Question 1

The circumference of a circle is 88 cm. Using pi = 22/7, find the area of the circle in square centimetres.

Accepted Answer

Introduction / Context: This problem connects two standard formulas for a circle: one for circumference and one for area. We are given the circumference and must find the area. To do this, we first determine the radius from the circumference and then substitute that radius into the area formula. This tests multi step reasoning with basic circle geometry. Given Data / Assumptions: Circumference C = 88 cm. Formula for circumference: C = 2 * pi * r. Formula for area: A = pi * r^2. Use pi = 22/7 as standard approximation. Concept / Approach: Step one is to solve for the radius r using C = 2 * pi * r. Once r is known, step two is to compute A = pi * r^2. Because pi is given as 22/7, calculations can be simplified by using fraction arithmetic. The key is to perform operations carefully and keep track of units, ending with square centimetres for area. Step-by-Step Solution: Given C = 88 cm and C = 2 * pi * r.Substitute pi = 22/7: 88 = 2 * (22/7) * r.Compute 2 * (22/7) = 44/7.So 88 = (44/7) * r.Multiply both sides by 7 to clear denominator: 88 * 7 = 44 * r.Compute 88 * 7 = 616, so 616 = 44 * r.Therefore r = 616 / 44 = 14 cm.Now area A = pi * r^2 = (22/7) * 14^2.Compute 14^2 = 196, so A = (22/7) * 196.196 / 7 = 28, hence A = 22 * 28 = 616 sq cm. Verification / Alternative check: We can check by reversing the process. With radius r = 14 cm, circumference C should be 2 * pi * r = 2 * (22/7) * 14 = (44/7) * 14 = 44 * 2 = 88 cm, which matches the given value. The area from this radius is (22/7) * 196 = 616 sq cm, confirming that the computation is consistent at every step. Why Other Options Are Wrong: 308 sq cm would be obtained if we mistakenly used r^2 = 98 instead of 196. 154 sq cm and 77 sq cm are fractions of the correct area and arise from missing multiplication steps. 484 sq cm might come from squaring 22 instead of using the radius properly. None of these fit both the circumference and area formulas simultaneously for a single radius value. Common Pitfalls: Typical errors include confusing diameter with radius, or using C = pi * r instead of 2 * pi * r. Some learners also forget to square the radius when computing the area, which halves the correct value. Careful substitution and stepwise solving, as shown, minimise these mistakes. Always verify by recomputing the circumference from the found radius. Final Answer: 616 sq cm

Question 2

In an equilateral triangle, the radius of the circumscribed circle (circumcircle) is 14 cm. What is the radius of the inscribed circle (incircle) in centimetres?

Accepted Answer

Introduction / Context: This geometry question is about the relationship between the circumradius and inradius of an equilateral triangle. An equilateral triangle has all sides equal and all internal angles equal to 60 degrees. For such a triangle, the circumradius and inradius are directly related by simple formulas, which allows us to find one when the other is known. Given Data / Assumptions: Triangle is equilateral. Circumradius R = 14 cm. Side length of equilateral triangle is denoted by a. Standard formulas: R = a / (sqrt(3)) and r = a / (2 * sqrt(3)), where r is inradius. Alternatively, r = R / 2 for an equilateral triangle. Concept / Approach: For an equilateral triangle, the circumradius R and inradius r have a fixed ratio. Since R = a / sqrt(3) and r = a / (2 * sqrt(3)), dividing r by R gives r / R = 1/2. Thus r = R / 2. With R known as 14 cm, we can directly find r by halving R. This approach avoids having to compute the side length explicitly, though we can still do so as a check. Step-by-Step Solution: Given circumradius R = 14 cm.For an equilateral triangle, r = R / 2.So inradius r = 14 / 2 = 7 cm.Therefore, the radius of the inscribed circle is 7 cm. Verification / Alternative check: We can verify by computing the side length. From R = a / sqrt(3), we have a = R * sqrt(3) = 14 * sqrt(3). Then r = a / (2 * sqrt(3)) = (14 * sqrt(3)) / (2 * sqrt(3)) = 14 / 2 = 7 cm. This alternative method matches the earlier result and confirms that r = 7 cm is correct. Why Other Options Are Wrong: 14 cm would make inradius equal to circumradius, which is not true for an equilateral triangle. 28 cm and 35 cm are larger than the circumradius and cannot be the inradius. 21 cm would imply r = (3/2) * R, which contradicts the known geometric relation. Only 7 cm correctly reflects the relationship between inradius and circumradius in an equilateral triangle. Common Pitfalls: A frequent mistake is to confuse inradius and circumradius formulas, or to apply formulas that belong to general triangles instead of the special equilateral case. Some learners attempt to compute side length and mis-handle the square roots. Remembering the simple relation r = R / 2 for equilateral triangles is a fast and reliable way to answer such questions. Always ensure that inradius is smaller than circumradius, which helps catch obvious errors. Final Answer: 7 cm

Question 3

What is the measure, in degrees, of each exterior angle of a regular polygon that has 9 sides?

Accepted Answer

Introduction / Context: This question concerns the exterior angles of a regular polygon. A regular polygon has all sides and all interior angles equal. The sum of all exterior angles of any convex polygon is always 360 degrees. Therefore, for a regular polygon, each exterior angle can be found by dividing 360 by the number of sides. Here, the polygon has 9 sides, so we apply this standard rule. Given Data / Assumptions: Polygon is regular with n = 9 sides. All exterior angles of a regular polygon are equal. Sum of all exterior angles of any convex polygon = 360 degrees. Concept / Approach: The key concept is that the sum of the exterior angles, taken one at each vertex, is 360 degrees irrespective of the number of sides. For a regular polygon, each exterior angle is equal, so each is simply 360 / n. Substituting n = 9 gives the measure of each exterior angle. No advanced geometry is needed, just a clear understanding of this property. Step-by-Step Solution: Number of sides n = 9.Sum of all exterior angles = 360 degrees.Each exterior angle in a regular polygon = 360 / n.So each exterior angle = 360 / 9.Compute 360 / 9 = 40 degrees.Therefore, each exterior angle of a regular 9 sided polygon is 40 degrees. Verification / Alternative check: We can also relate interior and exterior angles. For any regular polygon, interior angle + exterior angle at a vertex = 180 degrees. If each exterior angle is 40 degrees, then each interior angle would be 180 - 40 = 140 degrees. Using the interior angle formula, interior angle = ((n - 2) * 180) / n = (7 * 180) / 9 = 140 degrees, which matches. This confirms that 40 degrees is correct for the exterior angle. Why Other Options Are Wrong: 36 degrees would be correct for a regular polygon with 10 sides, not 9. 30 degrees corresponds to 12 sides, 24 degrees to 15 sides, and 60 degrees to 6 sides. None of these match n = 9. Only 40 degrees gives a sum of 9 * 40 = 360 degrees, which satisfies the basic property of exterior angles. Common Pitfalls: A common mistake is to use formulas meant for interior angles when the question asks about exterior angles. Some learners also incorrectly add 360 and divide by (n - 2). Remember that the sum of exterior angles is always 360 degrees, independent of n. Keeping this rule in mind and directly dividing by the number of sides ensures a quick and accurate solution. Final Answer: 40

Question 4

In right angled triangle DEF, the right angle is at E. If angle D is 30 degrees and side EF = 6√3 cm, what is the length of side DE in centimetres?

Accepted Answer

Introduction / Context: This question is about a special right angled triangle with angles 30 degrees, 60 degrees, and 90 degrees. Such a triangle has well known side length ratios. We are given one side of the triangle and a specific angle, and we must find another side using these ratios. This is a standard trigonometry and geometry concept often tested in aptitude exams. Given Data / Assumptions: Triangle DEF is right angled at E, so angle E = 90 degrees. Angle D = 30 degrees, so angle F = 60 degrees. Side EF = 6√3 cm. We need to find DE in cm. Triangle is therefore a 30-60-90 triangle. Concept / Approach: In a 30-60-90 right triangle, the sides opposite these angles follow the ratio 1 : √3 : 2. The smallest side, opposite 30 degrees, is k. The side opposite 60 degrees is k√3, and the hypotenuse opposite 90 degrees is 2k. Here, angle D is 30 degrees, and the side opposite angle D is EF. Thus EF corresponds to k. Using EF, we can determine k and then find DE, which is opposite 60 degrees and equals k√3. Step-by-Step Solution: Angle D = 30 degrees, angle E = 90 degrees, angle F = 60 degrees.In triangle DEF, side opposite angle D (30 degrees) is EF.Given EF = 6√3 cm.In a 30-60-90 triangle, let the side opposite 30 degrees be k, opposite 60 degrees be k√3, and hypotenuse be 2k.So EF corresponds to k and k = 6√3.Side opposite 60 degrees is DE = k√3 = 6√3 * √3 = 6 * 3 = 18 cm.Therefore DE = 18 cm. Verification / Alternative check: Having k = 6√3, hypotenuse DF would be 2k = 12√3. We can verify with the Pythagoras theorem. EF^2 = (6√3)^2 = 36 * 3 = 108. DE^2 = 18^2 = 324. Sum = 108 + 324 = 432. Hypotenuse squared DF^2 should be (12√3)^2 = 144 * 3 = 432, which matches. This confirms that DE = 18 cm fits the 30-60-90 triangle perfectly. Why Other Options Are Wrong: 12√3 cm and 18√3 cm treat DE as hypotenuse or misassign the roles of sides in the 30-60-90 ratio. A value of 12 cm comes from confusing which side is opposite which angle. 9 cm is too short and does not satisfy Pythagoras when combined with EF = 6√3 and a consistent hypotenuse. Only 18 cm leads to a valid 30-60-90 triangle relation and a correct Pythagoras check. Common Pitfalls: Students often mix up which side corresponds to which angle in a 30-60-90 triangle. Remember that the side opposite 30 degrees is the smallest, opposite 60 degrees is medium, and opposite 90 degrees is the largest. Another error is assigning EF to the wrong part of the ratio. Carefully identifying the angle opposite each given side avoids these issues. Final Answer: 18 cm

Question 5

The length and breadth of a rectangle are 24 cm and 7 cm respectively. Calculate the perimeter of the rectangle in centimetres.

Accepted Answer

Introduction / Context: This problem involves finding the perimeter of a rectangle when its length and breadth are known. Perimeter is the distance around the boundary of the rectangle. The question reinforces the basic formula for the perimeter of a rectangle, which is commonly used in everyday contexts such as fencing a garden or framing a picture. Given Data / Assumptions: Length L = 24 cm. Breadth B = 7 cm. Rectangle has opposite sides equal and four right angles. Perimeter formula: P = 2(L + B). Concept / Approach: The perimeter of a rectangle is twice the sum of its length and breadth because each dimension appears on two opposite sides. Therefore P = 2(L + B). We add 24 and 7, then multiply by 2 to obtain the perimeter in centimetres. The process is simple and is a direct application of the formula. Step-by-Step Solution: Length L = 24 cm, breadth B = 7 cm.Perimeter P = 2(L + B).Compute L + B = 24 + 7 = 31.Now P = 2 * 31 = 62 cm.Hence the perimeter of the rectangle is 62 centimetres. Verification / Alternative check: Another way is to add all four sides separately. The rectangle has two lengths of 24 cm and two breadths of 7 cm. Adding gives 24 + 7 + 24 + 7 = (24 + 24) + (7 + 7) = 48 + 14 = 62 cm. This matches the result from the formula and confirms our answer is correct. Why Other Options Are Wrong: 124 is obtained if someone multiplies 62 by 2 again, misinterpreting the formula. 48 is only the sum of the two lengths, ignoring the breadths. 96 could result from incorrectly using 4 * 24 or 4 * some average value, and 68 is a near miss from simple addition mistakes. None of these match the correct formula and calculation for perimeter. Common Pitfalls: Many learners confuse area and perimeter formulas. Area of a rectangle is L * B, while perimeter is 2(L + B). Another common mistake is to add 2L + B instead of 2(L + B). Visualising the rectangle and tracing its boundary can help remember that each side length is counted twice. Always double check that you have included both length and breadth twice in the perimeter calculation. Final Answer: 62

Question 6

What are the coordinates of the reflection of the point (-0.5, 6) in the x-axis?

Accepted Answer

Introduction / Context: This question tests understanding of reflections of points across the coordinate axes in the Cartesian plane. Reflecting a point across the x-axis changes its vertical position while keeping the horizontal coordinate the same. Recognising how the sign of the y coordinate changes under such a reflection is the key idea here. Given Data / Assumptions: Original point is P(-0.5, 6). Reflection is to be taken in the x-axis. In Cartesian coordinates, x-axis reflection flips the sign of the y coordinate but leaves the x coordinate unchanged. Concept / Approach: When a point (x, y) is reflected across the x-axis, its image becomes (x, -y). The x coordinate remains the same because the point stays directly above or below its original horizontal position. The y coordinate changes sign because the point moves to the opposite side of the x-axis by the same vertical distance. Applying this rule to the given point yields the answer directly. Step-by-Step Solution: Given point P has coordinates (-0.5, 6).Reflection in x-axis sends (x, y) to (x, -y).Here x = -0.5 and y = 6.So the reflected point P' has coordinates (x, -y) = (-0.5, -6).Therefore the reflection of (-0.5, 6) in the x-axis is (-0.5, -6). Verification / Alternative check: Visually, the original point is 0.5 units to the left of the origin and 6 units above the x-axis. Its reflection across the x-axis should be directly below at the same horizontal position, that is 6 units below the x-axis. This gives coordinates (-0.5, -6), matching our algebraic result. This geometric reasoning confirms the correctness of the answer. Why Other Options Are Wrong: (0.5, -6) changes both coordinates, which corresponds to a reflection across the origin combined with a horizontal flip, not just across the x-axis. (-6, -0.5) and (6, -0.5) swap the roles of x and y, which would be a reflection across the line y = x or similar, not across the x-axis. (-0.5, 6) is the original point and shows no reflection at all. Only (-0.5, -6) correctly applies the rule (x, y) to (x, -y). Common Pitfalls: Students often confuse reflection across the x-axis with reflection across the y-axis. Reflection across the y-axis would change (x, y) to (-x, y). It is useful to remember that x-axis reflection flips the sign of y, and y-axis reflection flips the sign of x. Careful reading of the axis name and the coordinates helps avoid such mix ups. Final Answer: (-0.5, -6)

Question 7

The areas of two similar triangles ΔABC and ΔPQR are 36 square centimetres and 9 square centimetres respectively. If side PQ = 4 cm, what is the length of side AB in centimetres?

Accepted Answer

Introduction / Context: This question applies the properties of similar triangles to relate side lengths and areas. When two triangles are similar, the ratio of their areas is equal to the square of the ratio of any corresponding sides. Given the areas of two similar triangles and one side length in the smaller triangle, we can determine the corresponding side length in the larger triangle by using square roots of the area ratio. Given Data / Assumptions: Triangles ABC and PQR are similar. Area of ΔABC = 36 sq cm. Area of ΔPQR = 9 sq cm. Side PQ = 4 cm, and PQ corresponds to side AB. We need the length of AB. Concept / Approach: For similar triangles, area ratio = (side ratio)^2. If we let k be the ratio of corresponding sides AB / PQ, then (Area of ABC) / (Area of PQR) = k^2. From this, we can solve for k by taking the square root of the area ratio. Once k is known, AB = k * PQ. The idea is to move from area ratio back to side ratio by square rooting. Step-by-Step Solution: Area of ΔABC = 36 sq cm.Area of ΔPQR = 9 sq cm.Area ratio = 36 / 9 = 4.If side ratio AB / PQ = k, then area ratio = k^2.So k^2 = 4, which means k = sqrt(4) = 2 (taking the positive value because lengths are positive).Given PQ = 4 cm, so AB = k * PQ = 2 * 4 = 8 cm.Therefore, AB = 8 cm. Verification / Alternative check: We can verify by checking that the area scale factor matches. If all linear dimensions of ΔABC are twice those of ΔPQR, then its area should be 2^2 = 4 times as large. Since area of ΔPQR is 9 sq cm, area of ΔABC should be 9 * 4 = 36 sq cm, which is exactly the given value. This confirms that the side length AB must indeed be 8 cm when PQ is 4 cm. Why Other Options Are Wrong: 16 cm would correspond to a side ratio of 4, giving an area ratio of 16, not 4. 12 cm would give a side ratio of 3 and area ratio 9, inconsistent with the given 36 and 9. 6 cm implies a side ratio of 1.5 and area ratio 2.25, which does not match 4. 10 cm gives a non integer ratio that would not yield the given area ratio. Only 8 cm produces a side ratio of 2 and an area ratio of 4, consistent with the problem data. Common Pitfalls: A frequent mistake is to treat the area ratio directly as the side ratio, forgetting to take the square root. Another error is to invert the ratio of areas or to mix up which triangle is larger. Always identify which triangle has the larger area, compute the area ratio correctly, and then take the square root to get the scale factor for side lengths. Careful reading and methodical algebra prevent such errors. Final Answer: 8

Question 8

If the perimeter of a semicircle is 108 cm, find the radius of the semicircle in centimetres. (Use pi = 22/7.)

Accepted Answer

Introduction / Context: This question involves the perimeter of a semicircle, which consists of a curved arc plus a straight diameter. Given the perimeter of the semicircle, we must work backwards to find the radius. This requires knowing and correctly applying the formula for the perimeter of a semicircle, using the given value of pi. It is a good test of algebraic manipulation with geometric formulas. Given Data / Assumptions: Perimeter of semicircle P = 108 cm. Radius of semicircle = r cm (unknown). Perimeter of a semicircle = pi * r + 2 * r (arc length plus diameter). Use pi = 22/7. Concept / Approach: We set up the equation P = pi * r + 2r, plug in P = 108 and pi = 22/7, and then solve for r. To simplify, we combine terms in r by expressing them with a common denominator. After we have an equation of the form coefficient * r = 108, we can isolate r by division. The key is arranging and simplifying the fractional expression correctly. Step-by-Step Solution: Perimeter P = pi * r + 2r.Given P = 108 and pi = 22/7.So 108 = (22/7) * r + 2r.Write 2 as 14/7 so that both terms have denominator 7: 2r = (14/7) * r.Therefore 108 = (22/7) * r + (14/7) * r = (36/7) * r.So (36/7) * r = 108.Multiply both sides by 7: 36 * r = 108 * 7.Compute 108 * 7 = 756, so 36 * r = 756.Therefore r = 756 / 36 = 21 cm. Verification / Alternative check: Using r = 21 cm, we can recompute the perimeter to verify. Arc length of semicircle = pi * r = (22/7) * 21 = 22 * 3 = 66 cm. Diameter = 2r = 2 * 21 = 42 cm. Total perimeter P = 66 + 42 = 108 cm, which exactly matches the given perimeter. This confirms that r = 21 cm is consistent with the problem statement. Why Other Options Are Wrong: If r = 42 cm, the perimeter would be (22/7) * 42 + 84 = 132 + 84 = 216 cm, not 108 cm. For r = 28 cm, perimeter would be (22/7) * 28 + 56 = 88 + 56 = 144 cm. For r = 56 cm, perimeter would be far larger. A radius of 14 cm gives perimeter (22/7) * 14 + 28 = 44 + 28 = 72 cm. None of these equal 108 cm, so these options are incorrect. Common Pitfalls: A common error is to forget the straight diameter term and use only pi * r, which underestimates the perimeter. Others mistakenly use pi * 2r (full circumference) instead of pi * r, which overestimates it. Learners also sometimes mismanage the fraction 22/7, leading to incorrect algebra. Writing the formula clearly and handling fractions step by step avoids these mistakes. Final Answer: 21

Question 9

If the length of each side of an equilateral triangle is 8 cm, what is the exact value of its area in square centimetres?

Accepted Answer

Introduction / Context: This problem tests the standard mensuration formula for the area of an equilateral triangle. In competitive exams, equilateral triangles appear frequently because their symmetry allows neat formulas involving the square of the side and the square root of 3. Here, you are given the side length and are asked to find the exact area, not a decimal approximation, so remembering the correct formula is essential. Given Data / Assumptions: The triangle is equilateral, so all three sides are equal and each interior angle is 60 degrees. Each side of the triangle has length 8 cm. The required answer is the area in square centimetres, preferably in simplest radical form involving √3. We assume standard Euclidean geometry and no rounding unless needed for checking. Concept / Approach: The area A of an equilateral triangle with side length a is given by the well known formula A = (√3 / 4) * a^2. This comes from dropping an altitude, which forms two congruent right triangles, and then applying Pythagoras theorem. Once you know this formula, you simply substitute the side length and simplify. There is no need for trigonometry or coordinate geometry here, only direct substitution and squaring of the side length. Step-by-Step Solution: Let the side length of the equilateral triangle be a = 8 cm.Use the formula for area: A = (√3 / 4) * a^2.Compute a^2 = 8^2 = 64.Substitute into the formula: A = (√3 / 4) * 64.Simplify 64 / 4 = 16, so A = 16√3 sq cm. Verification / Alternative check: To verify, you can derive the formula quickly. Drop an altitude from one vertex. This altitude splits the opposite side into two segments of length 4 cm each and creates two right triangles. The hypotenuse is 8 cm and one leg is 4 cm. Using Pythagoras theorem, height h satisfies h^2 + 4^2 = 8^2, so h^2 = 64 − 16 = 48 and h = 4√3. Area is then (1 / 2) * base * height = (1 / 2) * 8 * 4√3 = 16√3 sq cm, matching the direct formula. Why Other Options Are Wrong: 32√3 sq cm is double the correct area and would correspond to treating 8 as the height as well as the side. The value 16 sq cm ignores √3 and is far too small for a triangle with sides of length 8 cm. The value 32 sq cm comes from incorrectly using (1 / 2) * 8 * 8. The value 64√3 sq cm is four times the correct area and arises from omitting the division by 4 in the formula. Common Pitfalls: A common mistake is to confuse the area formula of an equilateral triangle with that of a general triangle and write (1 / 2) * a^2 instead of (√3 / 4) * a^2. Some students also forget to square the side, using (√3 / 4) * 8 rather than (√3 / 4) * 8^2. Another error is to treat 8 as both base and height, which is not correct because the altitude is shorter than the side. Careful substitution prevents these errors. Final Answer: The area of the equilateral triangle is 16√3 sq cm.

Question 10

The diagonals of a rhombus measure 8 cm and 6 cm respectively. What is the area of this rhombus in square centimetres?

Accepted Answer

Introduction / Context: This question focuses on the special area formula for a rhombus in terms of its diagonals. A rhombus is a quadrilateral with all sides equal, and its diagonals have the useful property of being perpendicular bisectors of each other. In many exam problems, diagonals are provided instead of base and height, and you are expected to apply the appropriate formula quickly and accurately. Given Data / Assumptions: The figure is a rhombus with four equal sides. The lengths of its diagonals are 8 cm and 6 cm. The diagonals intersect at right angles and bisect each other. We must compute the area in square centimetres. Standard Euclidean geometry formulas for quadrilaterals apply. Concept / Approach: The area A of a rhombus can be expressed in multiple ways. In terms of base and height, it is A = base * height. In terms of its diagonals d1 and d2, the more convenient formula is A = (1 / 2) * d1 * d2. This formula results from the fact that the diagonals divide the rhombus into four congruent right triangles, each with legs equal to half of each diagonal. Using this relation, we can directly substitute the diagonal lengths and compute the area in one simple step. Step-by-Step Solution: Let the diagonals be d1 = 8 cm and d2 = 6 cm.Use the area formula for a rhombus: A = (1 / 2) * d1 * d2.Substitute the values: A = (1 / 2) * 8 * 6.Compute the product: 8 * 6 = 48.So A = (1 / 2) * 48 = 24 sq cm. Verification / Alternative check: To verify, imagine the diagonals intersecting at right angles and dividing the rhombus into four right triangles. Each triangle has legs 4 cm and 3 cm (half of each diagonal). The area of one such triangle is (1 / 2) * 4 * 3 = 6 sq cm. Since there are four congruent triangles, total area = 4 * 6 = 24 sq cm. This alternative reasoning perfectly matches the formula based method, confirming the result is correct. Why Other Options Are Wrong: Option 96 sq cm is obtained by multiplying the diagonals directly without taking half, which overestimates the area by a factor of 2. Option 60 sq cm and 48 sq cm may come from random combinations of 8 and 6 with approximate heights or mistaken formulas. Option 36 sq cm corresponds to using 6 * 6 instead of 8 * 6, which ignores one diagonal entirely. None of these match the correct formula A = (1 / 2) * d1 * d2. Common Pitfalls: A frequent mistake is to forget the factor of 1 / 2 and simply multiply the diagonals. Some test takers confuse the rhombus formula with that of a rectangle or square, for which area is side squared or length times breadth. Others may incorrectly think that diagonals must be equal, which is true for squares but not for all rhombuses. Recognizing the shape and recalling the specific formula avoids such confusion. Final Answer: The area of the rhombus is 24 sq cm.

Question 11

The diagonal of a square measures 10 cm. What is the area of this square in square centimetres?

Accepted Answer

Introduction / Context: This question tests the relation between the side and the diagonal of a square and then uses that relation to compute the area. In a square, all angles are right angles, and the diagonal splits the square into two congruent right triangles. Recognizing the 45 degree angles and the Pythagoras relation makes this a quick formula based problem suitable for aptitude exams. Given Data / Assumptions: The figure is a square, so all sides are equal and all angles are right angles. The diagonal of the square measures 10 cm. We are asked to find the area of the square in square centimetres. Standard right triangle relationships apply. No rounding is necessary because the numbers are simple. Concept / Approach: If the side of a square is s, then the diagonal d is given by d = s√2. This comes from applying Pythagoras theorem to the right triangle formed by two sides and the diagonal. Once we know the diagonal, we can solve for the side using s = d / √2. The area A of the square is then s^2. There is also a direct formula A = d^2 / 2, which is derived by substituting s = d / √2 into s^2. Step-by-Step Solution: Given diagonal d = 10 cm.Use the direct formula for area of a square in terms of diagonal: A = d^2 / 2.Compute d^2 = 10^2 = 100.So A = 100 / 2 = 50 sq cm.Therefore, the area of the square is 50 square centimetres. Verification / Alternative check: Alternatively, compute the side length first. From d = s√2, we have s = d / √2 = 10 / √2. Rationalising the denominator gives s = (10√2) / 2 = 5√2. The area A = s^2 = (5√2)^2 = 25 * 2 = 50 sq cm. This method uses explicit side length but arrives at the same area, confirming the correctness of the result. Why Other Options Are Wrong: Option 100 sq cm corresponds to taking area as d^2 instead of d^2 / 2, which is incorrect. Option 200 sq cm is even larger and may come from using 2 * d^2. Option 25 sq cm is half of the correct answer and might come from dividing incorrectly. Option 75 sq cm is an arbitrary intermediate that does not follow from any valid formula relating side and diagonal of a square. Common Pitfalls: Students sometimes confuse the relationship between side and diagonal for squares and rectangles. Another typical error is to forget that Pythagoras theorem applies and assume the diagonal equals the sum of sides instead of the root of the sum of squares. Using the correct formula A = d^2 / 2 for squares resolves these mistakes quickly. Final Answer: The area of the square is 50 sq cm.

Question 12

The area of an equilateral triangle is 64√3 sq cm. What is the length of each side of this triangle in centimetres?

Accepted Answer

Introduction / Context: This problem reverses the standard equilateral triangle area formula. Instead of being given the side and asked for the area, you are given the area and asked to find the side length. This reverse application is very common in competitive exams and ensures that you really understand the formula and can manipulate it algebraically, not just plug numbers in one direction. Given Data / Assumptions: The triangle is equilateral, so all sides are equal and all angles are 60 degrees. The area A of the triangle is 64√3 sq cm. We are required to find the side length in centimetres. Standard equilateral triangle properties and formulas apply. No rounding is necessary because the answer is an exact integer. Concept / Approach: The area of an equilateral triangle with side length a is A = (√3 / 4) * a^2. Given A, we solve this equation for a. First we isolate a^2 by dividing both sides by (√3 / 4), and then we take the positive square root because side lengths are positive. Once we have a, we can verify by substituting back into the formula to check that the area matches 64√3 sq cm. Step-by-Step Solution: Given A = 64√3 sq cm and A = (√3 / 4) * a^2.Set (√3 / 4) * a^2 = 64√3.Divide both sides by √3: (1 / 4) * a^2 = 64.Multiply both sides by 4: a^2 = 256.Take the positive square root: a = 16 cm. Verification / Alternative check: To verify, compute the area again using a = 16 cm. Using A = (√3 / 4) * a^2, we get A = (√3 / 4) * 16^2 = (√3 / 4) * 256. Simplifying gives A = 64√3 sq cm, which matches the given area exactly. This confirms that the side length 16 cm is correct and consistent with the formula and the provided data. Why Other Options Are Wrong: Option 8 cm would give area (√3 / 4) * 8^2 = 16√3, which is much smaller than 64√3. Option 16√3 cm is far too large and would lead to area that grows with the square of that expression, not 64√3. Option 8√3 cm also produces a different area. Option 12 cm gives area (√3 / 4) * 144 = 36√3, which again does not match 64√3. Only 16 cm satisfies the equation derived from the given area. Common Pitfalls: Some students forget to divide by √3 and instead cancel incorrectly, leading to wrong values for a^2. Others might misapply the formula and use A = (1 / 2) * a^2, which is not correct for an equilateral triangle. Another typical mistake is to take the square root of 64 only and ignore the factor of 4 from the denominator. Working step by step and treating the equation carefully avoids these errors. Final Answer: The length of each side of the equilateral triangle is 16 cm.

Question 13

The total area of the four vertical walls of a cuboid (its lateral surface area) is 448 sq cm. If the length of the cuboid is 18 cm and its height is 8 cm, what is its breadth in centimetres?

Accepted Answer

Introduction / Context: This problem uses the concept of lateral surface area of a cuboid, which is the total area of its four vertical walls without counting the top and bottom faces. These questions simulate real life scenarios such as painting walls or putting wallpaper, where only the side surfaces matter. Given the lateral area, length, and height, we need to find the missing breadth of the cuboid. Given Data / Assumptions: The figure is a cuboid with length, breadth, and height. The total area of the four walls (lateral surface area) is 448 sq cm. Length l = 18 cm. Height h = 8 cm. We are asked to find the breadth b in centimetres. Concept / Approach: The lateral surface area of a cuboid is given by the formula LSA = 2 * h * (l + b), because there are two walls of size l by h and two walls of size b by h. Using this formula, we can substitute the known values of LSA, l, and h, and then solve the resulting linear equation for b. Once we find b, we can quickly check whether it fits the original lateral surface area when substituted back into the formula. Step-by-Step Solution: Given LSA = 448 sq cm, l = 18 cm, and h = 8 cm.Formula: LSA = 2 * h * (l + b).Substitute: 448 = 2 * 8 * (18 + b).Compute 2 * 8 = 16, so 448 = 16 * (18 + b).Divide both sides by 16: 18 + b = 448 / 16 = 28.Therefore, b = 28 − 18 = 10 cm. Verification / Alternative check: To verify, recompute the lateral surface area using l = 18 cm, b = 10 cm, and h = 8 cm. LSA = 2 * h * (l + b) = 2 * 8 * (18 + 10) = 16 * 28 = 448 sq cm, which matches the given value exactly. This confirms that the breadth 10 cm is correct and consistent with the problem conditions. Why Other Options Are Wrong: Option 9 would give l + b = 27 and LSA = 2 * 8 * 27 = 432 sq cm, which is too small. Option 8 produces LSA = 2 * 8 * 26 = 416 sq cm, still not 448. Option 7 leads to LSA = 2 * 8 * 25 = 400 sq cm, and option 12 gives LSA = 2 * 8 * 30 = 480 sq cm, which is larger than required. Only b = 10 cm yields the correct lateral surface area of 448 sq cm. Common Pitfalls: Students sometimes confuse lateral surface area with total surface area and mistakenly use the formula 2(lb + bh + hl). Others may forget the factor of 2 when counting both pairs of opposite walls, or they might mix up the roles of height and breadth. Setting up the correct formula at the start and substituting values carefully helps avoid these errors. Final Answer: The breadth of the cuboid is 10 cm.

Question 14

A rectangle has a diagonal of length 51 cm and one of its sides is 24 cm. What is the area of this rectangle in square centimetres?

Accepted Answer

Introduction / Context: This question involves a rectangle where one side and the diagonal are known. By using the Pythagoras theorem on the right triangle formed by the two sides and the diagonal, we can find the length of the missing side. Once both side lengths are determined, the area of the rectangle is straightforward to calculate as the product of its length and breadth. This type of problem checks understanding of right triangles and basic mensuration. Given Data / Assumptions: The figure is a rectangle with right angles at each corner. The diagonal of the rectangle measures 51 cm. One side (breadth or length) measures 24 cm. We need to find the area in square centimetres. Standard Pythagoras theorem applies: diagonal^2 = side1^2 + side2^2. Concept / Approach: In a rectangle, the diagonal acts as the hypotenuse of a right triangle formed by the length and breadth. If we let the unknown side be x, then 51^2 = 24^2 + x^2. Solving this equation gives the value of x. After finding x, the area A of the rectangle is simply A = 24 * x. Performing these steps in the correct order ensures an accurate and efficient solution. Step-by-Step Solution: Let the known side be 24 cm and the unknown side be x cm.Using Pythagoras theorem: diagonal^2 = side1^2 + side2^2.So 51^2 = 24^2 + x^2.Compute squares: 51^2 = 2601 and 24^2 = 576, so 2601 = 576 + x^2.Thus x^2 = 2601 − 576 = 2025, so x = √2025 = 45 cm.Area A = 24 * 45 = 1080 sq cm. Verification / Alternative check: We can quickly check that 24, 45, and 51 form a Pythagorean triple. Calculating 24^2 + 45^2 gives 576 + 2025 = 2601, which is indeed 51^2. This confirms that the triangle is right angled and that the sides are consistent with the diagonal. Multiplying 24 by 45 again gives 1080 sq cm, so there is no arithmetic mistake in the area computation. Why Other Options Are Wrong: Option 540 sq cm is exactly half of the correct area and might result from multiplying 24 by 22.5 or from some misread. Option 810 sq cm results from mixing up numbers incorrectly such as using 30 instead of 45. Option 360 sq cm is far too small and would correspond to an unknown side of only 15 cm, which is inconsistent with Pythagoras theorem. Option 720 sq cm also does not satisfy the length needed to match the diagonal of 51 cm. Common Pitfalls: Common mistakes include squaring the diagonal incorrectly, forgetting to subtract 24^2 from 51^2, or taking the square root of an incorrect intermediate result. Some students also confuse which side is given and incorrectly write 24^2 = 51^2 + x^2, which reverses the roles of hypotenuse and legs. Keeping diagonal as the largest side in the Pythagoras equation and working stepwise helps avoid such issues. Final Answer: The area of the rectangle is 1080 sq cm.

Question 15

The perimeter of a rectangle is 90 cm and its breadth is 20 cm. What is the area of this rectangle in square centimetres?

Accepted Answer

Introduction / Context: This question combines the concept of perimeter of a rectangle with the calculation of its area. Often in aptitude tests you are not directly given both length and breadth, but instead given the perimeter and one dimension. You must first deduce the missing dimension using the perimeter formula, and then use both dimensions to compute the area. This checks your ability to manipulate simple linear equations and recall basic rectangle formulas. Given Data / Assumptions: The figure is a rectangle. Perimeter P = 90 cm. Breadth b = 20 cm. Length l is unknown and must be found. We are asked for the area A in square centimetres. Concept / Approach: The perimeter of a rectangle is given by P = 2 * (l + b). Using the known values of P and b, we can set up an equation and solve for l. Once l is known, the area is A = l * b. This is a straightforward two step process: first find the missing side from the perimeter, then find the area as a product of length and breadth. Step-by-Step Solution: Given P = 90 cm and b = 20 cm.Use the perimeter formula: P = 2 * (l + b).Substitute: 90 = 2 * (l + 20).Divide both sides by 2: 45 = l + 20.Solve for l: l = 45 − 20 = 25 cm.Area A = l * b = 25 * 20 = 500 sq cm. Verification / Alternative check: Verify that the length and breadth give the stated perimeter. With l = 25 cm and b = 20 cm, perimeter P = 2 * (25 + 20) = 2 * 45 = 90 cm, which matches the given value. This confirms that the side lengths are correct. Multiplying 25 by 20 again gives 500 sq cm, so there are no arithmetic mistakes in calculating the area. Why Other Options Are Wrong: Option 400 would correspond to an area with side 20 and length 20, which would give perimeter 80, not 90. Option 250 could result from multiplying 25 by 10 instead of 25 by 20. Option 450 may arise from miscalculating length as 22.5. Option 300 is also inconsistent with both the perimeter condition and the correct area formula. Only 500 sq cm satisfies all given conditions. Common Pitfalls: A common error is to confuse the perimeter formula with the area formula and attempt to use P = l * b. Another mistake is forgetting to divide by 2 when solving 90 = 2 * (l + 20). Some students also misread the breadth or accidentally add an extra step, such as subtracting 20 twice. Careful algebra and awareness of which formula applies help avoid these issues. Final Answer: The area of the rectangle is 500 sq cm.

Question 16

A circle is inscribed in a square whose diagonal measures 6√2 cm. What is the area of this inscribed circle (in square centimetres)?

Accepted Answer

Introduction / Context: This problem combines properties of squares and circles. A circle inscribed in a square touches all four sides, so its diameter is equal to the side length of the square. The information is given in terms of the square diagonal instead of its side, requiring you to use the Pythagoras relation between diagonal and side before moving on to the circle. This two step structure is very typical in geometry based aptitude questions. Given Data / Assumptions: A square has a diagonal of length 6√2 cm. A circle is inscribed in this square, touching all four sides. The diameter of the circle equals the side length of the square. We are asked to find the area of the circle in square centimetres. We use the standard formulas for squares and circles in Euclidean geometry. Concept / Approach: If the side of the square is s, then its diagonal d is given by d = s√2. We are given d and need to find s, so s = d / √2. Once we know s, we know the diameter of the circle because the circle just fits inside the square. Thus the radius r is s / 2. The area of the circle is then A = π * r^2. We will carefully follow these steps: find s, then find r, then find A. Step-by-Step Solution: Given diagonal d = 6√2 cm.For a square, d = s√2, so side s = d / √2 = (6√2) / √2 = 6 cm.The inscribed circle has diameter equal to the side of the square, so diameter = 6 cm, and radius r = 6 / 2 = 3 cm.Area of the circle A = π * r^2 = π * 3^2 = 9π sq cm.Thus the required area of the inscribed circle is 9π square centimetres. Verification / Alternative check: To double check, visualise the square of side 6 cm. The circle fits exactly inside, touching the midpoint of each side. The radius is clearly half the side length, which is 3 cm. Using any common approximation for π, such as 3.14 or 22/7, gives an area around 28.3 sq cm, but since the question keeps π symbolic, the exact expression 9π is correct. There are no other hidden steps, so the answer is reliable. Why Other Options Are Wrong: Option 6π would correspond to a radius √6, which is not consistent with the geometry here. Option 3π is far too small and would mean a radius of √3. Option 9√2π introduces an unnecessary √2 factor and does not arise from any correct combination of diagonal and radius. Option 12π is larger than 9π and does not match the radius 3 cm. Only 9π correctly reflects the square side and circle radius relation. Common Pitfalls: Students often forget that the diagonal of a square is side times √2, not side times 2. Another mistake is to treat the diagonal directly as the diameter of the circle, which would give a radius of 3√2 instead of 3. Some may also mix up radius and diameter, using d^2 instead of r^2 in the area formula. Keeping track of each geometric step prevents these errors. Final Answer: The area of the inscribed circle is 9π sq cm.

Question 17

Consider the region in the coordinate plane enclosed by the vertical lines x = -4 and x = 2 and the horizontal lines y = -2 and y = 3. What is the area of this region in square units?

Accepted Answer

Introduction / Context: This problem asks you to find the area of a rectangular region defined in the coordinate plane by two vertical lines and two horizontal lines. It checks your understanding of how to interpret algebraic equations like x = constant and y = constant as vertical and horizontal boundaries, and then how to compute the area of the rectangle formed by those boundaries. Such coordinate geometry questions are common in aptitude and school level exams. Given Data / Assumptions: The region is bounded on the left by the line x = -4. It is bounded on the right by the line x = 2. It is bounded below by the line y = -2. It is bounded above by the line y = 3. The shape enclosed by these four lines is a rectangle, and we must find its area in square units. Concept / Approach: The distance between two vertical lines x = a and x = b is |b − a|, and the distance between two horizontal lines y = c and y = d is |d − c|. These give the width and height of the rectangle in the coordinate plane. The area of the rectangle is then simply width * height. We will compute the horizontal length along the x axis and the vertical length along the y axis, and then multiply them to obtain the area. Step-by-Step Solution: Horizontal distance between x = -4 and x = 2 is |2 − (-4)| = |2 + 4| = 6 units.Vertical distance between y = -2 and y = 3 is |3 − (-2)| = |3 + 2| = 5 units.These distances form the sides of a rectangle with width 6 and height 5.Area of the rectangle = width * height = 6 * 5.Therefore, the area is 30 square units. Verification / Alternative check: You can visualise this region by sketching the lines on graph paper. Mark x = -4 and x = 2 as two vertical lines, and y = -2 and y = 3 as two horizontal lines. Their intersection points are the corners of a rectangle at (-4, -2), (-4, 3), (2, -2), and (2, 3). Counting grid squares or calculating the lengths along each axis both confirm that the rectangle is 6 units wide and 5 units tall, giving an area of 30 square units. Why Other Options Are Wrong: Option 12 likely comes from multiplying 6 by 2 instead of 6 by 5. Option 24 might arise from mixing one of the distances or misreading a boundary value. Option 15 is only half of the correct area and would correspond to width 3 and height 5 or similar confusion. Option 60 doubles the correct area and might result from mistakenly using 12 and 5 as dimensions. None of these match the correct rectangle formed by the given lines. Common Pitfalls: Learners sometimes misinterpret negative coordinates and treat -4 and 2 as a distance of 2, rather than 6. Another common mistake is to subtract the smaller from the larger without taking absolute value, but here that still gives the correct positive value if done correctly. Some also forget that area must be positive and might be unsure which value to subtract from which. Thinking in terms of absolute lengths between coordinates helps avoid these issues. Final Answer: The area of the enclosed region is 30 square units.

Question 18

The length and breadth of a rectangle are increased by 40% and 70% respectively. By what percentage does the area of the rectangle increase?

Accepted Answer

Introduction / Context: This question examines how the area of a rectangle changes when both its length and breadth are increased by certain percentages. It highlights that percentage changes in two dimensions combine multiplicatively, not additively. Understanding this is important in many real life situations, such as resizing photographs, scaling blueprints, or calculating land area changes when both directions are altered. Given Data / Assumptions: Original length of the rectangle is L and original breadth is B. Length is increased by 40 percent, so the new length is 140 percent of L. Breadth is increased by 70 percent, so the new breadth is 170 percent of B. We are asked to find the percentage increase in the area of the rectangle. All calculations use basic percentage and multiplication rules. Concept / Approach: The original area is A = L * B. After the changes, new length is 1.4L and new breadth is 1.7B. So the new area A_new = 1.4L * 1.7B. The factor by which the area changes is A_new / A = (1.4 * 1.7). The percentage increase is then (A_new / A − 1) * 100 percent. It is important to multiply the individual factors for length and breadth rather than simply adding the percentage increases, because area depends on both dimensions together. Step-by-Step Solution: Original area A = L * B.New length = L_new = 1.4L (since 40 percent increase means multiply by 1.4).New breadth = B_new = 1.7B (since 70 percent increase means multiply by 1.7).New area A_new = L_new * B_new = 1.4L * 1.7B = (1.4 * 1.7) * L * B.Compute 1.4 * 1.7 = 2.38, so A_new = 2.38 * A.Percentage increase in area = (2.38 − 1) * 100 percent = 1.38 * 100 percent = 138 percent. Verification / Alternative check: Take simple numbers to verify. Suppose L = 10 units and B = 10 units. Then original area A = 100 sq units. New length = 1.4 * 10 = 14 units and new breadth = 1.7 * 10 = 17 units. New area A_new = 14 * 17 = 238 sq units. The increase in area is 238 − 100 = 138 sq units. As a percentage of 100, this is exactly 138 percent. This numerical example supports the algebraic calculation and confirms that the percentage increase is 138 percent. Why Other Options Are Wrong: Option 118 percent might come from incorrectly adding 40 percent and 70 percent and then subtracting a small value, but it does not match the correct multiplicative change. Option 110 percent could be from simply adding 40 percent and 70 percent, which gives 110 percent, but this ignores the fact that area scales with the product of both sides. Option 128 percent is another incorrect combination of the numbers. Option 90 percent arises from subtracting one of the percentages instead of combining them properly. None of these match the true 2.38 area factor. Common Pitfalls: The most common mistake is to simply add the two percentage increases and conclude that the area increases by 40 percent + 70 percent = 110 percent. This is wrong because the new area is based on both new length and new breadth. Another pitfall is to forget to convert percentages to decimal multipliers before multiplying. Careful step by step work with multipliers rather than raw percentages prevents these errors. Final Answer: The area of the rectangle increases by 138 percent.

Question 19

In triangle ABC, X and Y are the mid points of sides AB and AC respectively. If BC + XY = 12 units, what is the value of BC − XY (in units)?

Accepted Answer

Introduction / Context: This geometry problem makes use of the mid point theorem in triangles. When a segment joins the mid points of two sides of a triangle, it has a special relationship with the third side. Here, that segment is XY, and the third side is BC. You are given the sum BC + XY and asked to compute BC − XY, which requires understanding the proportional relation between BC and XY and then using simple algebra. Given Data / Assumptions: ABC is a triangle. X is the mid point of side AB. Y is the mid point of side AC. Segment XY is drawn joining X and Y. BC + XY = 12 units. We must find BC − XY in units. Concept / Approach: The mid point theorem states that the segment joining the mid points of two sides of a triangle is parallel to the third side and has length equal to half of that third side. So in triangle ABC, XY is parallel to BC and XY = (1 / 2) * BC. Using this relationship, we can express XY in terms of BC and substitute into the equation BC + XY = 12. Solving this simple equation gives the value of BC, and then XY can be found and subtracted to get BC − XY. Step-by-Step Solution: From the mid point theorem, XY is parallel to BC and XY = (1 / 2) * BC.Let BC = b. Then XY = b / 2.Given BC + XY = 12, substitute: b + (b / 2) = 12.Combine like terms: (3b / 2) = 12.Solve for b: b = (2 / 3) * 12 = 8 units.Then XY = b / 2 = 8 / 2 = 4 units, so BC − XY = 8 − 4 = 4 units. Verification / Alternative check: Check that BC and XY satisfy the original condition. With BC = 8 units and XY = 4 units, BC + XY = 8 + 4 = 12 units, which matches the given sum. Also, XY = 4 is indeed half of BC = 8, confirming that the mid point theorem relation is satisfied. Finally, BC − XY = 8 − 4 = 4 units, so the computed value is consistent with all conditions in the problem. Why Other Options Are Wrong: Option 8 units would mean BC − XY equals the same value as BC, which is impossible unless XY were zero, contradicting the mid point theorem. Option 6 units would correspond to some incorrect algebra leading to a different value of BC. Option 2 units and 10 units similarly do not satisfy both the sum condition and the mid segment relation XY = BC / 2. Only 4 units works when you test it back in the original equation. Common Pitfalls: Some students incorrectly assume XY = BC rather than BC / 2. Others might misinterpret BC + XY as BC * XY or set up the equation with reversed roles. Mistakes in solving (3b / 2) = 12, such as multiplying by 2 and dividing incorrectly, can also occur. Remembering the exact statement of the mid point theorem and doing the algebra step by step avoids these issues. Final Answer: The value of BC − XY is 4 units.

Question 20

In a triangle, the point of intersection of all three medians is called which special point of the triangle?

Accepted Answer

Introduction / Context: This is a conceptual question about special points in a triangle. A triangle has several important points associated with it, including the centroid, orthocentre, circumcentre, and incentre. Each is defined by a different type of line set, such as medians, altitudes, perpendicular bisectors, or angle bisectors. Here, you must recall which special point is formed by the intersection of the three medians of a triangle. Given Data / Assumptions: We are dealing with a general triangle, not restricted to being right, isosceles, or equilateral. A median of a triangle is a line segment joining a vertex to the mid point of the opposite side. There are three medians, one from each vertex. These three medians intersect at a single point. We must identify the name of this common point of intersection. Concept / Approach: By definition, the centroid of a triangle is the point where all three medians intersect. This point has several key properties: it always lies inside the triangle, it divides each median in the ratio 2 : 1 from vertex to midpoint, and it can be interpreted as the centre of mass of a triangle made from uniform material. Other special points, like the orthocentre, circumcentre, and incentre, are defined using different sets of lines and should not be confused with the centroid. Step-by-Step Solution: Recall that a median connects a vertex to the mid point of the opposite side.There are three such medians in a triangle.All three medians are concurrent; they meet at a single point.This common point of intersection is called the centroid of the triangle.Hence, the required special point is the centroid. Verification / Alternative check: To further confirm, recall the definitions of the other special points. The orthocentre is the intersection of the three altitudes of a triangle. The circumcentre is the intersection of the perpendicular bisectors of the sides and is the centre of the circumcircle. The incentre is the intersection of the angle bisectors and is the centre of the incircle. None of these mention medians. Only the centroid is defined in terms of medians, so this matches the requirement exactly. Why Other Options Are Wrong: Orthocentre is wrong because it involves altitudes, not medians. Incentre is incorrect because it is formed by angle bisectors, not medians. Circumcentre refers to perpendicular bisectors and may lie inside or outside the triangle depending on the type of triangle. Excentre is formed by intersection of internal and external angle bisectors and is not associated with medians at all. Therefore, these options do not match the definition given in the question. Common Pitfalls: Learners often confuse these special centres because their names sound similar. Another common issue is mixing up altitudes and medians, since both go from a vertex to the opposite side but in different ways. Remember that medians go to the mid point, altitudes go perpendicular to the side, perpendicular bisectors start from the side and go outward, and angle bisectors split angles. Keeping the definitions straight helps in quickly identifying the centroid as the intersection of medians. Final Answer: The point where all three medians of a triangle intersect is called the centroid.

Area Questions