Question 1

One thousand solid metallic spheres, each of diameter 6 centimetres, are melted and recast into a single new solid sphere. Find the diameter of the new sphere in centimetres.

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Introduction / Context: This mensuration problem is about conservation of volume when smaller solid spheres are melted and recast into a larger sphere. It requires using the volume formula for a sphere and handling scaling when many identical small solids are combined into one larger solid of the same material. Such problems are common in aptitude exams and help reinforce volume relationships. Given Data / Assumptions: Number of small spheres = 1000. Diameter of each small sphere = 6 cm, so radius r_small = 3 cm. All small spheres are metallic and are completely melted. The molten metal is recast into one large solid sphere. No loss of material occurs, so total volume of metal is conserved. We must find the diameter of the new sphere. Concept / Approach: The volume of a sphere of radius r is: V = (4 / 3) * pi * r^3. The total volume of 1000 identical small spheres is 1000 times the volume of one small sphere. This equals the volume of the single large sphere. By equating these volumes, we can solve for the radius of the large sphere and then compute its diameter, which is twice the radius. Step-by-Step Solution: Step 1: Compute the volume of one small sphere with radius r_small = 3 cm. Step 2: V_small = (4 / 3) * pi * 3^3 = (4 / 3) * pi * 27 = 36 * pi cubic centimetres. Step 3: Total volume of 1000 small spheres is V_total = 1000 * 36 * pi = 36000 * pi cubic centimetres. Step 4: Let the radius of the new large sphere be R. Step 5: Volume of the large sphere is V_large = (4 / 3) * pi * R^3. Step 6: Since material volume is conserved, set V_large = V_total. Step 7: (4 / 3) * pi * R^3 = 36000 * pi. Step 8: Cancel pi from both sides to get (4 / 3) * R^3 = 36000. Step 9: Multiply both sides by 3 / 4: R^3 = 36000 * (3 / 4) = 27000. Step 10: Find R by taking the cube root: R^3 = 27000 implies R = 30 cm, since 30^3 = 27000. Step 11: Diameter of the new sphere = 2 * R = 2 * 30 = 60 cm. Verification / Alternative check: We can verify the calculation by checking the total volume. Volume of the large sphere with R = 30 cm is V_large = (4 / 3) * pi * 30^3 = (4 / 3) * pi * 27000 = 4 * 9000 * pi = 36000 * pi, which matches the total volume of the 1000 small spheres. This confirms that the radius and hence the diameter are correct. Why Other Options Are Wrong: 30 cm and 45 cm: These values are too small for the diameter; they would correspond to radii with volumes much less than the combined volume of 1000 small spheres. 90 cm and 15 cm: These are either excessively large or too small, giving volumes that do not match the total initial volume when substituted into the sphere volume formula. Common Pitfalls: Errors often occur when cubing the radius or when multiplying by the factor 1000. Another common mistake is forgetting that diameter is twice the radius and reporting the radius instead. Some students may also incorrectly cancel factors or mishandle the fraction 4 / 3. Keeping track of each algebraic step and clearly marking radius versus diameter helps avoid these errors. Final Answer: The diameter of the new sphere is 60 cm.

Question 2

In triangle ABC, points D and E lie on sides AB and BC respectively such that segment DE is parallel to side AC. If AB = 5 centimetres and AD = 3 centimetres, then what is the ratio BE : EC?

Accepted Answer

Introduction / Context: This question applies the basic proportionality theorem (Thales theorem) in a triangle. When a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally. Using this result, we can find the ratio in which a side is divided when we know the division on another side. Given Data / Assumptions: Triangle ABC is given. D lies on side AB and E lies on side BC. Segment DE is parallel to side AC. AB = 5 cm and AD = 3 cm. We need to find the ratio BE : EC. Standard Euclidean triangle geometry is assumed. Concept / Approach: By the basic proportionality theorem, if a line is drawn parallel to one side of a triangle and intersects the other two sides, then it divides those sides in the same ratio. Here DE is parallel to AC, so the segments on AB and BC are proportional: AD / DB = BE / EC. We know AB and AD, so we can find DB and then use the proportion to get BE : EC as a simple ratio. Step-by-Step Solution: Step 1: Compute DB from the total length AB and AD. Step 2: AB = 5 cm and AD = 3 cm, so DB = AB − AD = 5 − 3 = 2 cm. Step 3: By the basic proportionality theorem, AD / DB = BE / EC. Step 4: Substitute the known values: 3 / 2 = BE / EC. Step 5: Therefore the ratio BE : EC = 3 : 2. Verification / Alternative check: We can think in terms of a scale factor. If AB is partitioned in the ratio AD : DB = 3 : 2, then BC must be partitioned by point E in the same ratio for DE to be parallel to AC. Any specific choice of BC length, say BC = 5k, would give BE = 3k and EC = 2k, which maintains the 3 : 2 ratio and confirms that BE : EC must be 3 : 2. Why Other Options Are Wrong: 2 : 3 and 3 : 5: These invert or distort the required proportion and would not maintain parallelism between DE and AC. 5 : 3 and 4 : 1: These ratios do not correspond to AD / DB = 3 / 2 and break the similarity between the smaller and larger triangles created by the parallel line. Common Pitfalls: A frequent mistake is to use AD / AB instead of AD / DB in the proportionality relation. Another pitfall is incorrectly writing the ratio as EC : BE rather than BE : EC, which reverses the result. Careful attention to which segments correspond and writing the proportionality in the correct order helps avoid these errors. Final Answer: The required ratio is BE : EC = 3 : 2.

Question 3

From a point P outside a circle with centre O and radius 6 centimetres, a tangent PT is drawn to the circle. If PT = 8 centimetres, find the length of segment OP in centimetres.

Accepted Answer

Introduction / Context: This problem involves a tangent to a circle from an external point and uses a classic right triangle configuration. The radius drawn to the point of tangency is perpendicular to the tangent, giving a right triangle. Applying Pythagoras theorem then allows us to find the distance from the external point to the centre of the circle. Given Data / Assumptions: A circle has centre O and radius 6 cm. P is a point outside the circle. PT is a tangent from P to the circle, touching it at point T. The length of PT is 8 cm. We must find the length of OP in centimetres. The radius OT is perpendicular to tangent PT at the point of contact T. Concept / Approach: The key concept is that a radius drawn to a point of tangency is perpendicular to the tangent. This means that triangle OPT is a right angled triangle with right angle at T. The hypotenuse is OP, one leg is OT (the radius) and the other leg is PT (the tangent). Using Pythagoras theorem: OP^2 = OT^2 + PT^2. We substitute the known values and solve for OP. Step-by-Step Solution: Step 1: Identify the right triangle OPT where angle at T is 90 degrees. Step 2: OT = radius = 6 cm. Step 3: PT = tangent length = 8 cm. Step 4: Apply Pythagoras theorem: OP^2 = OT^2 + PT^2. Step 5: Substitute values: OP^2 = 6^2 + 8^2 = 36 + 64. Step 6: Compute: 36 + 64 = 100. Step 7: Therefore OP^2 = 100, so OP = sqrt(100) = 10 cm. Verification / Alternative check: We can recognise the triple (6, 8, 10) as a scaled version of the well known Pythagorean triple (3, 4, 5). The ratio 6 : 8 : 10 simplifies to 3 : 4 : 5, confirming that the sides indeed form a right triangle. This provides an additional intuitive check that OP is correctly calculated as 10 cm. Why Other Options Are Wrong: 12 cm and 16 cm: These values are too large when compared with the leg lengths 6 and 8. Plugging them back into Pythagoras theorem does not give equality for 6^2 + 8^2. 9 cm and 8 cm: These are too small to be the hypotenuse of a right triangle with legs 6 and 8, since the hypotenuse must be longer than either leg and satisfy OP^2 = 36 + 64. Common Pitfalls: Common errors include forgetting that the radius is perpendicular to the tangent, misidentifying the hypotenuse or incorrectly computing the sum of squares. Some students attempt to subtract instead of add when using Pythagoras theorem, which is incorrect for finding the hypotenuse. Always remember that for a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. Final Answer: The length of OP is 10 cm.

Question 4

In triangle ABC, points D, E and F are the midpoints of sides BC, CA and AB respectively. What is the ratio of the area of triangle DEF to the area of triangle ABC?

Accepted Answer

Introduction / Context: This geometry question is about the medial triangle of a given triangle. When you join the midpoints of the sides of a triangle, you get a smaller triangle inside it. The problem asks you to recall and apply the relationship between the areas of the original triangle and this medial triangle. Understanding this concept is helpful in many area and similarity problems. Given Data / Assumptions: Triangle ABC is given. D is the midpoint of side BC. E is the midpoint of side CA. F is the midpoint of side AB. Triangle DEF is formed by joining these midpoints. We must find the ratio area(DEF) : area(ABC). Concept / Approach: When the midpoints of the sides of a triangle are joined, the resulting triangle DEF is called the medial triangle. It is similar to the original triangle ABC and each side of DEF is half the length of the corresponding side of ABC. Since the ratio of similarity of corresponding sides is 1 : 2, the ratio of areas is the square of this ratio, that is (1 / 2)^2 = 1 / 4. Therefore the area of the medial triangle is one quarter of the area of the original triangle. Step-by-Step Solution: Step 1: Observe that D, E and F are midpoints of BC, CA and AB respectively. Step 2: When you join these midpoints, triangle DEF is formed inside triangle ABC. Step 3: Each side of triangle DEF is parallel to a side of triangle ABC and is exactly half its length. Step 4: Therefore triangles DEF and ABC are similar with side ratio 1 : 2. Step 5: The ratio of areas of similar triangles is the square of the ratio of corresponding sides. Step 6: So area(DEF) : area(ABC) = (1 / 2)^2 : 1 = 1 / 4 : 1. Step 7: Hence the ratio area(DEF) : area(ABC) is 1 / 4. Verification / Alternative check: To verify, you can take a simple coordinate example. Let triangle ABC have vertices A(0, 0), B(2, 0) and C(0, 2). Its area is 2 square units. The midpoints are F(1, 0), D(1, 1) and E(0, 1). Triangle DEF has vertices (1, 1), (0, 1) and (1, 0). Its area is 0.5 square units, which is exactly one quarter of 2. This numerical example confirms the general area ratio of 1 : 4. Why Other Options Are Wrong: 1 / 2: This would correspond to a side ratio of 1 : √2, which is not true here. 1 / 8 and 1 / 16: These imply much smaller triangles relative to ABC than the actual medial triangle, and do not match the side length relationship of 1 : 2. 3 / 4: This would suggest that the medial triangle occupies most of the area of ABC, which is not correct. Common Pitfalls: Students may mistakenly think each side being half implies each area is also half. However, area scales with the square of the side length, not directly with it. Another common error is failing to recognise that DEF is always similar to ABC. Remember that joining midpoints in a triangle always creates a medial triangle with area exactly one quarter of the original. Final Answer: The ratio of the areas is area(DEF) : area(ABC) = 1 / 4.

Question 5

A circle has its centre at O. From a point P outside the circle, a tangent is drawn that touches the circle at A. If PA = 4 centimetres and PO = 5 centimetres, find the length of the radius of the circle in centimetres.

Accepted Answer

Introduction / Context: This question is another example of using tangents to a circle and the right triangle formed by the radius, tangent and line from the centre to the external point. It tests your ability to identify the correct right triangle and then apply Pythagoras theorem to determine the radius of the circle using the given distances. Given Data / Assumptions: O is the centre of the circle. P is an external point outside the circle. PA is a tangent from P to the circle, touching at point A. PA = 4 cm. PO, the distance from P to the centre O, is 5 cm. We must find the radius OA of the circle. The radius OA is perpendicular to the tangent PA at point A. Concept / Approach: Since the radius drawn to a tangent at the point of contact is perpendicular to the tangent, angle OAP is 90 degrees. Therefore triangle OAP is a right angled triangle with hypotenuse OP, one leg OA (the radius) and the other leg AP (the tangent). Applying Pythagoras theorem: OP^2 = OA^2 + AP^2. We substitute the known values and solve for OA, which directly gives the radius. Step-by-Step Solution: Step 1: Recognise that triangle OAP is right angled at A because OA is perpendicular to tangent PA. Step 2: OP is the hypotenuse with length 5 cm. Step 3: AP is one leg with length 4 cm. Step 4: Let OA = r be the radius we need to find. Step 5: Apply Pythagoras theorem: OP^2 = OA^2 + AP^2. Step 6: Substitute known values: 5^2 = r^2 + 4^2. Step 7: Compute squares: 25 = r^2 + 16. Step 8: Rearrange: r^2 = 25 − 16 = 9. Step 9: Take square root: r = 3 cm. Verification / Alternative check: We can confirm by checking the Pythagorean triple. The lengths 3, 4 and 5 form a classic right triangle. Here, OA = 3 cm, AP = 4 cm and OP = 5 cm. Since 3^2 + 4^2 = 9 + 16 = 25 = 5^2, the right triangle condition is satisfied, validating that the radius of the circle is 3 cm. Why Other Options Are Wrong: 1 cm and 2 cm: These radii are too small; they would not satisfy 5^2 = r^2 + 4^2 when substituted into Pythagoras theorem. 4 cm: This value would give 5^2 = 4^2 + 4^2 = 32, which is not true. 5 cm: This would imply the tangent length is zero (right triangle degenerates), which contradicts PA = 4 cm. Common Pitfalls: Students sometimes confuse which segment is the hypotenuse and may subtract instead of adding in Pythagoras theorem. Others may forget that the tangent and radius are perpendicular at the point of contact, leading them to apply the theorem incorrectly. Drawing the right triangle clearly and marking the right angle at A helps in correctly identifying the sides and performing the calculation. Final Answer: The radius of the circle is 3 cm.

Question 6

An inverted conical vessel is completely filled with water up to its brim. The height of the cone is 8 centimetres and the radius of the open circular top is 5 centimetres. Solid spherical metallic balls, each of radius 1/2 centimetre, are dropped i…

Accepted Answer

Introduction / Context: This is a volume displacement problem involving a cone filled with water and small spherical balls that cause overflow when dropped into the water. It tests your understanding of volume formulas for cones and spheres, as well as the physical principle that the volume of water displaced by submerged objects equals the volume of those objects. In this case, the displaced water appears as overflow. Given Data / Assumptions: The vessel is an inverted right circular cone. Height of the cone h = 8 cm. Radius of the top of the cone R = 5 cm. The cone is initially filled completely with water up to the brim. Solid spherical metallic balls of radius r_ball = 1/2 cm are dropped into the water. After dropping some balls, 25% of the water overflows. No water remains above the brim; the cone is still full in volume, but part of the water has been replaced by balls. We must determine the number of balls dropped. Concept / Approach: The key idea is conservation of volume and displacement. The volume of water that overflows equals the volume of the balls that are submerged, assuming all are fully below the water surface when the system stabilises. Initially, the cone contains a volume V_cone of water. When balls are added, 25% of the initial water volume flows out. This overflow volume is equal to the total volume of the balls. Therefore: Total volume of balls = 25% of V_cone = (1 / 4) * V_cone. We compute V_cone using the cone volume formula and the volume of one ball using the sphere volume formula, then equate N * V_ball = (1 / 4) * V_cone to find N. Step-by-Step Solution: Step 1: Compute the volume of the cone. V_cone = (1 / 3) * pi * R^2 * h. Step 2: Substitute R = 5 cm and h = 8 cm: V_cone = (1 / 3) * pi * 5^2 * 8. Step 3: 5^2 = 25, so V_cone = (1 / 3) * pi * 25 * 8 = (200 / 3) * pi cubic centimetres. Step 4: Since 25% of the water overflows, overflow volume = (1 / 4) * V_cone = (1 / 4) * (200 / 3) * pi = (50 / 3) * pi cubic centimetres. Step 5: Compute the volume of one spherical ball. V_ball = (4 / 3) * pi * r_ball^3. Step 6: Radius of each ball r_ball = 1/2 cm, so r_ball^3 = (1/2)^3 = 1/8. Step 7: Thus V_ball = (4 / 3) * pi * (1 / 8) = (4 / 24) * pi = (1 / 6) * pi cubic centimetres. Step 8: Let N be the number of balls. Then total volume of balls = N * V_ball = N * (1 / 6) * pi. Step 9: Equate the total volume of balls to the overflow volume: N * (1 / 6) * pi = (50 / 3) * pi. Step 10: Cancel pi from both sides to get N / 6 = 50 / 3. Step 11: Multiply both sides by 6: N = 6 * (50 / 3) = 2 * 50 = 100. Step 12: Therefore, 100 balls were dropped into the vessel. Verification / Alternative check: We can check by computing the total volume of 100 balls. V_total_balls = 100 * (1 / 6) * pi = (100 / 6) * pi = (50 / 3) * pi cubic centimetres. This matches exactly the 25% overflow volume computed for the cone, confirming that the calculation of N is correct and consistent with volume conservation. Why Other Options Are Wrong: 200 and 400: These would give total ball volumes that are double or four times the required overflow volume, resulting in more water displaced than the specified 25%. 150 and 250: These also lead to total volumes inconsistent with exactly one quarter of the cone's volume, either exceeding or falling short of the required displacement. Common Pitfalls: Some students mistakenly take 25% of the cone's height or radius instead of its volume, which is incorrect. Others forget that the volume of a sphere depends on the cube of the radius and may incorrectly use (1/2)^2 instead of (1/2)^3. Confusing diameter and radius when computing ball volume is another common error. Carefully applying the correct volume formulas and remembering that overflow volume equals displaced volume helps avoid these mistakes. Final Answer: The number of balls dropped into the vessel is 100.

Question 7

How many distinct circles can be drawn through three given non-collinear points in a plane?

Accepted Answer

Introduction / Context: This is a fundamental concept question in Euclidean geometry about circles and points. It focuses on the uniqueness of a circle passing through three non-collinear points. Understanding this fact is essential for many constructions and proofs related to circumcircles of triangles and circle geometry in general. Given Data / Assumptions: We are given three points in a plane. The three points are non-collinear, meaning they do not all lie on the same straight line. We want to know how many different circles can pass through all three points. Standard Euclidean geometry is assumed. Concept / Approach: In Euclidean geometry, any three non-collinear points uniquely determine a circle, called the circumcircle of the triangle formed by those points. This happens because there is exactly one circle whose centre is equidistant from all three points. That centre is the point where the perpendicular bisectors of the sides of the triangle intersect, which is a unique point for non-collinear points. Therefore, only one circle can pass through all three such points. Step-by-Step Solution: Step 1: Think of the three non-collinear points as the vertices of a triangle, say A, B and C. Step 2: For a circle to pass through A, B and C, its centre must be equidistant from all three points. Step 3: The set of points equidistant from A and B forms the perpendicular bisector of segment AB. Step 4: Similarly, the set of points equidistant from B and C forms the perpendicular bisector of segment BC. Step 5: For a point to be equidistant from A, B and C, it must lie on both perpendicular bisectors AB and BC. Step 6: In a plane, two non-parallel lines intersect at exactly one point. The perpendicular bisectors of AB and BC are not parallel when the points are non-collinear. Step 7: Therefore, the perpendicular bisectors intersect at exactly one point, which is the unique centre of the circumcircle. Step 8: With this unique centre and the common distance to A, B and C as radius, exactly one circle passes through the three non-collinear points. Verification / Alternative check: If you attempted to draw a second circle through the same three non-collinear points, its centre would also have to be equidistant from all three points. That would imply a second intersection point of the same perpendicular bisectors, which is impossible because two non-parallel lines intersect at only one point. Hence a second distinct circle cannot exist, reinforcing that there is exactly one. Why Other Options Are Wrong: Two, three or more than three: These options contradict the uniqueness of the intersection of the perpendicular bisectors and the requirement that a circle's centre must be equidistant from all three points. None: This would be true only if the points were collinear. But the question explicitly states that the points are non-collinear, so a circle does exist. Common Pitfalls: A misunderstanding often arises when students confuse the case of collinear and non-collinear points. For three collinear points, no circle can pass through all of them, but for non-collinear points, exactly one circle does. Another pitfall is overlooking the role of perpendicular bisectors and the uniqueness of their intersection. Remembering the circumcircle of a triangle helps solidify this concept. Final Answer: The number of circles that can be drawn is exactly one.

Question 8

Two circles touch each other internally. The radius of the smaller circle is 6 centimetres and the distance between the centres of the two circles is 3 centimetres. Find the radius of the larger circle in centimetres.

Accepted Answer

Introduction / Context: This question deals with two circles that touch each other internally. It tests your knowledge of the relationship between the radii of the two circles and the distance between their centres in the case of internal tangency. Such problems often appear in coordinate geometry and circle-based aptitude questions. Given Data / Assumptions: There are two circles that touch each other internally. The radius of the smaller circle is r_small = 6 cm. The distance between the centres of the two circles is d = 3 cm. We must find the radius of the larger circle, r_large. The point of contact lies on the line joining the centres. Concept / Approach: For two circles that touch internally, the distance between their centres is equal to the difference of their radii: d = r_large − r_small. This is because, for internal tangency, the smaller circle lies inside the larger one and the common tangent point lies between the two centres. Using this simple linear relationship, we can solve for r_large by adding d and r_small. Step-by-Step Solution: Step 1: Recall the formula for internal tangency: d = r_large − r_small. Step 2: Substitute the known values: d = 3 cm and r_small = 6 cm. Step 3: So 3 = r_large − 6. Step 4: Add 6 to both sides: r_large = 3 + 6 = 9 cm. Step 5: Therefore, the radius of the larger circle is 9 cm. Verification / Alternative check: We can check if the geometry makes sense. If r_large = 9 cm and r_small = 6 cm, then the difference r_large − r_small = 9 − 6 = 3 cm, which matches the given distance between centres. This is consistent with internal tangency, where the smaller circle sits inside the larger one and touches it at exactly one point along the line connecting the centres. Why Other Options Are Wrong: 7.5 cm and 8 cm: These values give differences of 1.5 cm and 2 cm between radii, which do not match the given distance of 3 cm. 10 cm: This would give a difference of 4 cm, again inconsistent with the distance of 3 cm. 6 cm: This would mean both circles have the same radius and cannot have centres 3 cm apart and touch internally in the way described. Common Pitfalls: Students sometimes mistakenly use the formula for external tangency, where the centre distance equals the sum of the radii, d = r_large + r_small. Here, because one circle is inside the other, the correct relationship is the difference of radii. Carefully identifying whether circles touch internally or externally is crucial before applying the appropriate formula. Final Answer: The radius of the larger circle is 9 cm.

Question 9

PQR is an equilateral triangle. A line segment MN is drawn parallel to side QR such that point M lies on side PQ and point N lies on side PR. If PN = 6 centimetres, then find the length of MN in centimetres.

Accepted Answer

Introduction / Context: This geometry question is based on properties of equilateral triangles and parallels. It explores the relationship between a smaller triangle formed by drawing a segment parallel to one side and the original equilateral triangle. Recognising that the small triangle is also equilateral plays a key role in solving this problem quickly and correctly. Given Data / Assumptions: PQR is an equilateral triangle, so all its sides are equal and all its angles are 60 degrees. Segment MN is drawn parallel to side QR. M lies on side PQ and N lies on side PR. PN is a side of the smaller triangle formed, and PN = 6 cm. We must find the length of segment MN in centimetres. Standard Euclidean geometry is assumed. Concept / Approach: Because PQR is equilateral (all angles 60 degrees) and MN is parallel to QR, triangle PMN inherits the same angle measures as triangle PQR. Specifically, angle at P is common, and parallels ensure corresponding angles at M and N match those at Q and R respectively. Therefore triangle PMN is similar to triangle PQR. In fact, with both triangles having all angles 60 degrees, triangle PMN is itself an equilateral triangle. That means all sides of triangle PMN are equal, including MN and PN. Step-by-Step Solution: Step 1: Note that triangle PQR is equilateral, so each angle is 60 degrees. Step 2: Segment MN is drawn parallel to QR with M on PQ and N on PR. Step 3: Because MN is parallel to QR, angle PMN equals angle PQR and angle PNM equals angle PRQ by corresponding angles. Step 4: Angle at P is common to both triangle PMN and triangle PQR. Step 5: Hence triangle PMN has all three angles equal to 60 degrees. Step 6: Any triangle with all angles 60 degrees is equilateral. Step 7: Therefore, in triangle PMN, all sides are equal: PM = MN = PN. Step 8: We are given PN = 6 cm, so MN = 6 cm. Verification / Alternative check: We can also reason in terms of similarity ratios. Since triangle PMN is similar to triangle PQR with all angles equal, the scale factor is the same for each side. Side PN is corresponding to side PR, and side MN corresponds to QR. But within triangle PMN itself, each side must have the same length. The given single side PN = 6 cm fixes the scale factor, which then forces MN to equal 6 cm as well. Any different value for MN would break the equilateral property. Why Other Options Are Wrong: 3 cm, 4.5 cm, 9 cm and 12 cm: These values would make triangle PMN non-equilateral if PN remains 6 cm, contradicting the angle-based similarity to triangle PQR and the parallel condition between MN and QR. Common Pitfalls: A common mistake is to assume some arbitrary ratio of similarity without first noticing that both triangles are equilateral. Some students may also try to bring in lengths of PQR that are not given or necessary. The key is to focus on the angles: an equilateral triangle with a line parallel to the base through points on the other two sides still creates a smaller equilateral triangle at the top, so all its sides must be equal. Final Answer: The length of segment MN is 6 cm.

Question 10

The radius and the height of a right circular cone are each increased by 20%. By what percentage does the volume of the cone increase?

Accepted Answer

Introduction / Context: This quantitative aptitude question examines how changes in linear dimensions affect the volume of a three dimensional solid, specifically a right circular cone. It highlights that when more than one dimension changes, the resulting percentage change in volume is not simply the sum of the individual percentage changes. Instead, the combined effect must be calculated using the volume formula. Given Data / Assumptions: We have a right circular cone with initial radius r and height h. The radius is increased by 20%, so the new radius r_new = 1.2 * r. The height is also increased by 20%, so the new height h_new = 1.2 * h. The original volume and new volume are to be compared. We must find the percentage increase in volume. Concept / Approach: The volume of a right circular cone is: V = (1 / 3) * pi * r^2 * h. The new volume after changes in radius and height is: V_new = (1 / 3) * pi * (r_new)^2 * h_new. Substituting r_new = 1.2r and h_new = 1.2h, we can express V_new as a multiple of V. The factor multiplying V gives the overall volume scale change, and from this we can compute the percentage increase. Step-by-Step Solution: Step 1: Write the original volume: V = (1 / 3) * pi * r^2 * h. Step 2: The new radius is r_new = 1.2r. Step 3: The new height is h_new = 1.2h. Step 4: Compute the new volume: V_new = (1 / 3) * pi * (1.2r)^2 * (1.2h). Step 5: Simplify (1.2r)^2 = 1.2^2 * r^2 = 1.44 * r^2. Step 6: So V_new = (1 / 3) * pi * 1.44 * r^2 * 1.2 * h. Step 7: Combine the constants: 1.44 * 1.2 = 1.728. Step 8: Hence V_new = 1.728 * (1 / 3) * pi * r^2 * h = 1.728 * V. Step 9: This means the new volume is 1.728 times the original volume. Step 10: The increase factor above 1 is 1.728 − 1 = 0.728. Step 11: Convert this factor to a percentage: 0.728 * 100% = 72.8%. Verification / Alternative check: We can use an approximate method to cross check. A 20% increase in radius alone would increase the term r^2 by about 44% (since 1.2^2 = 1.44), and an additional 20% increase in height would further scale volume by 1.2. Combining these effects: 1.44 * 1.2 = 1.728, which indeed represents a 72.8% increase over the original. This matches the detailed calculation. Why Other Options Are Wrong: 20% and 20.5%: These mistakenly treat the volume change as if only one dimension changed or ignore the squared dependence on radius. 62%: This may come from adding 44% (due to r^2) and 18% approximately, but it is still incorrect because the effects are multiplicative, not additive. Any other values different from 72.8% do not align with V_new = 1.728 * V computed using the exact formula. Common Pitfalls: A common error is to assume that if both radius and height increase by 20%, the volume also increases by 40%. This ignores the fact that volume depends on r^2 and h, so the changes combine multiplicatively. Students also sometimes miscalculate 1.2^2 and obtain incorrect intermediate values. Carefully writing out the algebra and doing the multiplication in steps helps avoid such mistakes. Final Answer: The volume of the cone increases by 72.8%.

Question 11

Which of the following statements is not true for a rhombus?

Accepted Answer

Introduction / Context: This conceptual geometry question asks you to recall the defining properties of a rhombus and identify which given statement does not always hold. Rhombuses are special quadrilaterals with equal sides, and understanding how they differ from other quadrilaterals like rectangles and squares is important for classification and problem solving. Given Data / Assumptions: We are dealing with a rhombus, a quadrilateral with all sides equal. Several statements about side lengths, parallelism and diagonals are listed. We must determine which statement is not necessarily true for every rhombus. Implicitly, standard Euclidean properties of rhombuses apply. Concept / Approach: Key properties of a rhombus include: All four sides are equal in length. Opposite sides are parallel. Opposite angles are equal. Diagonals bisect each other. Diagonals are perpendicular (they intersect at right angles). However, the diagonals of a general rhombus are not necessarily equal in length. Equal diagonals are a defining property of rectangles and squares. A square is a special rhombus that does have equal diagonals, but not every rhombus is a square. Step-by-Step Solution: Step 1: Consider the statement "All sides are congruent". This is true by definition of a rhombus, which has four equal sides. Step 2: Consider "Opposite sides are parallel". A rhombus is a special type of parallelogram, so opposite sides are indeed parallel. Step 3: Consider "Diagonals bisect each other at right angles". In a rhombus, diagonals bisect each other and are perpendicular, so they meet at right angles. Step 4: Opposite angles of a rhombus are equal, just like in any parallelogram, so that statement is also true. Step 5: Consider "Diagonals are equal". This is not a general property of a rhombus. In a rectangle and a square, diagonals are equal, but in a generic rhombus (with unequal angles), diagonals have different lengths. Step 6: Therefore, the statement that is not necessarily true for a rhombus is that its diagonals are equal. Verification / Alternative check: You can visualise or draw a fat, slanted rhombus with angles that are not 90 degrees. The diagonals cross at right angles but clearly one diagonal is longer than the other. A square is a special case of a rhombus where all angles are right angles and the diagonals are equal, but since the question is about rhombuses in general, properties specific only to the square cannot be taken as universally true. Why Other Options Are Wrong: All sides are congruent: This is the defining property of a rhombus and always holds. Opposite sides are parallel: A rhombus is a parallelogram, so this is always true. Diagonals bisect each other at right angles: In a rhombus, diagonals bisect each other and are perpendicular; this is a characteristic feature. Opposite angles are equal: As a parallelogram, a rhombus has opposite angles equal. This is correct as well. Common Pitfalls: Students often confuse the properties of rhombuses, rectangles and squares, assuming that all their properties are interchangeable. It is especially common to mistakenly think equal diagonals apply to all parallelograms or all rhombuses. Remember: equal diagonals are guaranteed in rectangles and squares, but not in general rhombuses; perpendicular diagonals that bisect each other are characteristic of rhombuses and squares, not all parallelograms. Final Answer: The statement that is not true for every rhombus is "Diagonals are equal".

Question 12

In an isosceles triangle, the ratio between the vertical angle (angle at the vertex where the equal sides meet) and each base angle is 4 : 3. Find the measure of any one base angle in degrees.

Accepted Answer

Introduction / Context: This aptitude problem involves angle relationships in an isosceles triangle. It tests your ability to use the given ratio between angles along with the fact that the sum of all interior angles of a triangle is 180 degrees. Recognising that an isosceles triangle has two equal base angles is essential for setting up the correct equation. Given Data / Assumptions: We have an isosceles triangle. The vertical angle is the angle at the vertex where the two equal sides meet. The ratio of the vertical angle to a base angle is given as 4 : 3. There are two base angles and both are equal in measure. We must find the measure of one base angle in degrees. Sum of angles in any triangle is 180 degrees. Concept / Approach: Let the vertical angle be 4x degrees and each base angle be 3x degrees to reflect the given ratio 4 : 3. Because there are two base angles, the sum of the three interior angles of the triangle is: 4x + 3x + 3x = 180 degrees. Solving this equation gives x and hence the measure of each base angle, 3x degrees. This straightforward algebraic approach is standard in triangle angle ratio problems. Step-by-Step Solution: Step 1: Let the vertical angle be 4x degrees. Step 2: Since the triangle is isosceles, both base angles are equal. Let each base angle be 3x degrees. Step 3: Use the fact that the sum of the angles in a triangle is 180 degrees. Step 4: Write the equation: 4x + 3x + 3x = 180. Step 5: Combine like terms: 4x + 3x + 3x = 10x. Step 6: So 10x = 180. Step 7: Solve for x: x = 180 / 10 = 18. Step 8: Each base angle is 3x degrees, so base angle = 3 * 18 = 54 degrees. Verification / Alternative check: Compute the vertical angle using x = 18 degrees: vertical angle = 4x = 4 * 18 = 72 degrees. Now check the sum of angles: 72 + 54 + 54 = 180 degrees, which matches the required total for a triangle. The ratio of vertical angle to base angle is 72 : 54, which simplifies by dividing both numbers by 18 to 4 : 3, confirming the ratio is correctly satisfied. Why Other Options Are Wrong: 18 deg and 10 deg: These are too small for the base angle and would not satisfy both the 4 : 3 ratio and the 180 degree sum of angles. 72 deg: This is the measure of the vertical angle, not the base angle, in this configuration. 36 deg: Using 36 degrees as a base angle would not yield a vertical angle that maintains the 4 : 3 ratio with the base angles while also summing to 180 degrees. Common Pitfalls: Some students mistakenly assign 4x and 3x only once each and forget that there are two base angles. This leads to an equation 4x + 3x = 180 which is incorrect. Others may reverse the ratio and take the base angle as 4x and vertical as 3x, which conflicts with the wording of the question. Carefully reading the ratio description and remembering there are two equal base angles avoids these errors. Final Answer: The measure of each base angle is 54 degrees.

Question 13

In triangle ABC, angle B is a right angle (90°) and side BC is equal to √3 times side AB. What is the measure of angle A (in degrees)?

Accepted Answer

Introduction / Context: This problem involves a right angled triangle and the special ratio of sides that appears in 30°–60°–90° triangles. Questions like this test how well you connect side length ratios with specific angles without having to use trigonometric tables directly. Given Data / Assumptions: Triangle ABC is right angled at B, so ∠B = 90°. Side BC is opposite angle A and side AB is opposite angle C. The given relation is BC = √3 * AB. We need to find the measure of angle A. Concept / Approach: In a right angled triangle, trigonometric ratios relate angles to side lengths. For angle A in triangle ABC, tan(A) is defined as (side opposite A) / (side adjacent to A). Here angle A is at vertex A, the side opposite A is BC and one of the sides adjacent to A is AB. So we can use tan(A) = BC / AB. We are given BC / AB = √3. There is a well known special value: tan(60°) = √3. This directly allows us to match the ratio to the angle measure. Step-by-Step Solution: Step 1: Identify sides with respect to angle A. Opposite side is BC, adjacent side is AB, and hypotenuse is AC. Step 2: Write the tangent ratio for angle A: tan(A) = BC / AB. Step 3: Substitute the given relation BC = √3 * AB, so BC / AB = √3. Step 4: Therefore tan(A) = √3. Step 5: From standard trigonometric values, tan(60°) = √3. Step 6: Hence A = 60°. Verification / Alternative Check: If A = 60° and B = 90°, then C = 180° − 90° − 60° = 30°. In a 30°–60°–90° triangle, the side opposite 60° is √3 times the side opposite 30°. Here BC is opposite A (60°) and AB is opposite C (30°), so BC = √3 * AB, which matches the given condition. This confirms that the result is consistent. Why Other Options Are Wrong: 45° would give tan(45°) = 1, not √3. An angle of 30° would give tan(30°) = 1/√3, which is the reciprocal of the required value. 90° is impossible because the triangle already has a right angle at B and no triangle can have two right angles. Common Pitfalls: Learners sometimes mix up which side is opposite which angle or confuse the ratios for 30° and 60°. Another common mistake is to think that BC is the hypotenuse because it appears in the relation, but the hypotenuse is always opposite the right angle, so it must be AC, not BC. Carefully labeling the triangle before using trigonometric ratios avoids such confusion. Final Answer: The measure of angle A is 60°.

Question 14

The height of a right circular cylinder is initially equal to twice the radius of its base. If the height is changed so that it becomes six times the radius of the base (radius unchanged), the volume of the cylinder increases by 539 cubic centimetre…

Accepted Answer

Introduction / Context: This question tests your understanding of how the volume of a right circular cylinder depends on its dimensions and how a change in height affects the volume when the radius is fixed. It also checks your ability to set up and solve an equation involving π and cubic measurements. Given Data / Assumptions: The cylinder has base radius r centimetres. Initial height h₁ = 2r. New height h₂ = 6r (radius unchanged). The increase in volume is 539 cubic centimetres. Use π = 22/7. We need the actual height (in centimetres) of the cylinder in its original configuration. Concept / Approach: The volume V of a right circular cylinder with radius r and height h is V = π * r^2 * h. Here we have two volumes with the same radius but different heights. The difference between these volumes is given. We express both volumes in terms of r, take the difference, equate it to 539, solve for r, and then compute the initial height 2r. Finally we match this value with one of the options. Step-by-Step Solution: Step 1: Write the initial volume: V₁ = π * r^2 * (2r) = 2π * r^3. Step 2: Write the new volume: V₂ = π * r^2 * (6r) = 6π * r^3. Step 3: The increase in volume is V₂ − V₁ = 6π * r^3 − 2π * r^3 = 4π * r^3. Step 4: Given that V₂ − V₁ = 539 cubic centimetres, we have 4π * r^3 = 539. Step 5: Substitute π = 22/7 to get 4 * (22/7) * r^3 = 539, so (88/7) * r^3 = 539. Step 6: Multiply both sides by 7: 88 * r^3 = 539 * 7 = 3773. Step 7: Divide by 88: r^3 = 3773 / 88 = 42.875. This equals 3.5^3, so r = 3.5 cm. Step 8: The original height is h₁ = 2r = 2 * 3.5 = 7 cm. Verification / Alternative Check: Compute the two volumes numerically. Using r = 3.5 cm and π = 22/7: V₁ = 2π * r^3 = 2 * (22/7) * 42.875 ≈ 268.5 cubic centimetres. V₂ = 6π * r^3 ≈ 805.5 cubic centimetres. The difference is 805.5 − 268.5 = 537 cubic centimetres if rounded early, but using exact fractions gives 539 cubic centimetres. This confirms that r = 3.5 cm and h₁ = 7 cm fit the given condition exactly. Why Other Options Are Wrong: A height of 5 cm or 9 cm or 11 cm would correspond to different radius values that do not satisfy the equation 4π * r^3 = 539 when π = 22/7. Only height 7 cm arises from a consistent positive real radius and preserves the exact volume difference of 539 cubic centimetres. Common Pitfalls: Common mistakes include using the same volume formula for both cases but forgetting that height changes from 2r to 6r, or incorrectly writing the volume difference as proportional to the difference in heights rather than recalculating each volume properly. Another pitfall is approximating π too early and losing accuracy, which can prevent you from recognising that r = 3.5 cm is a neat solution. Keeping π symbolic until the end helps avoid this. Final Answer: The original height of the cylinder is 7 cm.

Question 15

Two circles with centres P and Q touch each other externally at a point T. A straight line is drawn through T and meets the circle with centre P again at A and the circle with centre Q again at B. Which of the following statements about segments AP …

Accepted Answer

Introduction / Context: This geometry problem deals with two circles touching externally and a line passing through their common point of contact. It tests understanding of homothety (enlargement from a point), symmetry, and how chords and radii behave when circles touch each other externally. Given Data / Assumptions: Two circles have centres P and Q and touch externally at point T. A straight line through T meets the first circle (centre P) again at A and the second circle (centre Q) again at B. Segments AP and BQ join each point on the line back to the corresponding centre. We must identify the correct relation between AP and BQ. Concept / Approach: When two circles touch externally, there is a homothety (scaling transformation) centred at the point of contact T that maps one circle onto the other. This transformation sends centre P to centre Q and any point A on the first circle to a corresponding point B on the second circle along the same ray from T. Therefore, line segments TA and TB are collinear and their distances from T remain in a fixed ratio. Analysing the geometry with this symmetry shows that AP and BQ lie on the same straight line but in opposite directions, which makes them parallel (collinear). Lengths are related to the radii, but not necessarily equal unless the circles are congruent. Step-by-Step Solution: Step 1: Because the circles touch externally at T, line PQ passes through T and PT and QT are radii. Step 2: Consider any line through T. It cuts the circle with centre P at T and A, and the circle with centre Q at T and B. Step 3: There is a homothety with centre T that maps the first circle to the second, sending P to Q and A to B along the same line. Step 4: Under this transformation, segment AP in the first circle corresponds in direction to segment BQ in the second circle. Step 5: Vector analysis shows that AP and BQ are scalar multiples of each other with opposite sign, which means they lie on the same straight line but point in opposite directions. Step 6: Hence AP and BQ are parallel (collinear), although generally not equal in length. Verification / Alternative Check: A coordinate geometry approach confirms this. By placing T at the origin and P and Q on the x axis with appropriate distances, one can parametrise the line through T and find coordinates for A and B. Computing vectors AP and BQ shows that one is a negative scalar multiple of the other, which proves they are parallel for every possible line through T, not just a particular one. Why Other Options Are Wrong: AP = BQ would only hold if the radii of the two circles are equal, which is not stated. AP ⟂ BQ is incorrect because the vectors are collinear, not perpendicular. AP = 1⁄2 BQ is also not generally true; the scale factor depends on the ratio of the radii and is not fixed at one half. Common Pitfalls: Students often assume that because the circles touch, some special angle such as a right angle must appear between certain segments, but that is not the case here. Another common mistake is to focus only on one particular line through T instead of recognising that the relation must hold for every such line. Thinking in terms of homothety around the point of contact T helps reveal the true relationship between AP and BQ. Final Answer: The correct statement is that AP ∥ BQ (AP is parallel to BQ).

Question 16

Which of the following sets gives three possible side lengths of a valid triangle (all measurements in centimetres)?

Accepted Answer

Introduction / Context: This question checks understanding of the triangle inequality, which tells us which sets of three segments can form a triangle. It is a core concept in geometry and appears frequently in aptitude tests to quickly eliminate impossible side length combinations. Given Data / Assumptions: Each option lists three positive numbers representing possible side lengths of a triangle. All lengths are in centimetres. A valid triangle must satisfy the triangle inequality for every pair of sides. Concept / Approach: For three positive numbers a, b, c to form the sides of a triangle, all three triangle inequalities must hold simultaneously: a + b > c, b + c > a, and c + a > b. If even one of these fails or holds only as equality (a + b = c), the segments cannot form a non degenerate triangle. We check each option one by one using this rule and look for the set where all three inequalities are strictly satisfied. Step-by-Step Solution: Step 1: Option a: 2, 3, 6. Check 2 + 3 = 5, which is not greater than 6. So these cannot form a triangle. Step 2: Option b: 3, 4, 5. Check 3 + 4 = 7 > 5, 4 + 5 = 9 > 3, and 5 + 3 = 8 > 4. All three inequalities are satisfied, so this set can form a triangle. Step 3: Option c: 2.5, 3.5, 6. Here 2.5 + 3.5 = 6, which is equal to the third side, not greater. This would be a degenerate triangle, so it is not considered a valid triangle in most exam contexts. Step 4: Option d: 4, 4, 9. Check 4 + 4 = 8, which is less than 9, so these lengths also cannot form a triangle. Step 5: Therefore only option b gives a valid set of triangle side lengths. Verification / Alternative Check: The classic 3–4–5 set is well known as a Pythagorean triple, forming a right angled triangle. This supports our conclusion that 3 cm, 4 cm, and 5 cm can indeed form a triangle, and it is a useful example to remember for many geometry problems involving right triangles. Why Other Options Are Wrong: In option a, the sum of the two smaller sides is less than the largest side, so the segments cannot meet to close the triangle. In option c, the sum of the two smaller sides is exactly equal to the largest side, which would place all three points on a single straight line. In option d, again the two equal smaller sides sum to less than the largest side, violating the triangle inequality. Common Pitfalls: Students sometimes check only one inequality instead of all three, or they forget that equality is not enough for a proper triangle. Another mistake is to assume that if one pair of sides satisfies the condition, the others automatically do as well, which is not always true. Always check all three combinations carefully to be sure. Final Answer: The only possible set of triangle side lengths is 3 cm, 4 cm, 5 cm.

Question 17

In triangle ABC, AD is a median drawn to side BC and O is the centroid of the triangle. If AO = 10 cm, what is the length of OD (in centimetres)?

Accepted Answer

Introduction / Context: This question is about medians and the centroid of a triangle. Understanding how the centroid divides each median in a fixed ratio is a standard result in coordinate geometry and Euclidean geometry, and it appears frequently in exam problems on triangle properties. Given Data / Assumptions: Triangle ABC has median AD drawn from vertex A to side BC, so D is the midpoint of BC. O is the centroid of triangle ABC (point of intersection of all three medians). The length AO from vertex A to centroid O is 10 cm. We need the length of segment OD from centroid O to midpoint D. Concept / Approach: A key property of the centroid is that it divides every median of a triangle in the ratio 2 : 1, measured from the vertex to the opposite side. This means that for median AD, AO : OD = 2 : 1. If AO is known, we can express the full median length AD in terms of AO and then find OD as the remaining one third of the median. No angles or coordinates are required if this property is remembered. Step-by-Step Solution: Step 1: Use the centroid property: AO : OD = 2 : 1 along the median AD. Step 2: Let AD be the full length of the median. Then AO = (2/3) * AD and OD = (1/3) * AD. Step 3: Given AO = 10 cm, substitute: (2/3) * AD = 10. Step 4: Solve for AD: AD = (10 * 3) / 2 = 30 / 2 = 15 cm. Step 5: Now find OD: OD = (1/3) * AD = (1/3) * 15 = 5 cm. Verification / Alternative Check: We can also reason directly from the ratio AO : OD = 2 : 1. If AO is 10 cm and represents two equal parts, then each part is 5 cm. Therefore OD, which is one such part, must be 5 cm. This matches the calculation using the median length and confirms the result. Why Other Options Are Wrong: Option b, 20 cm, would make OD longer than AO, contradicting the 2 : 1 ratio from vertex to midpoint. Option c, 10 cm, would imply AO and OD are equal, which is again inconsistent with the centroid property. Option d, 30 cm, is even larger and clearly impossible because the entire median would need to be longer than 45 cm in that case, which does not match AO = 10 cm. Common Pitfalls: A common mistake is to invert the ratio and assume AO is one third of the median instead of two thirds. Another pitfall is to think that the centroid divides the medians in a 1 : 1 ratio like a midpoint, which is not correct. Remember that the centroid is closer to the base than to the vertex and always lies two thirds of the way from each vertex along its median. Final Answer: The length of OD is 5 cm.

Question 18

In triangle ABC, the incentre is I. If angle ABC = 90° and angle ACB = 70°, what is the measure of angle BIC (in degrees)?

Accepted Answer

Introduction / Context: This problem uses a standard result about the angles formed at the incentre of a triangle. The incentre is the point where the internal angle bisectors meet and is the centre of the inscribed circle. Knowing how angles at the incentre relate to the angles of the triangle is useful in many geometry questions. Given Data / Assumptions: I is the incentre of triangle ABC. Angle ABC = 90°. Angle ACB = 70°. We need to find the measure of angle BIC. Concept / Approach: In any triangle, if I is the incentre, the angle between the angle bisectors at vertices B and C (that is, angle BIC) is given by the formula: angle BIC = 90° + (A / 2), where A is the measure of angle at vertex A. This result comes from the fact that the sum of the angles around point I is 360° and each angle at the incentre is formed by halves of the original triangle angles. So our first step is to find angle A using the triangle angle sum, then apply the formula. Step-by-Step Solution: Step 1: Use the angle sum property of a triangle: A + B + C = 180°. Step 2: Substitute B = 90° and C = 70°: A + 90° + 70° = 180°. Step 3: Simplify: A + 160° = 180°, so A = 20°. Step 4: Use the incentre formula for angle BIC: angle BIC = 90° + (A / 2). Step 5: Substitute A = 20°: angle BIC = 90° + 20° / 2 = 90° + 10° = 100°. Verification / Alternative Check: We can check other incentre angle formulas to ensure consistency. For example, angle CIA = 90° + (B / 2) and angle AIB = 90° + (C / 2). Using B = 90° and C = 70°, we get angle CIA = 90° + 45° = 135° and angle AIB = 90° + 35° = 125°. The three angles at the incentre must sum to 360°, and indeed 100° + 135° + 125° = 360°, confirming that the formula and calculation are correct. Why Other Options Are Wrong: 115°, 110°, or 105° do not match the formula 90° + (A / 2) once A is correctly found to be 20°. Using any of those values would make the sum of angles at the incentre differ from 360°, contradicting the geometric structure of the triangle and its incentre. Common Pitfalls: A frequent mistake is to use the wrong vertex angle in the formula and write angle BIC = 90° + (B / 2), which would be incorrect. Another pitfall is forgetting to first compute angle A from the angle sum rule. Some students also confuse the incentre with the circumcentre or orthocentre, each of which has different angle relations. Carefully recalling that angle BIC uses the angle at vertex A helps avoid these errors. Final Answer: The measure of angle BIC is 100°.

Question 19

A rectangle is inscribed in a circle. The lengths of its two adjacent sides are 5 cm and 12 cm. What is the radius of the circle (in centimetres)?

Accepted Answer

Introduction / Context: This question tests your understanding of the relationship between a rectangle inscribed in a circle and the circle's diameter. It uses the Pythagorean theorem and the fact that the diagonal of a rectangle inscribed in a circle is always the diameter of that circle. Given Data / Assumptions: A rectangle is drawn inside a circle so that all four vertices lie on the circle. The adjacent sides of the rectangle measure 5 cm and 12 cm. We need to find the radius of the circle. Concept / Approach: For any rectangle inscribed in a circle, the diagonal of the rectangle passes through the centre of the circle and is equal to the diameter of the circle. Therefore, if we find the length of the diagonal using the Pythagorean theorem, half of this diagonal will give the radius. In a rectangle with sides a and b, the diagonal d satisfies d^2 = a^2 + b^2. Here the sides are 5 cm and 12 cm, which form a well known Pythagorean triple. Step-by-Step Solution: Step 1: Let the sides of the rectangle be a = 5 cm and b = 12 cm. Step 2: Use the Pythagorean theorem to find the diagonal d: d^2 = a^2 + b^2. Step 3: Compute a^2 = 5^2 = 25 and b^2 = 12^2 = 144. Step 4: Add them: d^2 = 25 + 144 = 169. Step 5: Take the square root: d = √169 = 13 cm. Step 6: The diagonal is the diameter of the circle, so diameter = 13 cm. Step 7: Radius r = diameter / 2 = 13 / 2 = 6.5 cm. Verification / Alternative Check: You can recognise that 5, 12, 13 is a standard Pythagorean triple, so any right triangle or rectangle with legs 5 and 12 automatically has a hypotenuse or diagonal of 13. Since the rectangle's diagonal matches the circle's diameter, the radius must be half of 13, which is 6.5 cm. This quick recognition reinforces the calculation above. Why Other Options Are Wrong: A radius of 6 cm or 8 cm or 8.5 cm would correspond to diameters of 12 cm, 16 cm, or 17 cm respectively, none of which match the necessary 13 cm diagonal created by the 5–12 sides. This would contradict the geometry of a rectangle inscribed in a circle, where the diagonal must equal the diameter. Common Pitfalls: Sometimes learners confuse the radius with the diagonal itself and forget to divide by 2, wrongly choosing 13 instead of 6.5. Others might calculate the diagonal incorrectly by adding side lengths instead of adding the squares of the side lengths. Always remember that the Pythagorean theorem uses squared lengths, and that the radius is half the diameter, not the full diagonal. Final Answer: The radius of the circle is 6.5 cm.

Question 20

Which of the following sets could be the measures of the three interior angles of a triangle?

Accepted Answer

Introduction / Context: This question checks whether you know the basic angle sum property of a triangle and can quickly verify if a set of three angles can be internal angles of a triangle. It is a simple but very common test of fundamental geometry knowledge. Given Data / Assumptions: Each option lists three angles in degrees. We assume these are interior angles of a triangle, if possible. A valid triangle must have all interior angles positive and their sum equal to 180°. Concept / Approach: The interior angles of any triangle in Euclidean geometry always add up to 180°. Therefore, to check if a set of three angles can represent the angles of a triangle, we simply add them and verify that the total is exactly 180° and that none of the angles is zero or negative. If the sum is greater than or less than 180°, or one angle is zero, the set cannot represent a triangle. Step-by-Step Solution: Step 1: Option a: 33°, 42°, 115°. Sum = 33° + 42° + 115° = 190°, which is greater than 180°, so this is not possible. Step 2: Option b: 40°, 70°, 80°. Sum = 40° + 70° + 80° = 190°, again greater than 180°, so not possible. Step 3: Option c: 30°, 60°, 100°. Sum = 30° + 60° + 100° = 190°, also greater than 180°, so this set is impossible as well. Step 4: Option d: 50°, 60°, 70°. Sum = 50° + 60° + 70° = 180°, which exactly matches the triangle angle sum. Step 5: All angles in option d are positive and less than 180°, so this set is valid for a triangle. Verification / Alternative Check: Because only one option yields a sum of exactly 180°, there is no ambiguity. Any small change, such as increasing or decreasing one angle while keeping the others fixed, would break the 180° total, so only 50°, 60°, and 70° can work together as the internal angles of a triangle. Why Other Options Are Wrong: Options a, b, and c all sum to 190°, which is too large. A triangle with such angles would have to "overlap" itself, which is impossible in flat Euclidean geometry. The angle sum property is strict: the sum must be exactly 180° for a standard triangle in a plane. Common Pitfalls: Learners sometimes assume that any three positive angles less than 180° must form a triangle without checking the sum. Another mistake is misadding the angles in haste, especially under exam pressure. Taking a few seconds to carefully sum the values and compare with 180° is enough to avoid such errors. Final Answer: The only possible set of angles of a triangle is 50°, 60°, 70°.

Volume and Surface Area Questions