' Exam, 2015]

Accepted Answer

Concept & Area Subtraction: The problem refers to semicircles formed on the sides of a square that cut into its area. The remaining area is the area of the square minus the area of these two semicircles (which combine to form one full circle). Area_{remaining} = Area_{square} - Area_{circle} Step-by-Step Solution: • Let the side of the square be a. The total area of the square is a^2. • Two semicircles are drawn with a diameter equal to the side of the square (a). Their radius is r = (a/2). • The combined area of these two semicircles is equal to the area of one full circle: \pi × ((a/2))^2 = (\pi a^2/4). • Substitute \pi = (22/7): Area = (22/7) × (a^2/4) = (11a^2/14). • The remaining area of the square is a^2 - (11a^2/14) = (3a^2/14). • We are given that the remaining area is 42 cm^2. So, (3a^2/14) = 42. • Solving for a^2: a^2 = 42 × (14/3) = 14 × 14 = 196. • Therefore, a = 14 cm. Since the diameter of the circle is equal to the side of the square, the diameter is also 14 cm. Exam Strategy & Shortcut: Memorize that when a full circle (or two semicircles) of maximum diameter is inscribed in a square, the circle occupies (11/14) of the square's area, and the remaining space is (3/14) of the area. Setting (3/14) × a^2 = 42 quickly yields a = 14. Common Pitfall: Taking "two equal circles" literally as two independent full circles inside the square, which would cause an impossible negative remaining area. Contextually, this phrasing in competitive exams implies semicircles on opposite edges. Final Answer: Therefore, the correct answer is 14 cm.

Question 20

A plate on square base made of brass is of length $x$ c and width 1 mm. The plate weights 4725gm. If 1 cubic cm of brass weights 8.4 gram. then the value of $x$ is [SSC—CHSL (10+2) Exam, 2015]

Accepted Answer

Concept & Density and Volume: The volume of a 3D object can be found by dividing its total mass by its density. For a rectangular prism or square-based plate, the volume is also the product of its length, breadth, and height/thickness. Volume = (Total\ Weight/Density) = Length × Breadth × Height Step-by-Step Solution: • Given: Total weight = 4725 gm, Density = 8.4 gm/cm^3. • First, find the total volume of the brass plate: Volume = (4725/8.4) = 562.5 cm^3. • The plate has a square base of length x cm. (Note: "c" in the question implies cm based on standard units). Since the base is square, both length and breadth are x cm. • The thickness (width) is given as 1 mm. We must convert this to cm to match the other units: 1 mm = 0.1 cm. • Set up the volume equation: Volume = Base Area × Thickness = x × x × 0.1. • Equate the two volume expressions: 0.1x^2 = 562.5. • Solve for x^2: x^2 = (562.5/0.1) = 5625. • Taking the square root gives x = 75. Exam Strategy & Shortcut: Instead of long division for (4725/8.4), recognize that you will eventually divide by 0.1 (the thickness). You can compute (4725/0.84) = 5625 directly. Knowing your square roots speeds this up: 75^2 = 5625. Common Pitfall: Forgetting to convert the thickness from millimeters to centimeters before multiplying. Using 1 instead of 0.1 will result in x^2 = 562.5, which does not yield a clean integer root. Final Answer: Therefore, the correct answer is 75.

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