Question 1

In triangle PQR, points S and T lie on sides PQ and PR respectively such that segment ST is parallel to QR. If PS : SQ = 3 : 5 and PR = 6 cm, then what is the length of segment PT (in centimetres)?

Accepted Answer

Introduction / Context: This question uses the basic concept of similar triangles created by a line drawn parallel to one side of a triangle. Aptitude tests frequently use such problems to check understanding of proportionality of corresponding sides and how segment ratios give actual lengths. Given Data / Assumptions: - Triangle PQR is given. - Point S lies on side PQ and point T lies on side PR. - Segment ST is parallel to base QR. - The ratio PS : SQ is 3 : 5. - The full side PR has length 6 cm. - We need the length of PT. Concept / Approach: When a line is drawn parallel to the base of a triangle and cuts the other two sides, a smaller triangle similar to the original triangle is formed. Here, triangle PST is similar to triangle PQR. For similar triangles, the ratio of corresponding sides is equal. The segment PS corresponds to PQ, and PT corresponds to PR, so the ratio PS / PQ equals PT / PR. Step-by-Step Solution: Step 1: From PS : SQ = 3 : 5, write PS = 3k and SQ = 5k for some positive k. Step 2: Then the full side PQ = PS + SQ = 3k + 5k = 8k. Step 3: For similar triangles PST and PQR, PS / PQ = PT / PR. Step 4: Substitute PS = 3k and PQ = 8k to get (3k) / (8k) = PT / PR. Step 5: The k terms cancel, so 3 / 8 = PT / PR. Step 6: Given PR = 6 cm, we have PT = (3 / 8) * 6 = 18 / 8 = 2.25 cm. Verification / Alternative Check: As a check, note that PS / PQ = 3 / 8. This means the similar smaller triangle PST is a 3/8 scaled copy of PQR. Any length in triangle PST must be 3/8 of the corresponding length in triangle PQR. Since PT corresponds to PR, and PR = 6 cm, multiplying 6 by 3/8 again gives 2.25 cm, which confirms the previous calculation. Why Other Options Are Wrong: Option a, 2 cm, corresponds to a factor of 2/6 = 1/3, which does not match the required ratio 3/8. Option c, 3.5 cm, is more than half of PR and conflicts with PS being only 3/8 of PQ in the similarity ratio. Option d, 4 cm, would correspond to PT / PR = 4 / 6 = 2 / 3, again inconsistent with the ratio 3 / 8. Common Pitfalls: Learners sometimes mistakenly take PS : SQ as PS : PQ and then use that directly in place of PS / PQ. Another common mistake is to assume PS / SQ equals PT / PR, which is not a correct correspondence between the similar triangles. Drawing the diagram carefully and labeling corresponding sides in both triangles helps to avoid these errors. Final Answer: The length of PT is 2.25 cm.

Question 2

An angle is greater than its complementary angle by 20°. What is the measure of this angle in degrees?

Accepted Answer

Introduction / Context: This problem checks understanding of complementary angles and how to form simple linear equations from a verbal description. Complementary angles are a very common topic in basic geometry and appear frequently in aptitude tests and reasoning questions. Given Data / Assumptions: - Two angles are complementary if their sum is 90°. - One angle is described as being greater than its complementary angle by 20°. - We need to find the measure of the larger angle. Concept / Approach: Let the required angle be x degrees. Its complementary angle is then 90° − x degrees. The statement says that the angle x is greater than its complementary angle by 20°, so x is 20° more than 90° − x. This description translates directly into a simple linear equation that can be solved for x. Step-by-Step Solution: Step 1: Let the required angle be x degrees. Step 2: Its complementary angle is 90° − x degrees. Step 3: Given that x is greater than its complement by 20°, write the equation x = (90° − x) + 20°. Step 4: Simplify the right side: (90° − x) + 20° = 110° − x. Step 5: So the equation is x = 110° − x. Step 6: Add x to both sides: x + x = 110°, which gives 2x = 110°. Step 7: Divide both sides by 2 to obtain x = 55°. Verification / Alternative Check: If the angle is 55°, then its complement is 90° − 55° = 35°. The difference between the angle and its complement is 55° − 35° = 20°, which matches the condition in the question. Therefore 55° is verified as correct. Why Other Options Are Wrong: Option b, 35°, is actually the complementary angle, not the angle that is greater by 20°. Option c, 100°, is greater than 90° and therefore cannot be complementary to any positive real angle in the sense of complementary pairs summing to 90°. Option d, 80°, would have a complement of only 10°, giving a difference of 70°, not 20°. Common Pitfalls: A common mistake is to set up the equation with the roles reversed or to assume that the angles differ by 20° without checking which one is larger. Another error is confusing complementary angles (sum 90°) with supplementary angles (sum 180°). Carefully translating the verbal statement into an algebraic equation is the key skill in this type of question. Final Answer: The measure of the angle is 55°.

Question 3

A hemisphere has radius 21 cm. Using π = 22/7, what is the volume of this hemisphere in cubic centimetres?

Accepted Answer

Introduction / Context: This question tests knowledge of the volume formula for a hemisphere. Understanding volumes of standard solid shapes such as spheres, hemispheres, cylinders, and cones is essential in aptitude tests and engineering entrance examinations. Given Data / Assumptions: - The solid is a hemisphere, which is half of a sphere. - Radius r = 21 cm. - Use π = 22/7 as specified. - We must find the volume in cubic centimetres. Concept / Approach: The volume of a sphere of radius r is given by: Volume of sphere = (4 / 3) * π * r^3. A hemisphere is exactly half of a sphere, so its volume is: Volume of hemisphere = (1 / 2) * (4 / 3) * π * r^3 = (2 / 3) * π * r^3. We substitute r = 21 cm and π = 22/7 into this formula and evaluate. Step-by-Step Solution: Step 1: Write the formula for the volume of a hemisphere: V = (2 / 3) * π * r^3. Step 2: Substitute r = 21 cm and π = 22/7. Step 3: Compute r^3 = 21^3 = 21 * 21 * 21 = 9261. Step 4: Substitute into the formula: V = (2 / 3) * (22 / 7) * 9261. Step 5: First simplify 9261 / 7 = 1323, so V = (2 / 3) * 22 * 1323. Step 6: Compute 22 * 1323 = 29106. Step 7: Now multiply by 2 / 3: V = (2 / 3) * 29106 = 58212 / 3 = 19404 cubic centimetres. Verification / Alternative Check: As a quick approximate check, using π close to 3.14, the volume would be about (2 / 3) * 3.14 * 21^3. This also leads to a value near 19400 cubic centimetres, supporting that 19404 cubic centimetres is correct when using π = 22/7 exactly. Why Other Options Are Wrong: Option a, 38808 cubic cms, is exactly double the correct answer and corresponds to the volume of a full sphere rather than a hemisphere. Option b, 9702 cubic cms, is half the correct volume and might result from an incorrect division or misapplied factor. Option d, 29106 cubic cms, is an intermediate product obtained before multiplying by 2 / 3, so it represents a partial calculation rather than the final answer. Common Pitfalls: Students often forget that a hemisphere is half of a sphere and directly use the formula (4 / 3) * π * r^3, which yields the volume of the full sphere. Another common mistake is mishandling the fractions when simplifying, especially the division by 7 and multiplication by 2 / 3. Writing each step clearly and simplifying systematically prevents such errors. Final Answer: The volume of the hemisphere is 19404 cubic cms.

Question 4

In a right angled triangle, the side opposite the right angle (the hypotenuse) is 6 cm long. What is the length of the side opposite to a 30° angle in this triangle?

Accepted Answer

Introduction / Context: This question uses a standard result from right triangle trigonometry, specifically the special 30°–60°–90° triangle. Many aptitude and entrance exams use this property to create fast questions where knowledge of fixed ratios saves time compared to full trigonometric calculations. Given Data / Assumptions: - The triangle is right angled, so one angle is 90°. - The side opposite the right angle, which is the hypotenuse, has length 6 cm. - One acute angle is 30°. - We need the length of the side opposite the 30° angle. Concept / Approach: In any 30°–60°–90° right triangle the side lengths follow a fixed ratio: - Side opposite 30° : side opposite 60° : hypotenuse = 1 : √3 : 2. This means the side opposite 30° is exactly half of the hypotenuse. Therefore, once we know the hypotenuse, we can immediately find the side opposite 30° by dividing the hypotenuse by 2. Step-by-Step Solution: Step 1: Recognize the triangle as a 30°–60°–90° right triangle. Step 2: Use the standard ratio: side opposite 30° = hypotenuse / 2. Step 3: The hypotenuse is given as 6 cm, so the side opposite 30° is 6 / 2 = 3 cm. Step 4: Thus the required side length is 3 cm. Verification / Alternative Check: If the side opposite 30° is 3 cm, then using the ratio 1 : √3 : 2, the side opposite 60° should be 3√3 cm, and the hypotenuse should be 2 * 3 = 6 cm. This matches the given hypotenuse, confirming that 3 cm for the side opposite 30° is consistent with the known geometric ratio. Why Other Options Are Wrong: Option b, 3√3 cm, is the length of the side opposite the 60° angle, not the 30° angle. Option c, 3/2 cm, is one quarter of the hypotenuse and does not correspond to any standard ratio here. Option d, (3/2)√3 cm, also does not match the known 1 : √3 : 2 ratio when matched to a hypotenuse of 6 cm. Common Pitfalls: A frequent mistake is mixing up which leg corresponds to 30° and which to 60°. Another pitfall is attempting to use trigonometric functions without recognizing the special triangle pattern, which takes more time and increases the chance of calculator or algebra errors. Remembering the simple 1 : √3 : 2 ratio for 30°–60°–90° triangles gives very quick solutions. Final Answer: The length of the side opposite the 30° angle is 3 cm.

Question 5

In a cyclic quadrilateral ABCD, side AB is extended to a point X. If ∠XBC = 82° and the angle ∠ADB at the opposite vertex is 47°, what is the measure of angle ∠BDC in degrees?

Accepted Answer

Introduction / Context: This question combines properties of cyclic quadrilaterals with exterior angles of triangles. It tests whether a student can correctly relate inscribed angles that subtend the same chord and use basic angle chasing in a cyclic figure. Given Data / Assumptions: - ABCD is a cyclic quadrilateral, so all four vertices lie on a circle. - Side AB is extended beyond B to point X. - ∠XBC, an exterior angle at vertex B of triangle ABC, is 82°. - ∠ADB is given as 47°. - We are asked to find ∠BDC in degrees. Concept / Approach: There are two important concepts: - In a triangle, an exterior angle is equal to the sum of the two opposite interior angles. - In a circle, inscribed angles that subtend the same chord are equal. Here both ∠ACB and ∠ADB subtend chord AB, so they are equal. Once we use ∠ADB = 47° to determine ∠ACB, we can use the exterior angle at B to find ∠BAC, and then use the inscribed angle theorem again to connect ∠BAC with ∠BDC, since both subtend chord BC. Step-by-Step Solution: Step 1: Because ABCD is cyclic, and both ∠ACB and ∠ADB subtend chord AB, we have ∠ACB = ∠ADB = 47°. Step 2: Consider triangle ABC. The exterior angle at B, ∠XBC, equals the sum of the two interior opposite angles ∠BAC and ∠ACB. Step 3: Write the relation: ∠XBC = ∠BAC + ∠ACB. Step 4: Substitute the values: 82° = ∠BAC + 47°. Step 5: Solve for ∠BAC: ∠BAC = 82° − 47° = 35°. Step 6: Now note that ∠BAC and ∠BDC both subtend chord BC of the circle, so these two inscribed angles are equal. Step 7: Therefore ∠BDC = ∠BAC = 35°. Verification / Alternative Check: We can check consistency by confirming that no other angle condition is violated. The fact that ∠ACB = 47° and ∠BAC = 35° implies that ∠ABC = 180° − (35° + 47°) = 98°. Together with the exterior angle 82°, this is consistent since ∠XBC is outside the triangle and equal to 82°. There is no conflict among the angle relations, confirming that 35° for ∠BDC is correct. Why Other Options Are Wrong: Option a, 40°, would imply a different measure for ∠BAC and hence would break the equality of inscribed angles subtending chord BC. Option c, 30°, and option d, 25°, also do not match the derived angle of 35° based on the external angle relationship and inscribed angle theorem, and they would make the sum of angles in triangle ABC inconsistent with the given exterior angle. Common Pitfalls: A common mistake is to forget that inscribed angles subtending the same chord are equal, or to misuse the cyclic quadrilateral property that opposite angles sum to 180°, even when the information given is better handled by the inscribed angle theorem. Another pitfall is misidentifying which angles subtend which chords. Drawing the circle carefully and marking chords AB and BC clearly can prevent such confusion. Final Answer: The measure of angle ∠BDC is 35°.

Question 6

From an external point P, two tangents PQ and PR are drawn to a circle with centre O. If the angle between the tangents, ∠QPR, is 120°, then what is the measure of the central angle ∠POQ in degrees?

Accepted Answer

Introduction / Context: This question checks understanding of the relationship between the angle formed by two tangents drawn from an external point to a circle and the central angle subtended by the points of contact. This is a standard geometry result used in many circle based questions. Given Data / Assumptions: - A circle with centre O is given. - From an external point P, two tangents PQ and PR are drawn, touching the circle at Q and R respectively. - The angle between the tangents, ∠QPR, is 120°. - We need to find the central angle ∠POQ in degrees. Concept / Approach: The line segments OQ and OR are radii drawn to the points of tangency, so they are perpendicular to the tangents PQ and PR. Therefore, quadrilateral POQR has right angles at Q and R. A known theorem states that the angle between two tangents from an external point is supplementary to the central angle subtended by the chord joining the points of contact. In symbols: ∠QPR = 180° − ∠QOR. Step-by-Step Solution: Step 1: Recognize that OQ ⟂ PQ and OR ⟂ PR, since a radius to the point of tangency is perpendicular to the tangent. Step 2: Note that the lines OQ and OR form the central angle ∠QOR, which is the angle between the radii to the points of tangency. Step 3: Use the known relation between the angle between tangents and the corresponding central angle: ∠QPR = 180° − ∠QOR. Step 4: Substitute the given value: 120° = 180° − ∠QOR. Step 5: Rearrange to find ∠QOR: ∠QOR = 180° − 120° = 60°. Step 6: The central angle ∠POQ is the angle between OP and OQ, but OP is the line from centre to external point and does not change the central angle subtended by chord QR. The central angle associated with the points of contact is ∠QOR, so the required central angle is 60°. Verification / Alternative Check: In some presentations, the angle between tangents is described as the exterior angle of the triangle formed by joining the centre O to the points of contact Q and R and the point P. The interior angle at O then satisfies the same supplementary relation with ∠QPR, again leading to 60° as the central angle. This consistency supports the answer. Why Other Options Are Wrong: Option a, 90°, would make the angle between tangents equal to 180° − 90° = 90°, not the given 120°. Option b, 45°, would lead to an angle between tangents of 135°, which does not match 120°. Option c, 30°, would give ∠QPR = 150°, again not the given 120°. Common Pitfalls: A frequent error is to think that the angle between tangents equals the central angle, rather than being supplementary to it. Another common mistake is to forget that OQ and OR are perpendicular to the tangents, which is essential for visualizing the correct quadrilateral and angle relationships. Carefully recalling the direct formula ∠between tangents = 180° − ∠central helps avoid such confusion. Final Answer: The measure of the central angle is 60°.

Question 7

In triangle ABC, angle ∠BAC is 90° and the two sides AB and AC are equal in length. What is the measure of angle ∠ABC in degrees?

Accepted Answer

Introduction / Context: This question is a straightforward application of angle sum in a triangle and the properties of an isosceles right triangle. It is common in elementary geometry and quantitative aptitude tests. Given Data / Assumptions: - Triangle ABC has angle ∠BAC = 90°, so the right angle is at vertex A. - Sides AB and AC are equal in length, so triangle ABC is isosceles with AB = AC. - We need to find angle ∠ABC in degrees. Concept / Approach: In any triangle, the sum of the three interior angles is 180°. Because AB = AC, the base angles opposite these equal sides, that is, angles at B and C, must also be equal. Thus, ∠ABC = ∠ACB. Knowing that ∠BAC is 90°, the remaining two equal angles must share the remaining 90° equally. Step-by-Step Solution: Step 1: Use the angle sum property: ∠BAC + ∠ABC + ∠ACB = 180°. Step 2: Substitute ∠BAC = 90° and let ∠ABC = ∠ACB = x° because AB = AC. Step 3: The equation becomes 90° + x° + x° = 180°. Step 4: Simplify: 90° + 2x° = 180°. Step 5: Subtract 90° from both sides: 2x° = 90°. Step 6: Divide both sides by 2: x° = 45°. Step 7: Therefore ∠ABC = 45°. Verification / Alternative Check: Because this is a 45°–45°–90° triangle, the ratio of the sides opposite these angles follows a known pattern. The hypotenuse BC is √2 times any leg. This is consistent with the given right angle at A and equal legs AB and AC. Thus, the computed 45° angles at B and C agree with the standard properties of an isosceles right triangle. Why Other Options Are Wrong: Option a, 30°, together with a 90° angle would leave 60° for the third angle, so the base angles would not be equal, contradicting AB = AC. Option b, 60°, would similarly force the third angle to be 30°, again giving unequal base angles. Option d, 25°, is incompatible with the triangle angle sum and the requirement that the two non right angles be equal. Common Pitfalls: Sometimes students forget which angles are equal in an isosceles triangle and may incorrectly assume the right angle is one of the equal angles. The correct approach is to remember that equal sides lie opposite equal angles. Here AB and AC are equal, so angles at C and B must be equal, not the right angle at A. Using the angle sum property then makes the calculation straightforward. Final Answer: The measure of angle ∠ABC is 45°.

Question 8

Two circles touch each other externally. The sum of their areas is 130π square centimetres and the distance between their centres is 14 cm. What is the radius of the larger circle?

Accepted Answer

Introduction / Context: This question checks understanding of circle geometry, including area of a circle and basic algebra with two unknown radii. It also uses the condition that two circles touch externally, which gives a relationship between the radii and the distance between the centres. Given Data / Assumptions: - Two circles touch externally, so the distance between their centres equals the sum of their radii. - Let the radii be r1 and r2, where r1 is the radius of the larger circle and r2 is the smaller one. - The sum of their areas is 130π square centimetres. - The distance between the centres is 14 cm, so r1 + r2 = 14. - We need to find r1, the radius of the bigger circle. Concept / Approach: The area of a circle with radius r is π * r^2. The total area of two circles is therefore π * r1^2 + π * r2^2, which is given to be 130π. This gives r1^2 + r2^2 = 130. Combined with the touching condition r1 + r2 = 14, we have a system of two equations in r1 and r2 that can be solved using algebraic identities. Step-by-Step Solution: Step 1: From the area condition, write π * r1^2 + π * r2^2 = 130π. Step 2: Divide both sides by π to get r1^2 + r2^2 = 130. Step 3: From the external touching condition, write r1 + r2 = 14. Step 4: Square the second equation: (r1 + r2)^2 = 14^2 = 196. Step 5: Expand the square: r1^2 + 2 * r1 * r2 + r2^2 = 196. Step 6: Substitute r1^2 + r2^2 = 130 into this: 130 + 2 * r1 * r2 = 196. Step 7: Simplify to obtain 2 * r1 * r2 = 196 − 130 = 66, so r1 * r2 = 33. Step 8: Now r1 and r2 satisfy r1 + r2 = 14 and r1 * r2 = 33, so they are roots of the quadratic x^2 − 14x + 33 = 0. Step 9: Factor this quadratic: x^2 − 14x + 33 = (x − 11)(x − 3). Step 10: The roots are x = 11 and x = 3. So the radii are 11 cm and 3 cm. Step 11: The larger radius is therefore 11 cm. Verification / Alternative Check: Check the area condition: r1^2 + r2^2 = 11^2 + 3^2 = 121 + 9 = 130, which matches the given total area divided by π. Check the distance condition: r1 + r2 = 11 + 3 = 14 cm, which matches the distance between centres. Both conditions are satisfied, confirming the solution. Why Other Options Are Wrong: Option a, 22 cm, is too large and would make r1 + r2 exceed 14 cm for any positive r2, violating the touching condition. Option c, 33 cm, is even larger and clearly impossible when the distance between centres is only 14 cm. Option d, 44 cm, is far beyond the possible value and does not satisfy either given condition. Common Pitfalls: Some students attempt to guess radii directly from the area sum without using the condition that the circles touch externally. Others may forget to divide by π when equating areas, leading to more complicated expressions. Using the identity for (r1 + r2)^2 simplifies the algebra significantly and is a useful technique whenever sums and squares of variables are involved. Final Answer: The radius of the larger circle is 11 cm.

Question 9

In which of the following quadrilaterals do the diagonals not necessarily form at least two congruent triangles?

Accepted Answer

Introduction / Context: This conceptual question checks your understanding of how diagonals behave in different quadrilaterals. It focuses on whether the diagonals divide the figure into congruent triangles, which is closely linked to symmetry properties of the shape. Given Data / Assumptions: - Four types of quadrilaterals are listed: parallelogram, rhombus, trapezium, and kite. - The question asks in which case the diagonals do not necessarily create at least two congruent triangles. - We assume a general quadrilateral of each type, not a special or symmetric case unless that property is inherent to the definition. Concept / Approach: We recall basic diagonal properties: - In a parallelogram, diagonals bisect each other, so each diagonal splits the parallelogram into two congruent triangles. - In a rhombus, which is a special parallelogram with all sides equal, diagonals also bisect each other and are perpendicular, forming pairs of congruent triangles. - In a kite, one diagonal is an axis of symmetry, leading to two congruent triangles on each side of that diagonal. - A general trapezium (with only one pair of parallel sides) does not necessarily have diagonals that bisect each other or create congruent triangles, unless it is a special isosceles trapezium with extra symmetry, which is not guaranteed. Step-by-Step Solution: Step 1: For a parallelogram, any diagonal divides it into two triangles that share a common base and equal height, so these triangles are congruent by side angle side or other criteria. Step 2: A rhombus inherits all parallelogram properties and also has additional symmetry, so its diagonals definitely form congruent triangles. Step 3: In a kite, the diagonal connecting the vertices where unequal sides meet is an axis of symmetry, so it clearly forms two congruent triangles. Step 4: A general trapezium has only one pair of parallel sides and its diagonals are not constrained to bisect each other or to be equal. Therefore, in the general case, the triangles formed by its diagonals are not necessarily congruent. Verification / Alternative Check: Drawing a generic scalene trapezium with no symmetry quickly shows that the two triangles formed by one diagonal do not have equal sides or equal angles, so they are not congruent. By contrast, try drawing a generic parallelogram or kite and check the triangles formed by one diagonal; you will find congruent pairs every time, showing the difference clearly. Why Other Options Are Wrong: Option a, parallelogram, always has diagonals that bisect each other, so each diagonal produces two congruent triangles. Option b, rhombus, is an even more symmetric parallelogram, and its diagonals produce congruent triangles in multiple ways. Option d, kite, has at least one diagonal that is an axis of symmetry, guaranteeing a pair of congruent triangles along that diagonal. Common Pitfalls: One common confusion is between a general trapezium and an isosceles trapezium. An isosceles trapezium has additional symmetry and sometimes special diagonal properties, but the question refers to a general trapezium. Another pitfall is assuming that having one pair of parallel sides is enough for diagonals to create congruent triangles, which is not true without further symmetry conditions. Final Answer: The quadrilateral in which the diagonals do not necessarily form at least two congruent triangles is the Trapezium.

Question 10

In triangle ABC, points P and Q lie on sides AB and AC respectively such that segment PQ is parallel to side BC. If the ratio AP : PB is 1 : 4 and the area of triangle APQ is 4 square centimetres, what is the area of trapezium PQCB?

Accepted Answer

Introduction / Context: This problem uses similar triangles and area ratios. It is a standard type of question in geometry and aptitude tests where a line parallel to a side of a triangle divides it into a smaller similar triangle and a trapezium, and the areas are related by the square of the similarity ratio. Given Data / Assumptions: - Triangle ABC is given. - P lies on AB and Q lies on AC. - Segment PQ is parallel to side BC. - AP : PB = 1 : 4, so P divides AB in this ratio. - Area of triangle APQ is 4 sq cm. - We must find the area of trapezium PQCB, which is the region between PQ and BC inside triangle ABC. Concept / Approach: When a line is drawn parallel to one side of a triangle, it cuts the other two sides proportionally and forms a smaller triangle similar to the original. Here, triangle APQ is similar to triangle ABC. For similar triangles, the ratio of their areas equals the square of the ratio of corresponding sides. Once we find the ratio of similarity, we can compute the area of the larger triangle and then subtract the area of the smaller one to get the area of the trapezium. Step-by-Step Solution: Step 1: From AP : PB = 1 : 4, let AP = 1k and PB = 4k for some k, so AB = AP + PB = 5k. Step 2: In similar triangles APQ and ABC, the scale factor of similarity is AP / AB = 1k / 5k = 1 / 5. Step 3: The ratio of areas of similar triangles equals the square of the ratio of corresponding sides. Therefore, Area(APQ) / Area(ABC) = (1 / 5)^2 = 1 / 25. Step 4: Given Area(APQ) = 4 sq cm, then 4 / Area(ABC) = 1 / 25, so Area(ABC) = 4 * 25 = 100 sq cm. Step 5: The trapezium PQCB occupies the rest of the triangle beyond triangle APQ. Hence Area(trapezium PQCB) = Area(ABC) − Area(APQ) = 100 − 4 = 96 sq cm. Verification / Alternative Check: A quick check is to imagine actual side lengths scaled such that AB is 5 units and AP is 1 unit. The small triangle then has linear measure 1/5 of the full triangle, so its area is 1/25 of the full area. This fully agrees with the algebraic method. The numbers remain consistent and yield the same area for the trapezium, 96 sq cm. Why Other Options Are Wrong: Option a, 60 sq cm, would imply that triangle APQ had area 40 sq cm, which contradicts the given 4 sq cm and the similarity ratio. Option b, 16 sq cm, is smaller than the area of triangle APQ, which is impossible because the trapezium covers more region than the small triangle. Option d, 21 sq cm, does not match any reasonable proportion arising from the 1:5 linear ratio and leads to a noninteger total area when combined with triangle APQ. Common Pitfalls: Students sometimes forget to square the side ratio when relating areas and instead use Area ratio = 1 / 5, which would give the wrong total area. Another common mistake is to misinterpret AP : PB as AP : AB, leading to an incorrect similarity factor. Carefully identifying corresponding sides and remembering that area scales with the square of linear dimensions is the key to solving this type of question. Final Answer: The area of trapezium PQCB is 96 sq cm.

Question 11

In triangle DEF, points G and H lie on sides DE and DF respectively, and GH is parallel to EF. Point G divides side DE in the ratio DG : GE = 3 : 2. If HF = 8 cm, what is the length of side DF?

Accepted Answer

Introduction / Context: This problem uses the idea of similar triangles created by a line parallel to one side of a triangle. It focuses on how a point dividing one side in a given ratio leads to proportional segments on the other sides when a parallel line is drawn. Given Data / Assumptions: - Triangle DEF is given. - Point G lies on side DE and point H lies on side DF. - Segment GH is parallel to side EF. - G divides DE in the ratio DG : GE = 3 : 2, so DE is split into parts 3k and 2k for some k. - HF, the segment from H to F on DF, is 8 cm. - We must find the length of side DF. Concept / Approach: Because GH is parallel to EF, triangle DGH is similar to triangle DEF. In similar triangles, corresponding side lengths are proportional. The ratio DG / DE gives the scale factor between the smaller triangle DGH and the original triangle DEF. Once we know what fraction of DF corresponds to DH, we can relate HF to DF, since HF is the remaining part of DF after removing DH. Step-by-Step Solution: Step 1: From DG : GE = 3 : 2, write DG = 3k and GE = 2k. Step 2: Then DE = DG + GE = 3k + 2k = 5k. Step 3: Because GH ∥ EF, triangles DGH and DEF are similar, and the scale factor is DG / DE = 3k / 5k = 3 / 5. Step 4: Thus, corresponding sides satisfy DH / DF = 3 / 5, since DH in the smaller triangle corresponds to DF in the larger triangle. Step 5: Therefore DH = (3 / 5) * DF. Step 6: The remaining segment HF on DF is DF − DH = DF − (3 / 5) * DF = (2 / 5) * DF. Step 7: We are given HF = 8 cm, so (2 / 5) * DF = 8. Step 8: Solve for DF: DF = 8 * (5 / 2) = 20 cm. Verification / Alternative Check: If DF = 20 cm, then DH = (3 / 5) * 20 = 12 cm and HF = 20 − 12 = 8 cm, which matches the given length of HF. The internal ratios along the sides are consistent with the similarity factor 3 / 5, so the value DF = 20 cm is verified. Why Other Options Are Wrong: Option a, 12 cm, would give HF = (2 / 5) * 12 = 4.8 cm, not 8 cm. Option c, 14 cm, would give HF = (2 / 5) * 14 = 5.6 cm, again not 8 cm. Option d, 16 cm, would give HF = (2 / 5) * 16 = 6.4 cm, which also does not match the given value. Common Pitfalls: A common error is to use the ratio 3 : 2 incorrectly, for example using GE / DE instead of DG / DE, or to assume that HF corresponds directly to GH rather than to part of DF. Another pitfall is forgetting that the part of DF in the smaller triangle is DH, not HF. Careful labeling of the diagram and clear use of corresponding sides in similar triangles helps to avoid these mistakes. Final Answer: The length of DF is 20 cm.

Question 12

The total surface area of a hemisphere is 462 square centimetres. Using π = 22/7, what is the curved surface area of this hemisphere in square centimetres?

Accepted Answer

Introduction / Context: This question tests understanding of the relationship between total surface area and curved surface area for a hemisphere. Such problems are very common in mensuration topics in aptitude and school examinations. Given Data / Assumptions: - The solid is a hemisphere. - Total surface area (TSA) = 462 sq cm. - π = 22/7 is to be used. - We need the curved surface area (CSA) of the hemisphere. Concept / Approach: For a hemisphere of radius r, the following formulas hold: - Curved surface area (CSA) = 2 * π * r^2. - Total surface area (TSA) = curved surface area + area of the circular base = 2 * π * r^2 + π * r^2 = 3 * π * r^2. Given TSA, we can first find r by solving 3 * π * r^2 = 462, then compute CSA = 2 * π * r^2. Step-by-Step Solution: Step 1: Use the TSA formula: 3 * π * r^2 = 462. Step 2: Substitute π = 22/7 to get 3 * (22/7) * r^2 = 462. Step 3: Simplify the constant factor: 3 * 22 / 7 = 66 / 7. Step 4: The equation becomes (66 / 7) * r^2 = 462. Step 5: Multiply both sides by 7: 66 * r^2 = 462 * 7 = 3234. Step 6: Divide by 66: r^2 = 3234 / 66 = 49. Step 7: Therefore r = 7 cm. Step 8: Now compute CSA = 2 * π * r^2 = 2 * (22/7) * 7^2. Step 9: Evaluate: 7^2 = 49, so CSA = 2 * (22/7) * 49 = 2 * 22 * 7 = 308 sq cm. Verification / Alternative Check: We can verify by recomputing TSA using r = 7 cm: TSA = 3 * π * r^2 = 3 * (22/7) * 49 = 3 * 22 * 7 = 462 sq cm, which matches the given total surface area. Hence the radius and curved surface area are consistent with the original information. Why Other Options Are Wrong: Option a, 616 sq cms, is equal to 4 * π * r^2 for r = 7 and would correspond to the surface area of a full sphere, not a hemisphere. Option b, 154 sq cms, is only half of the base area when r = 7 and has no direct meaning in this context. Option d, 462 sq cms, is the total surface area, not just the curved part, so it includes the circular base as well. Common Pitfalls: Students sometimes confuse TSA and CSA for hemispheres and may use TSA = 2 * π * r^2 instead of 3 * π * r^2. Others attempt to guess the radius instead of solving the equation systematically. Remembering that for hemispheres TSA = 3 * π * r^2 and CSA = 2 * π * r^2 helps to quickly set up the correct equations. Final Answer: The curved surface area of the hemisphere is 308 sq cms.

Question 13

Triangles ABC and DEF are similar. If side AB is 10 cm and the corresponding side DE is 6 cm, what is the ratio of the perimeter of triangle ABC to the perimeter of triangle DEF?

Accepted Answer

Introduction / Context: This question tests the basic property of similar triangles that the ratio of their perimeters is the same as the ratio of any pair of corresponding sides. It is a common and direct result used in many geometry and mensuration problems. Given Data / Assumptions: - Triangle ABC is similar to triangle DEF. - Side AB in triangle ABC corresponds to side DE in triangle DEF. - AB = 10 cm. - DE = 6 cm. - We are asked to find the ratio of Perimeter(ABC) to Perimeter(DEF). Concept / Approach: For two similar triangles, corresponding sides are in proportion, and all linear dimensions scale by the same factor. Therefore: Perimeter(ABC) / Perimeter(DEF) = AB / DE = BC / EF = AC / DF. This means the perimeter ratio can be found directly from any pair of corresponding sides. We just compute AB / DE and express it as a simplified ratio. Step-by-Step Solution: Step 1: Note that AB corresponds to DE and AB = 10 cm, DE = 6 cm. Step 2: Compute the ratio AB / DE = 10 / 6. Step 3: Simplify the fraction 10 / 6 by dividing numerator and denominator by 2: 10 / 6 = 5 / 3. Step 4: Because the triangles are similar, the perimeter ratio must equal this side ratio. Step 5: Therefore Perimeter(ABC) : Perimeter(DEF) = 5 : 3. Verification / Alternative Check: If we imagine that each side of triangle ABC is scaled down by a factor of 3 / 5, we would obtain triangle DEF. That means every linear measure, including the perimeter, is multiplied by 3 / 5 when going from ABC to DEF, so the reverse ratio from ABC to DEF is 5 : 3. This confirms that the ratio 5 : 3 is correct and consistent with the idea of similarity. Why Other Options Are Wrong: Option b, 3 : 5, is the inverse ratio and would apply if DE were 10 cm and AB were 6 cm, not the other way around. Option c, 25 : 9, is the square of the linear ratio 5 : 3 and would apply to area ratios rather than perimeter ratios. Option d, 9 : 25, is the inverse of the area ratio and does not match any direct linear measurement relationship between the triangles. Common Pitfalls: One common mistake is confusing perimeter ratios with area ratios and squaring the side ratio by habit. Another error is inverting the ratio by writing DE : AB instead of AB : DE. Carefully identifying which triangle is in the numerator and remembering that perimeters scale linearly while areas scale with the square of the scale factor helps to avoid these issues. Final Answer: The ratio of the perimeters Perimeter(ABC) : Perimeter(DEF) is 5:3.

Question 14

In triangle ABC, X and Y are the midpoints of sides AB and AC respectively. Segment XY is the line joining these midpoints and is parallel to side BC. If BC + XY = 12 units, what is the length of side BC (in units)?

Accepted Answer

Introduction / Context: This problem involves the mid segment of a triangle, which is a standard concept in geometry. The mid segment theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and has half its length. Here that property is combined with a simple linear equation involving BC and XY. Given Data / Assumptions: - Triangle ABC is given. - X is the midpoint of AB and Y is the midpoint of AC. - Segment XY joins these midpoints, so XY ∥ BC by the mid segment theorem. - It is given that BC + XY = 12 units. - We need to find the length of BC. Concept / Approach: By the mid segment theorem: XY = (1 / 2) * BC. The condition BC + XY = 12 gives a simple equation involving BC and XY. Substituting XY in terms of BC into this equation allows us to solve for BC directly. Once BC is known, XY can also be found if needed, but the question only asks for BC. Step-by-Step Solution: Step 1: By the midpoint or mid segment theorem, XY = (1 / 2) * BC. Step 2: Use the given relation BC + XY = 12. Step 3: Substitute XY from Step 1 into this relation: BC + (1 / 2) * BC = 12. Step 4: Combine like terms: BC + (1 / 2) * BC = (3 / 2) * BC. Step 5: So (3 / 2) * BC = 12. Step 6: Solve for BC by multiplying both sides by 2 / 3: BC = 12 * (2 / 3) = 24 / 3 = 8 units. Verification / Alternative Check: If BC = 8 units, then XY = (1 / 2) * 8 = 4 units. Now check the given condition: BC + XY = 8 + 4 = 12 units, which matches the information in the question. Hence BC = 8 units is consistent and correct. Why Other Options Are Wrong: Option a, 10 units, would give XY = 5 units and BC + XY = 10 + 5 = 15, not 12. Option c, 6 units, would give XY = 3 units, and BC + XY = 6 + 3 = 9, which again does not match 12. Option d, 4 units, would give XY = 2 units, and BC + XY = 4 + 2 = 6, still not equal to 12. Common Pitfalls: Some students mistakenly think the mid segment equals the third side rather than half of it, leading to BC = XY or BC − XY = 12 instead of BC + XY = 12. Others may forget that both BC and XY must be positive, which restricts possible solutions. Carefully applying the mid segment theorem and forming the correct equation avoids these issues. Final Answer: The length of side BC is 8 units.

Question 15

In an isosceles triangle ABC, sides AB and AC are equal and BC is the unequal base. AD is the median to side BC, meeting BC at D. Ray DP is the angle bisector of angle ∠ADB and PQ is drawn parallel to BC, meeting AC at Q. What is the measure of angl…

Accepted Answer

Introduction / Context: This is a more advanced geometry question involving an isosceles triangle, a median to the base, an angle bisector, and a line parallel to the base. It tests deep understanding of symmetry and angle relationships inside triangles, and is typical of higher level aptitude and Olympiad style geometry problems. Given Data / Assumptions: - Triangle ABC is isosceles with AB = AC and base BC unequal to these sides. - AD is the median to BC, so D is the midpoint of BC. - DP is the angle bisector of angle ∠ADB, so P lies on side AB and DP divides ∠ADB into two equal angles. - PQ is drawn parallel to BC and meets AC at Q. - We must find the measure of angle ∠PDQ in degrees. Concept / Approach: Because AB = AC and AD is the median to the base BC, in an isosceles triangle this median is also an altitude and an angle bisector of the vertex angle at A. Thus AD is a line of symmetry that splits the triangle into two congruent right triangles. The constructions involving the angle bisector at D and the line parallel to BC create a symmetric configuration around AD that leads to right angles at D inside the smaller quadrilateral formed by P, D, and Q. Step-by-Step Solution: Step 1: Note that in isosceles triangle ABC with AB = AC and median AD to base BC, AD is perpendicular to BC and also bisects angle ∠BAC. Step 2: Thus triangles ABD and ACD are congruent right triangles with AD as common altitude and BD = DC. Step 3: At point D on BC, consider angle ∠ADB formed by segments DA and DB. Ray DP is constructed as the angle bisector of ∠ADB, so it lies symmetrically between DA and DB. Step 4: Because of the overall symmetry of the figure, a corresponding construction on the other side with line through P parallel to BC meeting AC at Q ensures that D is the midpoint of segment PQ and that segment PQ is parallel to BC. Step 5: When this is worked out carefully using coordinate geometry or reflections, one finds that P and Q are symmetric with respect to the vertical line AD, and that the quadrilateral PDQ is arranged so that angle ∠PDQ is a right angle. Step 6: Detailed coordinate calculations (for example, by placing B at (−1, 0), C at (1, 0), and A at (0, h)) confirm that P and Q have equal y coordinates and opposite x coordinates, and D lies at the origin, making ∠PDQ = 90°. Verification / Alternative Check: Using a specific numerical example, choose coordinates B(−1, 0), C(1, 0), and A(0, 5). Then D is (0, 0) as the midpoint of BC. By constructing the angle bisector at D and the parallel through P to BC, one obtains P(−5/6, 5/6) and Q(5/6, 5/6). Vectors DP and DQ are then perpendicular, since their dot product is zero, proving that ∠PDQ = 90°. This coordinate verification is consistent across any choice of height h, so the right angle is a geometric consequence of the given configuration, not an artifact of a particular drawing. Why Other Options Are Wrong: Option a, 130°, and option d, 45°, do not match the orthogonal arrangement of DP and DQ that arises from the symmetry of the isosceles triangle and the parallel line construction. Option c, 180°, would imply that P, D, and Q are collinear, which contradicts the fact that PQ is parallel to BC and passes above the base, creating a genuine interior angle at D rather than a straight angle. Common Pitfalls: Because the configuration is complex, it is easy to misplace points P and Q or misinterpret which segments are parallel. Some students attempt to guess an angle based on rough sketches, which can be misleading. Others forget that in an isosceles triangle the median to the base is also an altitude and angle bisector, which is a powerful symmetry property. A careful construction or a coordinate geometry approach, treating AD as an axis of symmetry, provides a clear path to the correct right angle at D. Final Answer: The measure of angle ∠PDQ is 90°.

Question 16

An angle is smaller than its supplementary angle by 30°. What is the measure of this angle in degrees?

Accepted Answer

Introduction / Context: This problem checks understanding of supplementary angles and how to translate a verbal relationship into a simple algebraic equation. Supplementary angle problems are very common in basic geometry and aptitude tests. Given Data / Assumptions: - Two angles are supplementary if their sum is 180°. - One of these angles is smaller than its supplementary angle by 30°. - We are asked for the measure of this smaller angle. Concept / Approach: Let the smaller angle be x degrees. Its supplementary angle must then be 180° − x degrees. The statement says that the smaller angle is 30° less than the supplementary angle. That means: x = (180° − x) − 30°. We can solve this linear equation for x to find the required angle. Step-by-Step Solution: Step 1: Let the smaller angle be x°. Step 2: Its supplementary angle is 180° − x°. Step 3: According to the question, x° is 30° less than its supplementary angle, so write x = (180° − x) − 30°. Step 4: Simplify the right side: (180° − x) − 30° = 150° − x. Step 5: The equation is x = 150° − x. Step 6: Add x to both sides: x + x = 150°, so 2x = 150°. Step 7: Divide both sides by 2: x = 75°. Verification / Alternative Check: If the smaller angle is 75°, then its supplementary angle is 180° − 75° = 105°. The smaller angle is indeed 30° less than the supplementary angle since 105° − 75° = 30°. This confirms that 75° is correct. Why Other Options Are Wrong: Option b, 105°, is actually the larger supplementary angle, not the smaller one. Option c, 60°, would have a supplementary angle of 120°, giving a difference of 60°, not 30°. Option d, 30°, would result in a supplementary angle of 150°, with a difference of 120°, which does not match the given condition. Common Pitfalls: A common error is setting up the equation with the roles reversed or accidentally using 90° instead of 180° by confusing supplementary and complementary angles. Another pitfall is misinterpreting the phrase smaller than its supplementary angle by 30° and writing 180° − x = x − 30° instead of the correct relation. Carefully reading and translating the language into algebra ensures the correct equation. Final Answer: The measure of the smaller angle is 75°.

Question 17

A square is inscribed in a circle. If each side of the square is 14 cm, what is the area (in square centimetres) of the circle that just encloses this square?

Accepted Answer

Introduction / Context: The given problem connects the geometry of a square and a circle. A square is inscribed in a circle, which means that all four vertices of the square lie on the circle. The question asks for the area of the circle when the side length of the square is known. This tests understanding of the relationship between the diagonal of a square and the diameter of its circumcircle, as well as the basic circle area formula. Given Data / Assumptions: - The figure is a square inscribed in a circle.- Side of the square = 14 cm.- All vertices of the square lie on the circle.- The diagonal of the square is equal to the diameter of the circle.- We use the area formula for a circle: Area = π * r^2. Concept / Approach: The key concept is that for a square inscribed in a circle, the diagonal of the square becomes the diameter of the circle. Once the diameter is known, the radius is half of the diameter. Then, we use the standard area formula of a circle. We also use the Pythagorean relationship for the diagonal of a square: diagonal = side * √2. Substituting the given side into this relation gives us the diameter, from which we obtain the radius. Step-by-Step Solution: Step 1: Let the side of the square be s = 14 cm.Step 2: The diagonal of the square is given by diagonal = s * √2.Step 3: Substitute s = 14 cm to get diagonal = 14 * √2 cm.Step 4: Since the square is inscribed in the circle, the diagonal of the square equals the diameter of the circle. So diameter of the circle = 14 * √2 cm.Step 5: Radius r of the circle = (diameter) / 2 = (14 * √2) / 2 = 7 * √2 cm.Step 6: Area of the circle = π * r^2 = π * (7 * √2)^2.Step 7: Compute (7 * √2)^2 = 7^2 * (√2)^2 = 49 * 2 = 98.Step 8: Therefore, the area of the circle = 98π square centimetres. Verification / Alternative check: An alternative way is to note that the area of the circle will always be π times the square of half the diagonal of the square. Since the diagonal is 14√2, half of it is 7√2, which matches our radius. Squaring 7√2 again gives 98, so the area is 98π. The same result confirms that no computational mistake has been made. Why Other Options Are Wrong: - 49π: This would correspond to a radius of 7 cm instead of 7√2 cm, which ignores the √2 factor from the diagonal.- 77π: This does not match any correct radius obtained from the diagonal of the square and is not consistent with the geometry.- 121π: This would require r^2 = 121, that is r = 11, which is not related to the given side length 14 cm in this configuration.- 64π: This would correspond to r^2 = 64 and r = 8, again not matching the derived radius 7√2. Common Pitfalls: Students often mistakenly assume that the side of the square is the diameter of the circle instead of the diagonal. Another frequent error is forgetting to divide the diagonal by 2 to get the radius before applying the area formula. Some may also incorrectly square 7√2 and get 49√2 instead of 98, which arises from not squaring both the number and the square root properly. Careful handling of the relationship between side, diagonal, diameter, and radius avoids these issues. Final Answer: The area of the circle is 98π square centimetres.

Question 18

A wooden bowl is in the shape of a hollow hemisphere with an internal radius of 7 cm and a uniform thickness of 1 cm. Using π = 22/7, what is the total surface area (in square centimetres) of the bowl (inside, outside, and rim)?

Accepted Answer

Introduction / Context: This question deals with the surface area of a three dimensional solid, specifically a hollow hemisphere that is used as a wooden bowl. The bowl has an inner radius and a given thickness, so there is also an outer radius. The task is to find the total surface area, which includes the inner curved surface, the outer curved surface, and the circular rim at the open mouth of the bowl. Understanding the formulas for the surface areas of spheres and hemispheres is essential here. Given Data / Assumptions: - The bowl is a hollow hemisphere.- Internal radius r_in = 7 cm.- Thickness of the bowl = 1 cm.- Therefore, external radius r_out = 7 + 1 = 8 cm.- π is taken as 22/7.- Total surface area includes: inner curved surface area, outer curved surface area, and the area of the circular ring (rim) between the two radii. Concept / Approach: For a hemisphere, curved surface area = 2 * π * r^2. For a hollow hemisphere (bowl), there is an inner curved surface and an outer curved surface. In addition, the bowl has an exposed ring shaped surface at the mouth. The area of this ring is the difference between the areas of two circles: π * r_out^2 minus π * r_in^2. Therefore total surface area TSA is the sum of these three parts: 2 * π * r_in^2 + 2 * π * r_out^2 + π * (r_out^2 - r_in^2). Step-by-Step Solution: Step 1: Compute r_in^2 = 7^2 = 49.Step 2: Compute r_out^2 = 8^2 = 64.Step 3: Inner curved surface area = 2 * π * r_in^2 = 2 * π * 49.Step 4: Outer curved surface area = 2 * π * r_out^2 = 2 * π * 64.Step 5: Area of rim (ring) = π * (r_out^2 - r_in^2) = π * (64 - 49) = π * 15.Step 6: Total surface area TSA = 2 * π * 49 + 2 * π * 64 + π * 15.Step 7: Combine inside: 2 * 49 = 98, 2 * 64 = 128, so TSA = (98 + 128 + 15) * π = 241 * π.Step 8: Using π = 22/7, TSA = 241 * 22 / 7.Step 9: Calculate 241 * 22 = 5302. Then 5302 / 7 ≈ 757.43 square centimetres. Verification / Alternative check: We can check the reasonableness by comparing inner and outer curved surface areas. Inner curved area is 2 * π * 49, outer curved area is 2 * π * 64, which clearly makes the outer area larger. Adding a positive rim area increases the total further. Numerically these give approximately 2 * 3.14 * 49 ≈ 307.72 and 2 * 3.14 * 64 ≈ 401.92, and the rim is roughly 3.14 * 15 ≈ 47.1, which when summed is around 756.7. This is very close to 757.43 and confirms that the result is consistent. Why Other Options Are Wrong: - 710.29: This underestimates the total area and would correspond to omitting part of the surface, such as the rim or the difference between inner and outer areas.- 355.14: This is close to the order of a single curved surface only, not the total combined surfaces.- 908.91: This is too large, likely arising from double counting or using incorrect radii in the formula.- 531.42: This also underestimates the correct total and may result from including only inner or outer curved surface plus a small portion. Common Pitfalls: One common mistake is to consider only one curved surface area (either inner or outer) and ignore the other, which significantly changes the result. Another frequent error is neglecting the ring shaped rim at the open top, even though it is part of the exposed surface. Some students also confuse volume formulas with surface area formulas. Careful identification of every surface that is exposed to the air is crucial in such hollow solid problems. Final Answer: The total surface area of the wooden bowl is approximately 757.43 square centimetres.

Question 19

A chord of length 16 cm is drawn in a circle of radius 10 cm. What is the perpendicular distance (in centimetres) of this chord from the centre of the circle?

Accepted Answer

Introduction / Context: This problem examines the basic geometry of a circle involving chords and the perpendicular distance from the centre of the circle to a chord. When a perpendicular is drawn from the centre of a circle to a chord, it always bisects the chord. Using this fact along with the Pythagorean theorem allows us to compute the distance from the centre to the chord when the radius and the chord length are known. Given Data / Assumptions: - Radius of the circle r = 10 cm.- Length of the chord = 16 cm.- A perpendicular is drawn from the centre of the circle to the chord.- The perpendicular from the centre to the chord bisects the chord.- The triangle formed is a right angled triangle, so the Pythagorean theorem can be used. Concept / Approach: For a circle, the perpendicular from the centre to a chord divides the chord into two equal segments. This creates a right angled triangle whose hypotenuse is the radius of the circle and whose one leg is half the chord length. The other leg is the perpendicular distance from the centre to the chord, which we need to find. By applying the Pythagorean theorem (hypotenuse^2 = base^2 + height^2), we can solve for this distance. Step-by-Step Solution: Step 1: Let O be the centre of the circle and AB be the chord of length 16 cm.Step 2: Draw a perpendicular from O to AB, meeting it at point M. Then OM is the distance from the centre to the chord.Step 3: Since OM is perpendicular from the centre to the chord, it bisects AB. Therefore, AM = MB = 16 / 2 = 8 cm.Step 4: Consider right triangle OMA, where OA is the radius of the circle.Step 5: OA = r = 10 cm, AM = 8 cm, and OM is the unknown distance.Step 6: Apply Pythagorean theorem: OA^2 = OM^2 + AM^2.Step 7: Substitute values: 10^2 = OM^2 + 8^2.Step 8: This gives 100 = OM^2 + 64.Step 9: Rearranging, OM^2 = 100 - 64 = 36.Step 10: Therefore, OM = √36 = 6 cm. Verification / Alternative check: We can verify by checking that the values satisfy the Pythagorean relation. Using OM = 6 cm and AM = 8 cm, compute OM^2 + AM^2 = 36 + 64 = 100, which equals OA^2 = 10^2. This confirms that the triangle is consistent and that the distance 6 cm is correct. Any other distance value would break this equality and therefore would not fit the geometry of the circle and chord as given. Why Other Options Are Wrong: - 8 cm: This would suggest that AM is 8 and OM is also 8, leading to OA^2 = 8^2 + 8^2 = 128, which does not equal 10^2.- 4 cm: Using 4 cm would give OA^2 = 4^2 + 8^2 = 16 + 64 = 80, which is not 100.- 12 cm: This would give OA^2 = 12^2 + 8^2 = 144 + 64 = 208, far larger than 100.- 5 cm: This gives OA^2 = 5^2 + 8^2 = 25 + 64 = 89, which again does not match the radius squared, 100. Common Pitfalls: Many students forget that the perpendicular from the centre to a chord bisects the chord, and they mistakenly use the full chord length in the Pythagorean theorem. Others may confuse the positions of the sides in the right triangle and incorrectly assign the radius to a smaller side instead of the hypotenuse. Carefully drawing the diagram and marking all given lengths helps avoid these errors and leads to the correct calculation of the perpendicular distance. Final Answer: The perpendicular distance of the chord from the centre of the circle is 6 cm.

Question 20

In a circle, one angle subtended by a diameter at any point on the semicircle is asked. What is the measure (in degrees) of any angle in a semicircle?

Accepted Answer

Introduction / Context: This problem tests a fundamental property of circles related to angles formed by diameters. When an angle is subtended by a diameter of a circle at a point on the circle, that angle always lies in a semicircle. The question asks for the measure of such an angle, often referred to as the angle in a semicircle. This is a standard result in circle geometry and is frequently used in many competitive examinations and school level problems. Given Data / Assumptions: - We have a circle with a diameter drawn.- A point is chosen on the circumference forming a triangle with the endpoints of the diameter.- The angle at this point on the semicircle is called the angle in a semicircle.- The circle is a standard Euclidean circle with usual geometric properties. Concept / Approach: The important theorem here is: “The angle in a semicircle is a right angle.” In other words, if a triangle is drawn in a circle such that one side of the triangle is the diameter of the circle, then the angle opposite this diameter is always 90 degrees. This result comes from the properties of cyclic quadrilaterals or can be derived using the concept of central angles and inscribed angles. The measure of an inscribed angle is half the measure of the central angle subtending the same arc. Step-by-Step Solution: Step 1: Consider a circle with centre O and diameter AB.Step 2: Take a point C on the circle such that A, B, and C form a triangle ABC.Step 3: Since AB is a diameter, the central angle AOB subtending arc ACB is 180 degrees (a straight line).Step 4: The angle at the circumference subtending the same arc ACB is angle ACB, which is the angle in the semicircle.Step 5: By the inscribed angle theorem, the measure of an inscribed angle is half of the corresponding central angle.Step 6: Therefore, angle ACB = 180 degrees / 2 = 90 degrees.Step 7: Hence, any angle in a semicircle is a right angle. Verification / Alternative check: An alternative view is to think of triangle ABC as a right angled triangle with AB as the hypotenuse. This corresponds with the converse of the theorem stating that the locus of all points forming a right angle with a given segment as hypotenuse is a circle with that segment as diameter. Both results point to the same conclusion: the angle in a semicircle is always 90 degrees, and this is independent of the position of the third point C on the semicircle. Why Other Options Are Wrong: - 45°: This angle is too small and does not satisfy the semicircle property for all such points.- 60°: This angle would correspond to an equilateral triangle under special conditions, which is not guaranteed here.- 120°: This is an obtuse angle and cannot be the angle in a semicircle according to the established theorem.- 150°: This is far too large and does not correspond to the fixed relationship between the diameter and inscribed angle. Common Pitfalls: Some learners confuse angles in semicircles with other special angles like 60 degrees or 45 degrees which appear in equilateral or isosceles right triangles. Another common mistake is to think the angle value may vary depending on the position of the third point on the semicircle, but the theorem clearly states that it is always a right angle. Remembering that a diameter subtending an angle at the circle always produces a right angle helps avoid such confusion. Final Answer: The measure of any angle in a semicircle is 90°.

Volume and Surface Area Questions