Question 1

In triangle ABC, points D and E lie on sides AB and AC respectively, and DE is parallel to BC. If AD : DB = 1 : 2 and the area of triangle ABC is 45 square cm, what is the area (in square cm) of quadrilateral BDEC?

Accepted Answer

Introduction / Context: This question again uses similar triangles formed when a line segment is drawn parallel to one side of a triangle. It specifically asks for the area of the quadrilateral that lies between the smaller similar triangle at the top and the original base. Understanding how areas scale with side ratios is key here. Given Data / Assumptions: Triangle ABC has D on AB and E on AC. DE is parallel to BC. AD : DB = 1 : 2, so D divides AB internally in this ratio. Area of triangle ABC is 45 square cm. We must find the area of quadrilateral BDEC. Concept / Approach: Since DE ∥ BC, triangle ADE is similar to triangle ABC. If the ratio of corresponding sides AD : AB is known, then the ratio of their areas is the square of that side ratio. Once we know the area of the smaller triangle ADE, we subtract it from the area of ABC to get the area of quadrilateral BDEC, which is the remaining region between DE and BC. Step-by-Step Solution: Step 1: From AD : DB = 1 : 2, we have AB = AD + DB = 1 + 2 = 3 parts. Step 2: Therefore, AD / AB = 1 / 3. Step 3: Because DE ∥ BC, triangles ADE and ABC are similar. Step 4: The ratio of their areas equals the square of the ratio of corresponding sides: area(ADE) / area(ABC) = (AD / AB)^2 = (1 / 3)^2 = 1 / 9. Step 5: Given area(ABC) = 45 square cm, we have area(ADE) = (1 / 9) * 45 = 5 square cm. Step 6: The area of quadrilateral BDEC is the remaining area after removing triangle ADE from the whole triangle ABC. Step 7: So area(BDEC) = area(ABC) − area(ADE) = 45 − 5 = 40 square cm. Verification / Alternative check: We can also reason with fractions of area. Triangle ADE occupies 1 / 9 of the area of ABC, so the remaining portion, which is quadrilateral BDEC, must occupy 8 / 9 of the area. Then 8 / 9 of 45 is (8 * 45) / 9 = 8 * 5 = 40. This confirms the computed value of 40 square cm and shows the consistency of the approach. Why Other Options Are Wrong: 20 and 15 square cm are too small; they would correspond to a smaller fraction of the original triangle than the 8 / 9 that the quadrilateral actually represents. 30 square cm would correspond to 2 / 3 of the total area, which does not match the correct similarity ratio. 25 square cm does not arise from any simple fraction of 45 given the 1 : 2 division of AB, so it cannot be correct. Common Pitfalls: Sometimes students mistakenly use the ratio 1 : 2 directly for area instead of converting it to AD : AB and then squaring. Others forget that AB is the sum of AD and DB, leading to a wrong side ratio. Another error is to assume that the quadrilateral has the same area as triangle ADE or as triangle ABC, which is clearly not supported by the similarity relations. Always carefully compute the side ratio with respect to the entire side AB and then square it to get the area ratio. Final Answer: The area of quadrilateral BDEC is 40 square cm.

Question 2

In right angled triangle XYZ, angle Y is 90° and angle Z is 30°. If the hypotenuse ZX is 9 cm, what is the length (in cm) of side YZ?

Accepted Answer

Introduction / Context: This trigonometry question uses special angle values in a right angled triangle. When one acute angle is 30° and the other is 60°, the triangle is a 30 60 90 special triangle with fixed ratios between its sides. Knowing these ratios allows you to find any side given one of them, especially the hypotenuse. Given Data / Assumptions: Triangle XYZ is right angled at Y, so angle Y = 90°. Angle Z = 30°, so angle X = 60° (because the sum of angles in a triangle is 180°). The hypotenuse is ZX, and ZX = 9 cm. We must find the length of side YZ. YZ is a leg of the triangle adjacent to angle Z and opposite angle X. Concept / Approach: In a 30 60 90 triangle, the side opposite 30° is half the hypotenuse, the side opposite 60° is (√3 / 2) times the hypotenuse, and the hypotenuse is the longest side. Here, the hypotenuse is ZX and angle Z is 30°, so: Side opposite 30° (which is YX) = (1 / 2) * hypotenuse. Side opposite 60° (which is YZ) = (√3 / 2) * hypotenuse. We need the side opposite 60°, so we multiply the hypotenuse by √3 / 2. Step-by-Step Solution: Step 1: Recognize that the triangle is a 30 60 90 special triangle because one angle is 30° and the right angle is 90°. Step 2: Identify side roles: ZX is the hypotenuse opposite the right angle at Y. Step 3: Side YX is opposite angle Z = 30° and side YZ is opposite angle X = 60°. Step 4: In a 30 60 90 triangle, side opposite 60° = (√3 / 2) * hypotenuse. Step 5: Substitute hypotenuse ZX = 9 cm: YZ = (√3 / 2) * 9 = 9√3 / 2 cm. Step 6: So the length of YZ is 9√3 / 2 cm. Verification / Alternative check: We can also find the other leg YX. Side opposite 30° is half the hypotenuse, so YX = (1 / 2) * 9 = 4.5 cm. Now check Pythagoras theorem: (YX)^2 + (YZ)^2 should equal hypotenuse^2. Compute YX^2 = 4.5^2 = 20.25 and YZ^2 = (9√3 / 2)^2 = 81 * 3 / 4 = 243 / 4 = 60.75. Sum: 20.25 + 60.75 = 81, which equals 9^2 = 81, confirming the correctness of the side lengths. Why Other Options Are Wrong: 6√3 and 3√3 correspond to incorrect multiples of the hypotenuse and do not respect the special 30 60 90 triangle ratios. 18 cm would be twice the hypotenuse and cannot be a leg in a right triangle. 4.5 cm is the side opposite 30°, not the side opposite 60°, so it is the other leg, not YZ. Common Pitfalls: A frequent mistake is to mix up which side corresponds to which angle in the 30 60 90 triangle. Some students incorrectly take the side opposite 30° as (√3 / 2) times the hypotenuse and the side opposite 60° as half, which is backwards. Always remember: the smaller angle (30°) faces the smaller leg (half the hypotenuse), and the larger acute angle (60°) faces the longer leg (√3 / 2 times the hypotenuse). Final Answer: The length of YZ is 9√3 / 2 cm.

Question 3

Triangle XYZ is similar to triangle PQR. If the ratio of corresponding sides XY : PQ is 5 : 1 and the area of triangle PQR is 5 square cm, what is the area (in square cm) of triangle XYZ?

Accepted Answer

Introduction / Context: This geometry question focuses on the relationship between the side lengths and the areas of similar triangles. When two triangles are similar, the ratio of their areas is the square of the ratio of any pair of corresponding sides. This principle appears frequently in aptitude problems and helps solve area questions quickly. Given Data / Assumptions: Triangle XYZ is similar to triangle PQR. The ratio of corresponding sides XY : PQ is 5 : 1. The area of triangle PQR is 5 square cm. We need to find the area of triangle XYZ. Similarity implies all corresponding side ratios are 5 : 1. Concept / Approach: For two similar triangles, if the ratio of corresponding side lengths is k, then the ratio of their areas is k^2. Here, the scale factor from triangle PQR to triangle XYZ is 5 (since XY is five times PQ). Therefore, the area of triangle XYZ is 5^2 times the area of triangle PQR. This uses the fact that area scales with the square of linear dimensions. Step-by-Step Solution: Step 1: Note that XY : PQ = 5 : 1. Step 2: Let k be the linear scale factor from PQR to XYZ, so k = 5. Step 3: For similar triangles, area(XYZ) / area(PQR) = k^2 = 5^2 = 25. Step 4: Given area(PQR) = 5 square cm, compute area(XYZ) = 25 * 5 = 125 square cm. Step 5: Therefore, the area of triangle XYZ is 125 square cm. Verification / Alternative check: Imagine scaling triangle PQR by a factor of 5 in all directions. Every side becomes five times longer, so any base or height pair used to compute the area is multiplied by 5. Since area is roughly base * height / 2, both base and height are scaled, leading to a factor of 5 * 5 = 25 in area. This intuitive picture matches the algebraic rule that area scales with the square of the scale factor and confirms the answer of 125 square cm. Why Other Options Are Wrong: 120 and 100 square cm are close but not equal to 25 times 5; they may come from miscalculating 5^2 or the multiplication with area(PQR). 64 square cm corresponds to a scale factor of 8 between sides, which contradicts the given ratio 5 : 1. 75 square cm corresponds to an area ratio of 15, which is not a square number and therefore cannot arise from a consistent side ratio of 5 : 1 between similar triangles. Common Pitfalls: A common mistake is to multiply the area by the side ratio directly (using 5 instead of 5^2), which would give 25 square cm rather than 125. Another error is to misinterpret the ratio and treat it as PQ : XY rather than XY : PQ, which reverses the scaling direction. Always confirm which triangle is larger and square the correct linear scale factor when dealing with areas of similar figures. Final Answer: The area of triangle XYZ is 125 square cm.

Question 4

In a circle, the length of a chord is equal to the radius of the circle. What is the angle, in radians, subtended at the centre of the circle by this chord?

Accepted Answer

Introduction / Context: This question uses the relationship between a chord length, the radius of a circle and the central angle that subtends the chord. It is a standard trigonometric application in circle geometry and often appears in questions involving radians and arc lengths. Given Data / Assumptions: A circle has radius R. A chord of the circle has length equal to R. The chord subtends some central angle θ at the centre of the circle. We are asked to find θ in radians. We can use the formula for chord length in terms of radius and central angle. Concept / Approach: For a circle of radius R, the length L of a chord that subtends a central angle θ (in radians) is: L = 2R * sin(θ / 2) Here, we are given that L = R. Substituting and simplifying gives an equation involving sin(θ / 2), which we can solve to find θ. Since we want the principal (smaller) central angle associated with that chord, we will choose the acute value that fits the geometry of the circle. Step-by-Step Solution: Step 1: Use the chord length formula L = 2R * sin(θ / 2). Step 2: Given L = R, set R = 2R * sin(θ / 2). Step 3: Divide both sides by R (R is non zero): 1 = 2 * sin(θ / 2). Step 4: Solve for sin(θ / 2): sin(θ / 2) = 1 / 2. Step 5: The principal angle whose sine is 1 / 2 is π / 6 (30°). Step 6: Therefore θ / 2 = π / 6, so θ = 2 * (π / 6) = π / 3. Step 7: Thus, the central angle subtended by the chord is π / 3 radians. Verification / Alternative check: You can cross check by substituting θ = π / 3 back into the chord formula. Then θ / 2 = π / 6, so sin(θ / 2) = sin(π / 6) = 1 / 2. The chord length becomes L = 2R * (1 / 2) = R, which matches the given condition. This confirms that π / 3 is the correct central angle for a chord whose length equals the radius. Why Other Options Are Wrong: 1 radian is approximately 57.3°, and substituting θ = 1 into the formula does not give sin(θ / 2) = 1 / 2. π / 2 corresponds to 90°, which would give a chord longer than the radius (it would be √2 times the radius). π / 4 (45°) gives sin(π / 8), which is not 1 / 2, so that chord is shorter than R. 2π / 3 (120°) would correspond to sin(60°) = √3 / 2, giving a chord longer than the radius. Only π / 3 gives exactly sin(θ / 2) = 1 / 2 and hence chord length equal to the radius. Common Pitfalls: Some students mistakenly use the arc length formula s = Rθ instead of the chord length formula, which relates to the curved distance along the circle, not the straight line segment. Another pitfall is forgetting to divide the angle by 2 when using sin(θ / 2) in the chord formula. Carefully distinguishing between arc and chord and remembering the correct half angle relationship is crucial for solving this type of problem correctly. Final Answer: The central angle subtended by the chord is π / 3 radians.

Question 5

In triangle ABC, angle B is a right angle (∠B = 90°). Point D is the midpoint of hypotenuse AC. If AB = 6 cm and BC = 8 cm, what is the length (in cm) of segment BD?

Accepted Answer

Introduction / Context: This problem uses a beautiful property of right angled triangles involving the midpoint of the hypotenuse. It tests your understanding that in a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices. This geometric fact allows you to find distances quickly without heavy calculations. Given Data / Assumptions: Triangle ABC is right angled at B (∠B = 90°). AB = 6 cm, BC = 8 cm. AC is the hypotenuse. D is the midpoint of AC. We must find the length BD. Concept / Approach: In any right angled triangle, the midpoint of the hypotenuse is equidistant from all three vertices. That is: DA = DB = DC = (hypotenuse) / 2 So, if we first compute the hypotenuse AC using the Pythagoras theorem, then BD is simply half of AC. This property can be remembered as: the midpoint of the hypotenuse is the circumcentre of the right triangle, and the circumradius equals half the hypotenuse. Step-by-Step Solution: Step 1: Use the Pythagoras theorem in triangle ABC to find AC. Step 2: AC^2 = AB^2 + BC^2 = 6^2 + 8^2 = 36 + 64 = 100. Step 3: Therefore, AC = √100 = 10 cm. Step 4: D is the midpoint of AC, so AD = DC = AC / 2 = 10 / 2 = 5 cm. Step 5: By the right triangle midpoint property, BD = AD = DC = 5 cm. Step 6: Hence, the required length BD is 5 cm. Verification / Alternative check: We can verify using coordinates. Place B at (0, 0), A at (6, 0) and C at (0, 8). Then AC is the segment from (6, 0) to (0, 8). Its midpoint D is ((6 + 0) / 2, (0 + 8) / 2) = (3, 4). The distance from B to D is √(3^2 + 4^2) = √(9 + 16) = √25 = 5 cm, which matches the result BD = 5 cm and confirms the geometric property. Why Other Options Are Wrong: 4 cm and 8 cm do not equal half the hypotenuse and do not match the circumradius of the right triangle. 12 cm is larger than the hypotenuse itself, which is 10 cm, so it cannot be a distance from a vertex to a point on the hypotenuse. 10 cm is the length of the hypotenuse AC, not the distance from B to the midpoint of AC. Thus none of these values are consistent with the geometry of a right triangle and the midpoint property. Common Pitfalls: A frequent mistake is to treat BD as the median and try to apply the median length formula inconsistently, instead of recalling the special right triangle property. Another error is computing AC correctly but then forgetting that BD equals half of AC in this specific case. Remembering that the midpoint of the hypotenuse is the centre of the circumscribed circle around the right triangle helps to quickly recall that all three vertices are the same distance from this point. Final Answer: The length of BD is 5 cm.

Question 6

AB is a chord of a circle with centre O. ON is drawn perpendicular from O to chord AB. If AB = 20 cm and ON = 2√11 cm, what is the radius (in cm) of the circle?

Accepted Answer

Introduction / Context: This geometry question uses the relationship between a chord, the perpendicular distance from the centre to the chord and the radius of a circle. It is a direct application of the Pythagoras theorem to a right triangle formed by the radius, the perpendicular and half of the chord. Given Data / Assumptions: Circle has centre O. AB is a chord of the circle with length AB = 20 cm. ON is perpendicular from O to AB, so ON ⟂ AB. ON = 2√11 cm. We must find the radius r = OA of the circle. Because ON is perpendicular to AB, it bisects the chord AB at N. Concept / Approach: When a perpendicular is drawn from the centre of a circle to a chord, it bisects the chord. Hence, AN = NB = AB / 2. Considering triangle ONA, we get a right triangle with hypotenuse OA (the radius), one leg ON and the other leg AN. Applying the Pythagoras theorem gives: OA^2 = ON^2 + AN^2 We can compute both ON^2 and AN^2 from the given information and then take the square root to find OA. Step-by-Step Solution: Step 1: Since ON ⟂ AB, chord AB is bisected at N. Step 2: Therefore, AN = NB = AB / 2 = 20 / 2 = 10 cm. Step 3: Consider right triangle ONA with right angle at N. Step 4: ON = 2√11 cm and AN = 10 cm. Step 5: Apply Pythagoras theorem: OA^2 = ON^2 + AN^2. Step 6: Compute ON^2: (2√11)^2 = 4 * 11 = 44. Step 7: Compute AN^2: 10^2 = 100. Step 8: So OA^2 = 44 + 100 = 144. Step 9: Taking the square root, OA = √144 = 12 cm. Verification / Alternative check: We can confirm the value by checking triangle ONA numerically. If OA = 12, ON = 2√11 ≈ 6.64 and AN = 10, then ON^2 + AN^2 ≈ 44 + 100 = 144, while OA^2 = 144. The equality holds, confirming that OA is indeed 12 cm. This is consistent with the geometry of a circle and a chord at a given distance from the centre. Why Other Options Are Wrong: 10 cm would make the radius equal to half the chord, which is only possible if the chord is a diameter, in which case ON would be zero. 13 cm and 15 cm produce OA^2 values larger than 144 and do not satisfy OA^2 = ON^2 + AN^2 with the given lengths. 9 cm is smaller than half the chord length, which is impossible because the radius must be at least as long as the distance from the centre to any point on the chord plus the perpendicular distance from the centre to the chord does not exceed the radius. Common Pitfalls: A typical mistake is to forget that the perpendicular from the centre to a chord bisects the chord, leading to an incorrect value of AN. Another error is miscomputing (2√11)^2 or mishandling the addition of squares. Carefully identifying the right triangle and keeping the Pythagoras theorem in mind helps avoid these problems and makes such questions straightforward. Final Answer: The radius of the circle is 12 cm.

Question 7

The largest chord of a circle has length 20.4 cm. What is the radius of the circle?

Accepted Answer

Introduction / Context: This question relies on an important and very simple property of circles: the largest possible chord in any circle is its diameter. Once you recognize that, finding the radius from the length of the largest chord becomes an easy one step calculation. Given Data / Assumptions: The largest chord of the circle has length 20.4 cm. The largest chord in a circle is always the diameter. We must find the radius of the circle. Radius is half of the diameter. Concept / Approach: Any chord of a circle is a straight line segment joining two points on the circle. The chord that passes through the centre is called the diameter and is the longest chord. No other chord can be longer than the diameter. Therefore, if the question explicitly tells us that the largest chord has a certain length, that length must equal the diameter. Once we identify this, we simply divide by 2 to obtain the radius. Step-by-Step Solution: Step 1: Recognize that the largest chord in a circle is the diameter. Step 2: Therefore, the given length 20.4 cm is the diameter of the circle. Step 3: The radius r is half the diameter: r = diameter / 2. Step 4: Compute r = 20.4 / 2 = 10.2 cm. Step 5: Hence, the radius of the circle is 10.2 cm. Verification / Alternative check: Consider any circle with radius r. The diameter is 2r, and any other chord lies inside the circle, forming a right triangle when you draw a radius to its midpoint. This right triangle geometry ensures that all other chords are shorter than 2r. Therefore, if 20.4 cm is indeed the largest chord, it must be the diameter, and the radius must be exactly 10.2 cm, with no ambiguity. Why Other Options Are Wrong: If the radius were greater than 10.2 cm, then the diameter would be greater than 20.4 cm, which contradicts the statement that 20.4 cm is the largest chord. If the radius were less than 10.2 cm, the diameter would be shorter than 20.4 cm, meaning a chord of length 20.4 cm could not even exist in that circle. Saying the radius is greater than or equal to 10.2 cm or that it cannot be determined conflicts with the unique relationship between the largest chord and the diameter in a circle. Common Pitfalls: One mistake is to overthink the problem and look for complicated geometry when a simple property suffices. Another is to confuse the largest chord with a random chord and think there might be many possibilities for the radius. Remembering that only the diameter is the maximum chord length allows you to immediately convert chord length to radius by halving it. Final Answer: The radius of the circle is 10.2 cm.

Question 8

The area of a semicircle is 1925 cm². What is the perimeter (in cm) of this semicircle, including its diameter?

Accepted Answer

Introduction / Context: This problem combines the area formula and perimeter formula for a semicircle. It tests your ability to work backwards from area to radius and then use the radius to compute the total perimeter, which includes both the curved arc and the straight diameter. Given Data / Assumptions: The figure is a semicircle. Its area is 1925 cm². We are asked to find the perimeter (curved part + diameter) in cm. We will use π = 22 / 7 for calculations unless otherwise specified. Area of a full circle is πr²; area of a semicircle is (1 / 2) * πr². Perimeter of a semicircle is πr + 2r (half the circumference plus the diameter). Concept / Approach: First, we use the area of the semicircle to find the radius. The formula is: (1 / 2) * π * r² = 1925 We solve this equation for r. Once r is known, we compute the perimeter using: Perimeter = π * r + 2 * r This includes both the curved arc and the straight edge of the semicircle. Step-by-Step Solution: Step 1: Write the area equation for the semicircle: (1 / 2) * π * r² = 1925. Step 2: Substitute π = 22 / 7 to get (1 / 2) * (22 / 7) * r² = 1925. Step 3: Simplify (1 / 2) * (22 / 7) = 11 / 7, so (11 / 7) * r² = 1925. Step 4: Solve for r²: r² = 1925 * 7 / 11. Step 5: Compute 1925 / 11 = 175, so r² = 175 * 7 = 1225. Step 6: Therefore, r = √1225 = 35 cm. Step 7: Now compute the perimeter of the semicircle: Perimeter = πr + 2r. Step 8: Using π = 22 / 7, πr = (22 / 7) * 35 = 22 * 5 = 110 cm. Step 9: The diameter is 2r = 2 * 35 = 70 cm. Step 10: Total perimeter = 110 + 70 = 180 cm. Verification / Alternative check: We can double check by recomputing the area with r = 35 cm. Area of full circle is πr² = (22 / 7) * 35² = (22 / 7) * 1225 = 22 * 175 = 3850 cm². Half of this is 3850 / 2 = 1925 cm², which matches the given area. This confirms that r = 35 cm and thus perimeter 180 cm is consistent and correct. Why Other Options Are Wrong: 80 cm and 160 cm are significantly smaller than the correct perimeter and would arise from incorrect radius calculations or from omitting either the arc or the diameter. 360 cm is about double the correct value and could come from mistakenly using the full circumference 2πr plus 2r. 140 cm corresponds to πr alone for r = 20, which does not satisfy the area condition of 1925 cm². None of these match the consistent radius and area relationship found above. Common Pitfalls: Students sometimes forget to include the diameter when calculating the perimeter of a semicircle and only compute πr. Another common mistake is misusing the area formula, for example by setting πr² = 1925 instead of (1 / 2) * πr². Careful attention to whether the figure is a full circle or a semicircle and using both arc and diameter in perimeter calculations will help avoid these errors. Final Answer: The perimeter of the semicircle is 180 cm.

Question 9

In equilateral triangle ABC, side AB = 12 cm. O is the centroid of the triangle (intersection of medians). What is the length (in cm) of segment AO?

Accepted Answer

Introduction / Context: This problem involves an equilateral triangle and its centroid. In an equilateral triangle, several important line segments coincide: medians, angle bisectors, perpendicular bisectors and altitudes. The centroid is a key point used in geometry and physics (centre of mass), and its distance from a vertex can be expressed in terms of the side length. Given Data / Assumptions: Triangle ABC is equilateral. Side AB = 12 cm (so all sides are 12 cm). O is the centroid, the point where the three medians intersect. We must find the length AO. In any triangle, the centroid divides each median in the ratio 2 : 1 from the vertex. Concept / Approach: First, we find the length of a median in an equilateral triangle. The median is also the height. For an equilateral triangle of side s, the height h is: h = (√3 / 2) * s The centroid lies two thirds of the way from the vertex to the midpoint of the opposite side along the median. Therefore: AO = (2 / 3) * h We substitute s = 12 cm into these formulas to find AO. Step-by-Step Solution: Step 1: Let the side length s of the equilateral triangle be 12 cm. Step 2: Compute the height h of the equilateral triangle: h = (√3 / 2) * s. Step 3: Substitute s = 12 to get h = (√3 / 2) * 12 = 6√3 cm. Step 4: The centroid divides the median in the ratio 2 : 1 from the vertex, so AO = (2 / 3) * h. Step 5: Substitute h = 6√3: AO = (2 / 3) * 6√3 = 4√3 cm. Step 6: Therefore, AO = 4√3 cm. Verification / Alternative check: We can check the lengths of the segments on the median. If AO = 4√3 cm, then the remaining segment from O to the midpoint of BC is (1 / 3) * h = 2√3 cm. The total median length is AO + OM = 4√3 + 2√3 = 6√3, which matches the computed height. Also, using coordinate geometry with the triangle placed conveniently and computing distances confirms that AO equals 4√3 cm for side length 12 cm. Why Other Options Are Wrong: 2√3 cm is the distance from the centroid to the midpoint of a side (one third of the median), not from the vertex. 3√3 cm and 9√3 cm do not match the 2 : 1 division of the median and give an incorrect total height. 6√3 cm is the full median or height, not the vertex to centroid distance. Only 4√3 cm is exactly two thirds of the median length and fits the centroid division property. Common Pitfalls: Some students mistakenly think the centroid divides the median in half, rather than in the ratio 2 : 1. Others confuse the incenter or circumcenter with the centroid, even though in an equilateral triangle these points coincide but are approached through different formulas. Remember that in any triangle the centroid always splits each median into a longer segment (2 parts) from vertex to centroid and a shorter segment (1 part) from centroid to side midpoint. Final Answer: The length of AO is 4√3 cm.

Question 10

A rhombus has perimeter 34 cm and one of its diagonals has length 8 cm. What is the length (in cm) of the other diagonal?

Accepted Answer

Introduction / Context: This question is about a rhombus, a special quadrilateral in which all sides are equal. It uses the fact that the diagonals of a rhombus are perpendicular bisectors of each other. By combining this property with the given perimeter and one diagonal, you can find the length of the other diagonal using the Pythagoras theorem. Given Data / Assumptions: The figure is a rhombus. Perimeter = 34 cm. One diagonal (say d1) has length 8 cm. Let the other diagonal be d2 (unknown). All sides of the rhombus are equal; let each side be s. The diagonals intersect at right angles and bisect each other. Concept / Approach: Because the perimeter is 34 cm and the rhombus has four equal sides, each side length s is 34 / 4. The diagonals cross at the centre and are perpendicular, so each half diagonal forms the legs of a right triangle whose hypotenuse is a side of the rhombus. Specifically: s² = (d1 / 2)² + (d2 / 2)² We can substitute the known values for s and d1 and solve for d2. Step-by-Step Solution: Step 1: Compute the side length s from the perimeter: s = 34 / 4 = 8.5 cm. Step 2: The given diagonal d1 = 8 cm, so half of it is d1 / 2 = 4 cm. Step 3: Let the other diagonal be d2; then half of it is d2 / 2. Step 4: Use the right triangle formed by half diagonals and a side: s² = (d1 / 2)² + (d2 / 2)². Step 5: Substitute s = 8.5 and d1 / 2 = 4: (8.5)² = 4² + (d2 / 2)². Step 6: Compute (8.5)² = 72.25 and 4² = 16. Step 7: So 72.25 = 16 + (d2 / 2)², giving (d2 / 2)² = 72.25 − 16 = 56.25. Step 8: Take the square root: d2 / 2 = √56.25 = 7.5. Step 9: Therefore d2 = 2 * 7.5 = 15 cm. Verification / Alternative check: We can verify by recomputing the side length from the diagonals: using d1 = 8 and d2 = 15, half diagonals are 4 and 7.5. Then the side length s should be √(4² + 7.5²) = √(16 + 56.25) = √72.25 = 8.5 cm. Multiplying by 4 gives perimeter 4 * 8.5 = 34 cm, which matches the given perimeter. This consistency confirms that d2 = 15 cm is correct. Why Other Options Are Wrong: 7.5 cm is only half of the required diagonal. 30 cm and 22.5 cm would create sides that are much longer than 8.5 cm and lead to incorrect perimeter values. 8 cm is the length of the first diagonal and cannot be the second diagonal here without changing the given perimeter relationship and side length. None of these other values satisfy the right triangle relation between half diagonals and side length. Common Pitfalls: Some students mistakenly use the full diagonals directly in the Pythagoras relation without halving them, which leads to incorrect equations. Others forget that the diagonals of a rhombus are perpendicular, a property that is essential for forming the right triangle used in the calculation. Carefully halving the diagonals and using the side derived from the perimeter avoids these errors. Final Answer: The length of the other diagonal is 15 cm.

Question 11

Find the perimeter (in cm) of a semicircle whose radius is 14 cm. Include both the curved arc and the diameter.

Accepted Answer

Introduction / Context: This question asks for the perimeter of a semicircle when the radius is known. The perimeter of a semicircle consists of two parts: the curved half circumference and the straight diameter. Knowing how to combine these correctly is a basic but important skill in mensuration. Given Data / Assumptions: Radius r of the semicircle is 14 cm. We must find the total perimeter: curved arc length plus the diameter. We use π = 22 / 7 for calculations. Full circumference of a circle is 2πr; half of this gives the semicircular arc. Concept / Approach: The perimeter P of a semicircle is given by: P = (1 / 2) * (2πr) + 2r = πr + 2r The term πr is the length of the curved arc (half the full circumference) and 2r is the diameter. We simply substitute the given radius into this formula and compute the numerical value using the chosen approximation for π. Step-by-Step Solution: Step 1: Write the perimeter formula for a semicircle: P = πr + 2r. Step 2: Substitute r = 14 cm. Step 3: Compute πr with π = 22 / 7: πr = (22 / 7) * 14 = 22 * 2 = 44 cm. Step 4: Compute the diameter: 2r = 2 * 14 = 28 cm. Step 5: Add to get total perimeter: P = 44 + 28 = 72 cm. Step 6: Therefore, the perimeter of the semicircle is 72 cm. Verification / Alternative check: We can cross check by first computing the full circumference and then halving it. Full circumference is 2πr = 2 * (22 / 7) * 14 = 2 * 44 = 88 cm. Half of that is 44 cm, which matches the curved part we used above. Adding the diameter 28 cm again gives 44 + 28 = 72 cm, confirming the perimeter is correct. Why Other Options Are Wrong: 46 cm and 56 cm are too small and would result from forgetting part of the perimeter, such as using only the arc or only the diameter. 92 cm is larger than the full circumference of the circle and comes from adding 2πr and 2r rather than half the circumference. 144 cm is far too large and does not correspond to any simple misapplication of the formula with radius 14 cm, so it is inconsistent with the given dimensions. Common Pitfalls: A common error is to use only πr (curved part) and forget the diameter, or mistakenly use 2πr for the arc of the semicircle. Another pitfall is mixing units or approximations for π without tracking them carefully. Always remember that a semicircle's perimeter is πr plus 2r, not just the curved portion, and double check arithmetic when substituting the radius. Final Answer: The perimeter of the semicircle is 72 cm.

Question 12

The volume of a cube is 166.375 cm³. What is the length (in cm) of each side of the cube?

Accepted Answer

Introduction / Context: This mensuration question asks you to find the side length of a cube when its volume is given. Since all edges of a cube are equal, the volume formula is simple and involves the cube of the side length. This is a common type of problem in aptitude tests and basic geometry. Given Data / Assumptions: The solid is a cube. Volume of the cube is 166.375 cm³. Let the side length be s cm. The volume of a cube is given by V = s³. We must find s. Concept / Approach: We use the direct relationship between volume and side length for a cube: V = s³ So the side length s is the cube root of the volume: s = ∛V We either compute the cube root directly or test the provided options by cubing them to see which one gives the exact volume 166.375 cm³. Since the options are decimals with one decimal place, checking them is straightforward. Step-by-Step Solution: Step 1: Use the formula V = s³, where V = 166.375 cm³. Step 2: We need s such that s³ = 166.375. Step 3: Try s = 5.5 cm from the options. Step 4: Compute 5.5² = 30.25. Step 5: Then 5.5³ = 5.5 * 30.25 = 166.375. Step 6: Since this exactly matches the given volume, the side length s is 5.5 cm. Verification / Alternative check: We can quickly verify that no other option gives the correct volume. For example, 4.5³ = 91.125 cm³ (too small), 6.5³ = 274.625 cm³ (too large), 7.5³ = 421.875 cm³ (much too large) and 8.5³ = 614.125 cm³. Only 5.5³ equals 166.375 cm³. This confirms that 5.5 cm is the unique side length that matches the given volume. Why Other Options Are Wrong: Any side length smaller than 5.5 cm (such as 4.5 cm) produces a significantly smaller volume than 166.375 cm³ because volume grows rapidly with the cube of side length. Any side length larger than 5.5 cm (such as 6.5, 7.5 or 8.5 cm) produces much larger volumes. None of these equal the target volume, so they cannot be correct side lengths for this cube. Common Pitfalls: A common error is to mistakenly square the side instead of cubing it or to take the square root instead of the cube root when working backwards. Another pitfall is rounding intermediate steps too aggressively, which can hide the exact equality between a candidate side length and the given volume. When options are given, directly cubing each option is often the quickest and most reliable way to find the correct answer. Final Answer: The side length of the cube is 5.5 cm.

Question 13

The base area of a right pyramid is 57 square units and its vertical height is 10 units. What is the volume (in cubic units) of the pyramid?

Accepted Answer

Introduction / Context: This mensuration problem deals with the volume of a right pyramid. A right pyramid has its apex directly above the centroid of the base, and the standard volume formula applies directly using the base area and the vertical height. It is similar in spirit to the volume formula for cones and prisms but with a different constant factor. Given Data / Assumptions: The solid is a right pyramid. Base area A = 57 square units. Vertical height h = 10 units. We are asked to find the volume in cubic units. The volume formula for any pyramid is V = (1 / 3) * base area * height. Concept / Approach: The volume of a pyramid is one third of the volume of a prism that has the same base area and height. This leads to the simple formula: V = (1 / 3) * A * h We are given A and h directly, so we only need to substitute and simplify the numeric expression. There is no need to know the exact shape of the base as long as its area is known. Step-by-Step Solution: Step 1: Recall the volume formula: V = (1 / 3) * base area * height. Step 2: Substitute the given base area A = 57 square units and height h = 10 units. Step 3: V = (1 / 3) * 57 * 10. Step 4: Compute 57 * 10 = 570. Step 5: Divide by 3: V = 570 / 3 = 190 cubic units. Step 6: Therefore, the volume of the pyramid is 190 cubic units. Verification / Alternative check: We can think of constructing a prism with the same base area 57 and height 10, which would have volume 57 * 10 = 570 cubic units. A pyramid with the same base and height fits three times inside such a prism (assuming the appropriate arrangement), which is why the formula has the factor 1 / 3. Thus, 570 divided by 3 naturally gives 190 cubic units. This reasoning confirms the formula and the numerical answer. Why Other Options Are Wrong: 380 cubic units and 570 cubic units correspond to forgetting the 1 / 3 factor or using 2 / 3 instead. 540 cubic units and 600 cubic units are significantly larger than the prism volume of 570 cubic units when using the given base and height, so they cannot possibly be the volume of a pyramid whose volume is strictly less than that of the corresponding prism. Only 190 cubic units fits the correct formula and logical bounds for the volume. Common Pitfalls: A common mistake is to use the prism formula V = A * h and forget the 1 / 3 factor that applies specifically to pyramids. Another pitfall is confusion between slant height and vertical height; here, the problem explicitly gives the vertical height, which is what the formula requires. Carefully reading the wording and recalling that pyramids have one third the volume of a corresponding prism helps avoid these errors. Final Answer: The volume of the pyramid is 190 cubic units.

Question 14

In triangle PQR, points L and M lie on sides PQ and PR respectively such that segment LM is parallel to side QR. If PL = 2 cm, LQ = 6 cm and PM = 1.5 cm, then find the length of segment MR in centimetres.

Accepted Answer

Introduction / Context: This geometry question tests your understanding of similar triangles and the basic proportionality theorem (also known as Thales theorem). When a line segment is drawn parallel to one side of a triangle, it creates a smaller triangle inside the original triangle that is similar to it, and corresponding sides are in proportion. Using these proportional relationships, we can find unknown segment lengths such as MR in this problem. Given Data / Assumptions: Triangle PQR is given. L lies on side PQ and M lies on side PR. LM is parallel to QR. PL = 2 cm and LQ = 6 cm, so PQ = PL + LQ = 8 cm. PM = 1.5 cm. We are required to find MR in centimetres. All geometry is in a standard Euclidean plane. Concept / Approach: When a line segment joins two sides of a triangle and is parallel to the third side, the smaller triangle formed is similar to the original triangle. Here triangle PLM is similar to triangle PQR. Therefore, the ratios of corresponding sides are equal. In particular, PL / PQ = PM / PR. Once we compute PR, we get MR by subtracting PM from the total length PR since PR = PM + MR. Step-by-Step Solution: Step 1: Compute the full length of side PQ. PQ = PL + LQ = 2 + 6 = 8 cm. Step 2: Because LM is parallel to QR, triangles PLM and PQR are similar. Step 3: Write the similarity ratio: PL / PQ = PM / PR. Step 4: Substitute known values: 2 / 8 = 1.5 / PR. Step 5: Simplify the left side: 2 / 8 = 1 / 4. Step 6: So 1 / 4 = 1.5 / PR, hence PR = 1.5 * 4 = 6 cm. Step 7: Segment PR consists of PM and MR. Thus PR = PM + MR. Step 8: Substitute values: 6 = 1.5 + MR, so MR = 6 - 1.5 = 4.5 cm. Verification / Alternative check: We can check the proportionality again using PR = 6 cm and MR = 4.5 cm. Since PM = 1.5 cm, the ratio PM : PR = 1.5 : 6 = 1 : 4, which matches PL : PQ = 2 : 8 = 1 : 4. This confirms that the similarity ratio is satisfied and that the computed PR and MR are consistent with the given data and the parallel line condition. Why Other Options Are Wrong: 0.5 cm: This would make PR = 2.0 cm, contradicting the proportionality PL / PQ = PM / PR. 8 cm and 9 cm: These imply much larger values of PR and do not satisfy the similarity ratio between PL / PQ and PM / PR. 6 cm: This is the value of PR itself, not MR. Using MR = 6 cm would make PR = 7.5 cm, which again breaks the proportional relationship. Common Pitfalls: A common mistake is to assume that the ratio PL : LQ equals PM : MR directly without using the full sides. Another frequent error is to forget that PR is the sum of PM and MR, not just one of them. Some learners also mix up which sides correspond in the similar triangles. Always map corresponding vertices carefully and then apply the proportionality to whole sides before subtracting to obtain internal segments. Final Answer: The length of segment MR is 4.5 cm.

Question 15

In a circle with centre O, the radius is 5 centimetres and the length of chord AB is 8 centimetres. Find the perpendicular distance of chord AB from the centre O in centimetres.

Accepted Answer

Introduction / Context: This circle geometry problem involves finding the perpendicular distance from the centre of a circle to a chord. It uses the relationship between the radius, the chord length and the perpendicular distance. Such questions frequently appear in aptitude and school geometry exams to test your understanding of right triangles inside a circle. Given Data / Assumptions: A circle has centre O. The radius of the circle is 5 cm. AB is a chord of the circle. The length of chord AB is 8 cm. We must find the distance from O to chord AB, measured along the perpendicular from O to AB. Standard Euclidean geometry is assumed. Concept / Approach: The perpendicular drawn from the centre of a circle to a chord bisects the chord. This means the chord is cut into two equal segments. If the full chord length is 8 cm, each half is 4 cm. The radius from the centre to each endpoint of the chord is 5 cm. These three segments form a right triangle with the radius as hypotenuse, half chord as one leg and the perpendicular distance as the other leg. Applying the Pythagoras theorem allows us to find the perpendicular distance from O to AB. Step-by-Step Solution: Step 1: Let the perpendicular from O to AB meet AB at point M. Step 2: Since OM is perpendicular to AB, M is the midpoint of AB. Step 3: Half of the chord AB is AM = AB / 2 = 8 / 2 = 4 cm. Step 4: OA is a radius, so OA = 5 cm. Step 5: In right triangle OAM, OA is the hypotenuse, AM is one leg and OM is the other leg. Step 6: Apply Pythagoras theorem: OA^2 = OM^2 + AM^2. Step 7: Substitute values: 5^2 = OM^2 + 4^2, so 25 = OM^2 + 16. Step 8: Rearrange: OM^2 = 25 - 16 = 9. Step 9: Take square root: OM = 3 cm. Verification / Alternative check: We can confirm by reversing the calculation: if OM = 3 cm and AM = 4 cm, then OA^2 should equal 3^2 + 4^2 = 9 + 16 = 25, so OA = 5 cm, matching the given radius. Also, doubling AM gives AB = 8 cm, the given chord length. This consistency among radius, chord and distance validates the result. Why Other Options Are Wrong: 2 cm and 4 cm: These values do not satisfy the relation OA^2 = OM^2 + AM^2 when OA = 5 cm and AM = 4 cm. 5 cm: This would mean the chord collapses to a point (distance equal to radius), which is not possible for an 8 cm chord. 15 cm: This is larger than the radius and cannot be the distance from the centre to a chord lying inside the circle. Common Pitfalls: Students sometimes forget that the perpendicular from the centre bisects the chord and incorrectly use the full chord length instead of half. Others misidentify which segment is the hypotenuse in the right triangle, leading to wrong Pythagoras calculations. It is also easy to make sign errors when rearranging equations. Carefully drawing a diagram and labelling all segments helps avoid these issues. Final Answer: The perpendicular distance of chord AB from the centre O is 3 cm.

Question 16

The radius of a sphere and the radius of a right circular cylinder are both equal to r. If their volumes are equal, then what is the ratio of the height of the cylinder to its radius?

Accepted Answer

Introduction / Context: This mensuration question compares the volume of a sphere and the volume of a right circular cylinder when both share the same radius. It tests your ability to use standard volume formulas and relate them through algebra to find an unknown ratio, here the ratio of the cylinder's height to its radius. Given Data / Assumptions: A sphere has radius r. A right circular cylinder also has radius r. The volume of the sphere and the volume of the cylinder are equal. We must find the ratio (height of cylinder) : (radius of cylinder). pi is treated as a constant that will cancel out in the ratio. Concept / Approach: The volume of a sphere of radius r is: V_sphere = (4 / 3) * pi * r^3. The volume of a cylinder of radius r and height h is: V_cylinder = pi * r^2 * h. Since the volumes are equal, we set V_sphere = V_cylinder and solve for h in terms of r. From this expression we can directly read the ratio h : r. Step-by-Step Solution: Step 1: Write the equality of volumes: (4 / 3) * pi * r^3 = pi * r^2 * h. Step 2: Cancel the common factor pi from both sides. Step 3: Cancel r^2 on both sides of the equation, leaving (4 / 3) * r = h. Step 4: So h = (4 / 3) * r. Step 5: The ratio of height to radius is h : r = (4 / 3) * r : r. Step 6: Divide both parts of the ratio by r: h : r = 4 / 3 : 1 = 4 : 3. Verification / Alternative check: To verify, you can choose a convenient value for r, for example r = 3 units. Then h = (4 / 3) * 3 = 4 units. Volume of sphere = (4 / 3) * pi * 3^3 = (4 / 3) * pi * 27 = 36 * pi. Volume of cylinder = pi * 3^2 * 4 = pi * 9 * 4 = 36 * pi. The volumes match, confirming that the ratio 4 : 3 is correct for all r. Why Other Options Are Wrong: 3 : 1 and 2 : 1: These ratios give heights that are too large relative to the radius and do not satisfy the equality of volumes. 3 : 2: This would make h = 1.5 * r, which does not yield equal volumes when substituted into the formulas. 1 : 1: This implies a cylinder with height equal to radius, clearly not enough volume to match a sphere with the same radius. Common Pitfalls: A common mistake is to confuse the formulas for volume and surface area, or to forget to cancel out common factors like pi and r^2. Some students also mistakenly cross multiply incorrectly or simplify the ratio h : r without first solving for h. Working step by step and cancelling common factors early makes the algebra easier and less error prone. Final Answer: The ratio of the height of the cylinder to its radius is 4 : 3.

Question 17

In a triangle, the orthocentre is defined as the point where which set of line segments intersect?

Accepted Answer

Introduction / Context: This conceptual question in geometry checks your understanding of special points associated with a triangle. In particular, it focuses on the orthocentre, which is one of the four main centres of a triangle, along with the centroid, circumcentre and incentre. Knowing the definitions of these points is crucial for solving many advanced geometry problems. Given Data / Assumptions: We are considering an arbitrary triangle. Several important line segments can be drawn: medians, altitudes, perpendicular bisectors and angle bisectors. The question asks specifically about which set of segments intersect at the orthocentre. Standard triangle geometry in a Euclidean plane is assumed. Concept / Approach: Each notable triangle centre is associated with a specific type of segment: Centroid: intersection of the three medians. Orthocentre: intersection of the three altitudes. Circumcentre: intersection of the perpendicular bisectors of the sides. Incentre: intersection of the internal angle bisectors. The orthocentre is the point where all three altitudes meet, where an altitude is a perpendicular drawn from a vertex to the opposite side (or its extension). Step-by-Step Solution: Step 1: Recall that an altitude of a triangle is a line drawn from a vertex perpendicular to the opposite side or its extended line. Step 2: Every triangle has three altitudes, one from each vertex. Step 3: In Euclidean geometry, these three altitudes are concurrent, meaning they intersect at a single point. Step 4: This common point of intersection of the three altitudes is defined as the orthocentre of the triangle. Step 5: Therefore, the orthocentre is the point where the altitudes meet. Verification / Alternative check: To visualize this, draw any acute triangle and construct its three altitudes using a right angle ruler. You will see that all three lines intersect at a single point inside the triangle. For a right angled triangle, this point coincides with the right angled vertex, and for an obtuse triangle, the orthocentre lies outside the triangle. In all cases, however, it is always the intersection of the altitudes. Why Other Options Are Wrong: The medians meet: Their intersection is the centroid, not the orthocentre. The perpendicular bisectors of the sides meet: Their intersection point is the circumcentre. The angle bisectors meet: Their intersection point is the incentre. The internal and external bisectors meet: These describe various other points, not the orthocentre, and are not the standard definition of a single special centre like the orthocentre. Common Pitfalls: Students often confuse the different centres because each is defined by the concurrency of three lines. Mixing up medians and altitudes is especially common. A good way to remember is to associate each centre with its geometric role: centroid with balancing point, circumcentre with circumscribed circle, incentre with inscribed circle, and orthocentre with right angles (altitudes). Clear mental images and practice with diagrams can help keep these definitions straight. Final Answer: The orthocentre of a triangle is the point where the altitudes meet.

Question 18

ABCD is a cyclic trapezium in which side AD is parallel to side BC. If angle ABC is 70 degrees, then what is the measure of angle BCD in degrees?

Accepted Answer

Introduction / Context: This problem combines properties of cyclic quadrilaterals and trapeziums. It checks your understanding of angle relationships when a quadrilateral is both cyclic and a trapezium with a pair of parallel sides. Recognizing how opposite and adjacent angles relate in such special figures is important for solving geometry problems efficiently in competitive exams. Given Data / Assumptions: ABCD is a cyclic quadrilateral, that is, all four vertices lie on a circle. ABCD is also a trapezium with AD parallel to BC. Angle ABC = 70 degrees. We must find angle BCD (the interior angle at C). Angles are measured in degrees in standard Euclidean geometry. Concept / Approach: Two key concepts are used: In a cyclic quadrilateral, the sum of a pair of opposite interior angles is 180 degrees. In a trapezium, if a pair of opposite sides are parallel, the interior angles on the same side of a transversal sum to 180 degrees. With AD parallel to BC, consider the leg AB as a transversal. The angles at A and B on this transversal are supplementary. At the same time, in a cyclic quadrilateral, angle A and angle C are supplementary. Combining these facts will lead us to a relationship between angle B and angle C. Step-by-Step Solution: Step 1: Since AD is parallel to BC and AB is a transversal, angles DAB and ABC are interior angles on the same side of AB. Step 2: Therefore, angle DAB + angle ABC = 180 degrees. Step 3: Let angle DAB = A and angle BCD = C. We are given angle ABC = 70 degrees. Step 4: From Step 2, A + 70 = 180, so A = 110 degrees. Step 5: In a cyclic quadrilateral, opposite interior angles sum to 180 degrees. Thus angle A + angle C = 180 degrees. Step 6: Substitute A = 110 degrees into A + C = 180: 110 + C = 180. Step 7: Solve for C: C = 180 - 110 = 70 degrees. Step 8: Therefore, angle BCD = 70 degrees. Verification / Alternative check: We have found that angle A = 110 degrees and angle C = 70 degrees. Opposite angles in the cyclic quadrilateral are A and C, and B and D. A + C = 110 + 70 = 180 degrees, satisfying the cyclic condition. Similarly, B = 70 degrees and D = 110 degrees must be supplementary, which is consistent with AD parallel to BC and CD as a transversal. Thus all conditions of both a cyclic quadrilateral and trapezium are satisfied. Why Other Options Are Wrong: 110°: This would make angle A also 110°, causing A + C to exceed 180° and breaking the cyclic quadrilateral property. 80° and 60°: These do not satisfy the requirement that opposite angles sum to 180° together with the parallel line condition. 90°: This is a typical distractor but does not work with the 70 degree angle and the cyclic plus parallel constraints. Common Pitfalls: A common error is to assume that opposite angles in a trapezium are equal rather than supplementary, or to forget which pairs of angles are opposite in a cyclic quadrilateral. Another mistake is mixing up adjacent and opposite angles when applying the 180 degree sum. Drawing a clear diagram with all labels and angle marks can greatly reduce confusion and help in applying both sets of properties correctly. Final Answer: The measure of angle BCD is 70°.

Question 19

In triangle ABC, point G is the centroid. If the length of segment AG is equal to the length of side BC, then what is the measure of angle BGC in degrees?

Accepted Answer

Introduction / Context: This is a higher level geometry question involving special points in a triangle and their relationships. The centroid is the point where the medians intersect. Here, a special condition is given: the segment from a vertex to the centroid has the same length as the opposite side. You are asked to determine the angle formed at the centroid between the two segments to the other vertices, which requires using properties of medians, centroids and analytic geometry or vector reasoning. Given Data / Assumptions: Triangle ABC is given. G is the centroid of triangle ABC. Segment AG is a median from vertex A to side BC. The length AG is equal to the length of side BC. We must find the measure of angle BGC in degrees. Standard Euclidean geometry is assumed. Concept / Approach: The centroid of a triangle divides each median in the ratio 2 : 1, counted from the vertex to the midpoint of the opposite side. To handle the special length condition AG = BC, a convenient method is to use coordinate geometry. By placing the triangle symmetrically on the coordinate axes, we can express side lengths and the position of the centroid in terms of variables, then apply the condition AG = BC to fix the shape. Once coordinates of B, G and C are known, we can use the dot product to compute angle BGC. Step-by-Step Solution: Step 1: Place triangle ABC on a coordinate plane for simplicity. Step 2: Let B = (−a, 0) and C = (a, 0). Then BC is horizontal with length BC = 2a. Step 3: Place A at (0, h), so the triangle is symmetric about the y axis. Step 4: The centroid G is the average of the vertices: G = ((−a + a + 0) / 3, (0 + 0 + h) / 3) = (0, h / 3). Step 5: The length AG is the vertical distance from (0, h) to (0, h / 3), which is h − h / 3 = 2h / 3. Step 6: The side BC has length 2a. The given condition is AG = BC, so 2h / 3 = 2a. Step 7: Simplify to get h / 3 = a, hence h = 3a. Step 8: Substitute h = 3a into the coordinates. Then G = (0, a). Step 9: Now compute vectors GB and GC. GB = B − G = (−a − 0, 0 − a) = (−a, −a). GC = C − G = (a − 0, 0 − a) = (a, −a). Step 10: Use the dot product formula to find angle BGC. The dot product GB · GC = (−a)(a) + (−a)(−a) = −a^2 + a^2 = 0. Step 11: If the dot product of two non-zero vectors is zero, the angle between them is 90 degrees. Step 12: Therefore, angle BGC = 90°. Verification / Alternative check: You can choose a specific value of a, for example a = 1. Then B = (−1, 0), C = (1, 0), A = (0, 3), G = (0, 1). The vectors GB and GC are (−1, −1) and (1, −1) respectively. Plotting or calculating their slopes shows one has slope 1 and the other has slope −1, so they are perpendicular. The right angle at G confirms that angle BGC is 90 degrees, consistent with the general vector calculation. Why Other Options Are Wrong: 45° and 60°: These would not make the vectors GB and GC perpendicular and would contradict the length condition AG = BC in the coordinate setup. 120° and 150°: These obtuse angles also do not arise from the perpendicular relationship implied by the dot product being zero. Common Pitfalls: The main difficulty lies in correctly using the centroid properties and the condition AG = BC. Some students forget that the centroid divides each median in a 2 : 1 ratio or try to apply only pure Euclidean geometry without a clear structure, leading to complicated constructions. Another common error is miscomputing vector dot products or misplacing points in the coordinate plane. Carefully defining coordinates and going step by step with algebra keeps the reasoning clear and reliable. Final Answer: The measure of angle BGC is 90°.

Question 20

In a circle, a chord of length 5√2 centimetres subtends a right angle at the centre. Find the length of the radius of the circle in centimetres.

Accepted Answer

Introduction / Context: This problem involves a circle, a chord and the central angle subtended by that chord. It tests your understanding of the relationship between chord length, radius and central angle. Recognising how to connect these elements using trigonometry or right triangle geometry is a valuable skill in circle based questions. Given Data / Assumptions: There is a circle with centre O and radius R. A chord of the circle has length 5√2 centimetres. This chord subtends a right angle (90 degrees) at the centre O. We must find the radius R in centimetres. Standard Euclidean geometry and basic trigonometry can be used. Concept / Approach: The central angle at O is 90 degrees. The endpoints of the chord and the centre form an isosceles triangle with two sides of length R (the radii) and base equal to the chord. Because the central angle is 90 degrees, this triangle is also a right angled isosceles triangle. Using properties of right triangles, the chord length is related to the radius by a simple trigonometric formula: chord = 2 * R * sin(theta / 2). For theta = 90 degrees, sin(45 degrees) = √2 / 2, so chord = R * √2. We can then solve for R. Step-by-Step Solution: Step 1: Let the radius of the circle be R. Step 2: The chord subtends a central angle of 90 degrees at the centre O. Step 3: Use the chord length formula: chord length = 2 * R * sin(theta / 2). Step 4: Here theta = 90 degrees, so theta / 2 = 45 degrees. Step 5: sin(45 degrees) = √2 / 2. Step 6: Substitute into the formula: chord = 2 * R * (√2 / 2) = R * √2. Step 7: The chord is given as 5√2 cm, so R * √2 = 5√2. Step 8: Divide both sides by √2: R = 5 cm. Verification / Alternative check: Alternatively, you can think in terms of the right triangle formed by the radius lines and half the chord. If the chord length is 5√2, then half the chord is (5√2) / 2. The triangle with sides R, R and 5√2 has a right angle at the centre. Using Pythagoras in a right isosceles triangle, the hypotenuse (diameter) is 2R and each leg (radius) is R. The base (chord) corresponds to R√2. Setting R√2 = 5√2 again gives R = 5 cm, confirming the answer. Why Other Options Are Wrong: 2.5 cm and 3.5 cm: These radii are too small, producing chords shorter than 5√2 cm for a 90 degree central angle. 7.5 cm and 10 cm: These are too large, generating chord lengths greater than 5√2 cm for the same central angle. Common Pitfalls: Some learners attempt to apply Pythagoras theorem directly without a clear diagram and mix up the sides of the triangle. Others may incorrectly use the full diameter instead of the radius in formulas. A careful sketch and the use of the standard chord formula or properties of a 45-45-90 triangle help make the relationships clear and reduce errors. Final Answer: The radius of the circle is 5 cm.

Volume and Surface Area Questions