Question 1

Supplementary angles in a fixed ratio: Two angles are supplementary and are in the ratio 1 : 4. What is the measure of the smaller angle (in degrees)?

Accepted Answer

Introduction / Context: Supplementary angles sum to 180°. When given their ratio, distribute 180° proportionally to find each angle. Given Data / Assumptions: Angles in ratio 1 : 4. Sum = 180°. Concept / Approach: Let the angles be x and 4x. Then x + 4x = 180 ⇒ 5x = 180. Step-by-Step Solution: x = 180 / 5 = 36°Larger angle = 144° (check: 36 + 144 = 180) Verification / Alternative check: Ratio check: 36 : 144 = 1 : 4, as required. Why Other Options Are Wrong: 45° or 35° do not satisfy both the 1:4 ratio and 180° sum; 72° would be the smaller angle if the ratio were 2:3, not 1:4. Common Pitfalls: Confusing complementary (90°) with supplementary (180°). Final Answer: 36 degree

Question 2

Complement of an angle in degrees and minutes: Find the complement of 30° 20′.

Accepted Answer

Introduction / Context: Complementary angles sum to 90°. Subtract the given angle from 90° taking care with minutes conversion (60′ = 1°). Given Data / Assumptions: Angle = 30° 20′. Complement = 90° 00′ − 30° 20′. Concept / Approach: Borrow 1° as 60′ if needed, then subtract degrees and minutes separately. Step-by-Step Solution: 90° 00′ − 30° 20′ = 59° 40′ Verification / Alternative check: 59°40′ + 30°20′ = 90°00′, confirming complementarity. Why Other Options Are Wrong: 69°40′ exceeds 90° − 30°; 35°80′ uses an invalid minute format (80′ ≡ 1°20′). Common Pitfalls: Forgetting to convert 60′ to 1° when minutes exceed 59. Final Answer: 59°40′

Question 3

An angle is equal to one-third of its supplement. Find the numerical measure of the angle (in degrees).

Accepted Answer

Introduction / Context: Supplementary angles sum to 180°. If one angle is a fixed fraction of its supplement, we can set up a linear equation to compute the exact measure. Given Data / Assumptions: Let the required angle be x°. Its supplement is 180 − x. Given: x = (1/3) * (180 − x). Concept / Approach: Model the verbal statement as an equation and solve for x with simple algebra. Step-by-Step Solution: x = (1/3)(180 − x)3x = 180 − x4x = 180x = 45° Verification / Alternative check: The supplement is 135°. One-third of 135° is 45°, which matches x, so the condition holds. Why Other Options Are Wrong: 40°, 50°, 55° do not satisfy x = (1/3)(180 − x). Common Pitfalls: Using complement (sum 90°) instead of supplement (sum 180°), or incorrectly solving the proportion. Final Answer: 45°

Question 4

Six times the complement of an angle is 12° less than twice its supplement. Find the measure of the angle.

Accepted Answer

Introduction / Context: Complementary angles sum to 90°, and supplementary angles sum to 180°. The statement links linear expressions of the same unknown through these definitions. Given Data / Assumptions: Angle = x°. Complement = 90 − x; supplement = 180 − x. Condition: 6(90 − x) = 2(180 − x) − 12. Concept / Approach: Translate to an equation in x and solve. Keep complements and supplements distinct. Step-by-Step Solution: 6(90 − x) = 2(180 − x) − 12540 − 6x = 360 − 2x − 12540 − 6x = 348 − 2x540 − 348 = 6x − 2x192 = 4x ⇒ x = 48° Verification / Alternative check: Complement = 42°, supplement = 132°. LHS = 6*42 = 252; RHS = 2*132 − 12 = 264 − 12 = 252 ✓. Why Other Options Are Wrong: They fail the given linear relation when checked. Common Pitfalls: Mixing up 90 − x and 180 − x; arithmetic slips while simplifying; sign mistakes with the “12° less”. Final Answer: 48°

Question 5

In △ABC, if 2∠A = 3∠B = 6∠C, find the measure of ∠A.

Accepted Answer

Introduction / Context: Angle relations given as equalities allow expressing each angle in terms of a common parameter and then using the triangle sum property (180°). Given Data / Assumptions: 2A = 3B = 6C = k (some common value). A + B + C = 180°. Concept / Approach: Express A, B, C via k and substitute into A + B + C = 180° to solve for k, then A. Step-by-Step Solution: A = k/2, B = k/3, C = k/6A + B + C = k(1/2 + 1/3 + 1/6) = k(1) = 180°k = 180° ⇒ A = k/2 = 90°But check: 2A = 180°, 3B = 180°, 6C = 180° ⇒ A = 90°, B = 60°, C = 30°. Verification / Alternative check: Sum 90° + 60° + 30° = 180° ✓, and the proportional relations hold. Why Other Options Are Wrong: 60°, 120°, 30° do not match the derived A = 90°. Common Pitfalls: Confusing 2A = 3B = 6C with A:B:C directly; not converting each to k-based expressions first. Final Answer: 90°

Question 6

In △ABC, if A − B = 15° and B − C = 30°, find the measure of ∠A.

Accepted Answer

Introduction / Context: Angles in a triangle relate via differences given. Convert them to expressions in one variable, then apply A + B + C = 180° to solve. Given Data / Assumptions: A − B = 15° ⇒ A = B + 15°. B − C = 30° ⇒ C = B − 30°. A + B + C = 180°. Concept / Approach: Substitute the expressions for A and C into the sum of angles and solve for B, then get A. Step-by-Step Solution: (B + 15°) + B + (B − 30°) = 180°3B − 15° = 180°3B = 195° ⇒ B = 65°A = B + 15° = 80° Verification / Alternative check: C = B − 30° = 35°. Check sum: 80° + 65° + 35° = 180° ✓, and differences match the givens. Why Other Options Are Wrong: 75°, 85°, 65° are values of other angles or near-misses; only 80° satisfies all constraints for A. Common Pitfalls: Sign errors when expressing C, or forgetting to add all three angles to 180° in the final step. Final Answer: 80°

Question 7

In △ABC, AB is extended to D and AC is extended to E. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, find ∠BOC.

Accepted Answer

Introduction / Context: Extending AB and AC makes ∠CBD and ∠BCE external angles at B and C. The intersection of these external bisectors is the excenter opposite A. A known result links the angle ∠BOC at this excenter to ∠A of the triangle. Given Data / Assumptions: △ABC with internal angle A = 40°. O is the excenter opposite A (intersection of external bisectors at B and C). Concept / Approach: For the excenter opposite A, the angle between the lines OB and OC at O equals 90° − A/2. (For the incenter, it would be 90° + A/2.) Here we use the excenter formula. Step-by-Step Solution: ∠BOC = 90° − (A/2)= 90° − 20°= 70° Verification / Alternative check: Construct a simple acute triangle (e.g., A=40°, B=70°, C=70°) and verify with angle-chasing that the angle at the A-excenter equals 70°; synthetic geometry texts list this identity directly. Why Other Options Are Wrong: 60°, 65°, 75° correspond to using incorrect ±A/2 adjustments or confusing incenter vs excenter. Common Pitfalls: Applying the incenter formula ∠BOC = 90° + A/2 by mistake, or misidentifying external vs internal bisectors. Final Answer: 70°

Question 8

In the figure, DE ∥ BC in △ABC. Given AD = 1.7 cm, AB = 6.8 cm, AC = 9 cm. Find AE (in cm).

Accepted Answer

Introduction / Context: When a segment DE is drawn parallel to BC in △ABC, triangles ADE and ABC are similar (Basic Proportionality Theorem/Thales). Corresponding sides are proportional. Given Data / Assumptions: DE ∥ BC ⇒ △ADE ∼ △ABC. AD = 1.7, AB = 6.8, AC = 9. Need AE. Concept / Approach: Use the similarity ratio AD/AB = AE/AC to compute AE directly. Step-by-Step Solution: AD/AB = AE/AC1.7 / 6.8 = AE / 91.7/6.8 = 0.25 ⇒ AE = 9 * 0.25 = 2.25 cm Verification / Alternative check: Scale factor from big to small is 0.25 (since AD is one-fourth of AB); then AE must be one-fourth of AC, i.e., 2.25 cm. Why Other Options Are Wrong: They correspond to using wrong sides or inverting the ratio; only 2.25 cm respects the similarity scale factor consistently. Common Pitfalls: Using BD/BC or mixing sides from different triangles; misreading AD/AB as AB/AD. Final Answer: 2.25cm

Question 9

If in a triangle the bisector of one angle also bisects the opposite side, the triangle must be of which type?

Accepted Answer

Introduction / Context: The Angle Bisector Theorem states that an internal angle bisector divides the opposite side into segments proportional to the adjacent sides. If it bisects the opposite side exactly (creates equal segments), that imposes a specific equality on the adjacent sides. Given Data / Assumptions: Let △ABC have angle at A bisected by AD with D on BC, and BD = DC (opposite side bisected). Concept / Approach: By the Angle Bisector Theorem, BD/DC = AB/AC. If BD = DC, then BD/DC = 1, hence AB/AC = 1 and AB = AC. That means the triangle is isosceles with AB = AC (vertex at A). Step-by-Step Solution: Given: AD bisects ∠A and also BC (so BD = DC).Angle Bisector Theorem: BD/DC = AB/AC.Since BD = DC, AB/AC = 1 ⇒ AB = AC ⇒ Isosceles. Verification / Alternative check: Construct an isosceles triangle AB = AC; the angle bisector from A is also the median to BC, confirming the property. Why Other Options Are Wrong: Scalene forbids any equal sides; right triangle is about a 90° angle (not implied here); “circle” is not a triangle type. Common Pitfalls: Confusing angle bisector with perpendicular bisector; assuming any median is an angle bisector (true only in isosceles). Final Answer: Isosceles

Question 10

Two similar triangles have areas 81 cm² and 144 cm². If the largest side of the smaller triangle is 27 cm, find the largest side of the larger triangle.

Accepted Answer

Introduction / Context: For similar triangles, the ratio of areas equals the square of the ratio of corresponding side lengths. We can extract the linear scale factor by taking square roots of the area ratio. Given Data / Assumptions: Areas: 81 and 144 cm² (smaller to larger). Largest side (smaller) = 27 cm. Concept / Approach: If k is the linear scale (larger/smaller), then area ratio = k². Hence k = √(144/81). Multiply the smaller corresponding side by k to get the larger side. Step-by-Step Solution: k = √(144/81) = √(16/9) = 4/3Largest side (larger) = 27 * (4/3) = 36 cm Verification / Alternative check: (36/27)² = (4/3)² = 16/9; 81 * (16/9) = 144 ✓. Why Other Options Are Wrong: 48 cm uses a scale of 16/9 on length; 24 cm uses 8/9; 88 cm is unrelated to the scale ratio. Common Pitfalls: Using area ratio directly on sides without taking the square root first. Final Answer: 36 cm

Question 11

In the figure, ∠BAD = ∠CAD (AD is the angle bisector at A). Given AB = 4 cm, AC = 5.2 cm, and BD = 3 cm. Find BC.

Accepted Answer

Introduction / Context: With AD the internal bisector of ∠A, the Angle Bisector Theorem gives a proportional split of side BC. From one segment (BD) we can obtain the other (DC) and thus BC. Given Data / Assumptions: AB = 4, AC = 5.2. BD = 3, need BC. Concept / Approach: Angle Bisector Theorem: BD/DC = AB/AC ⇒ DC = BD * (AC/AB). Then BC = BD + DC. Step-by-Step Solution: DC = 3 * (5.2/4) = 3 * 1.3 = 3.9BC = BD + DC = 3 + 3.9 = 6.9 cm Verification / Alternative check: BD:DC = 3:3.9 = 30:39 = 10:13, and AB:AC = 4:5.2 = 40:52 = 10:13 ✓. Why Other Options Are Wrong: 9.6 and 9.3 add incorrect DC values; 3.9 is only DC, not BC. Common Pitfalls: Inverting AB/AC; forgetting to add BD to get BC; rounding prematurely. Final Answer: 6.9 cm

Question 12

A 15 m ladder rests against a window 9 m above the ground on one side of a street. Keeping the foot fixed, it is turned to the other side to reach a window 12 m high. Find the width of the street.

Accepted Answer

Introduction / Context: This is a classic two right-triangles setup sharing the same ladder length and foot position. Horizontal reaches to each wall add to the street width. Given Data / Assumptions: Ladder length L = 15 m. Heights: 9 m on one side, 12 m on the other. Concept / Approach: For each side, horizontal distance to the wall is the base of a right triangle with hypotenuse 15 and vertical leg equal to the window height. Sum the two bases to get the street width. Step-by-Step Solution: Side 1: x^2 + 9^2 = 15^2 ⇒ x = √(225 − 81) = √144 = 12 mSide 2: y^2 + 12^2 = 15^2 ⇒ y = √(225 − 144) = √81 = 9 mWidth = x + y = 12 + 9 = 21 m Verification / Alternative check: Both bases are positive and consistent with the Pythagorean theorem; the same hypotenuse is used in each orientation. Why Other Options Are Wrong: 30 m or 31 m double-count or miscompute the square roots; 12 m is just one side’s base. Common Pitfalls: Using 9 + 12 under the square root; forgetting to add bases from both walls. Final Answer: 21 m

Question 13

In △ABC, points D on AB and E on AC satisfy AD = 8 cm, BD = 12 cm, AE = 6 cm, EC = 9 cm. Find the ratio BC/DE.

Accepted Answer

Introduction / Context: If a line through points D on AB and E on AC is parallel to BC, similar triangles imply fixed ratios among segments. We can test proportionality from the given splits to confirm DE ∥ BC, then use the similarity ratio to relate BC and DE. Given Data / Assumptions: AB = AD + DB = 8 + 12 = 20. AC = AE + EC = 6 + 9 = 15. AD/AB = 8/20 = 2/5; AE/AC = 6/15 = 2/5. Concept / Approach: Since AD/AB = AE/AC, DE ∥ BC (converse of Basic Proportionality Theorem). Then △ADE ∼ △ABC with scale factor k = AD/AB = 2/5. Thus DE/BC = k, so BC/DE = 1/k = 5/2. Step-by-Step Solution: k = AD/AB = 2/5DE/BC = 2/5 ⇒ BC/DE = 5/2 Verification / Alternative check: Ratios from both sides match (2/5), confirming parallelism and the similarity scale. Why Other Options Are Wrong: Other ratios invert or distort the scale factor; only 5/2 follows from k = 2/5. Common Pitfalls: Assuming parallelism without checking the side ratios; flipping BC/DE by mistake. Final Answer: 5 2

Question 14

In an equilateral △ABC, let AD ⟂ BC. Which relation between AB and AD is true?

Accepted Answer

Introduction / Context: In an equilateral triangle of side s, the altitude AD has a standard length related to s. Substituting that value yields identities among AB and AD. Given Data / Assumptions: AB = s. Altitude AD = (√3/2)s. Concept / Approach: Compute AB² and AD², then test which given identity holds exactly without approximation. Step-by-Step Solution: AB² = s²AD² = (3/4)s²Check 3AB² = 3s² and 4AD² = 4*(3/4)s² = 3s² ⇒ 3AB² = 4AD² (true) Verification / Alternative check: Other listed relations evaluate to unequal expressions when substituting AB² and AD² as above. Why Other Options Are Wrong: They swap coefficients incorrectly; only the 3:4 relation matches altitude length in an equilateral triangle. Common Pitfalls: Using median length s/2 or height s*(√3/3) (which is the inradius) instead of the altitude length (√3/2)s. Final Answer: 3AB2 = 4AD2

Question 15

In △ABC, the internal angle bisectors of ∠B and ∠C meet at O (the incenter). If ∠A = 70°, find ∠BOC.

Accepted Answer

Introduction / Context: The incenter is the intersection of the internal angle bisectors. A standard identity links ∠BOC at the incenter with the opposite vertex angle ∠A of the triangle. Given Data / Assumptions: O is the incenter; A = 70°. Concept / Approach: For the incenter, the angle between lines OB and OC equals 90° + (A/2). This comes from the tangency geometry of the incircle and angle-bisector properties. Step-by-Step Solution: ∠BOC = 90° + (A/2) = 90° + 35° = 125° Verification / Alternative check: Test with a sample triangle (e.g., A=70°, B=60°, C=50°). Construct bisectors and measure angle at the incenter; it matches 125° per the identity. Why Other Options Are Wrong: 135°, 115°, 110° correspond to misapplied formulas (e.g., 90° ± A/2) or rounding errors; only 125° fits the incenter rule. Common Pitfalls: Confusing excenter and incenter results: excenter would use 90° − A/2 for the angle between external bisectors at B and C. Final Answer: 125°

Question 16

The complement of an angle exceeds the angle by 60°. Find the angle.

Accepted Answer

Introduction / Context: A complement is defined as 90° minus the angle. If the complement is 60° larger than the angle itself, set up and solve a linear equation. Given Data / Assumptions: Let the angle be x°. Complement = 90° − x. Given: (90° − x) = x + 60°. Concept / Approach: Rearrange to isolate x and compute the measure directly. Step-by-Step Solution: 90 − x = x + 6090 − 60 = x + x30 = 2x ⇒ x = 15° Verification / Alternative check: Complement = 75°; 75° exceeds 15° by 60°, satisfying the condition. Why Other Options Are Wrong: 25°, 30°, 35° fail the relation (90 − x) − x = 60. Common Pitfalls: Confusing complement with supplement; sign mistakes moving terms across the equality. Final Answer: 15°

Question 17

If two parallel lines are intersected by a transversal, then the bisectors of the two interior angles on the same side of the transversal are drawn. The figure enclosed by these four bisectors is a:

Accepted Answer

Introduction / Context: Two parallel lines cut by a transversal form two interior (co-interior) angles that are supplementary (sum to 180°). If we construct the angle bisectors for each pair of interior angles on both parallel lines, we examine the shape enclosed by these four bisectors. Given Data / Assumptions: Lines l₁ ∥ l₂ and a transversal t. Interior angles on the same side of t are supplementary. For each pair, we draw their internal bisectors. Concept / Approach: If two angles are supplementary (α and 180°−α), their bisectors are perpendicular because they are at α/2 and 90°−α/2 apart ⇒ total separation 90°. On each parallel line, the two bisectors formed with the transversal are perpendicular lines. Because corresponding interior angles on parallel lines are equal, the bisector lines on l₁ are parallel to their counterparts on l₂. Step-by-Step Reasoning: Interior angles: α and 180°−α ⇒ bisectors at α/2 and 90°−α/2Angle between these two bisectors = 90° (perpendicular)By parallelism, the corresponding bisectors on the other line are parallel to these, yielding two pairs of parallel lines at right anglesHence the enclosed quadrilateral has opposite sides parallel and all angles 90° ⇒ rectangle Verification / Alternative check: Choose α = 60° for concreteness. Bisectors at 30° and 60° from the transversal’s complement produce a rectangle in any coordinate construction consistent with l₁ ∥ l₂. Why Other Options Are Wrong: Trapezium: has only one pair of parallel sides. Square: would additionally require equal adjacent side lengths; not guaranteed. Circle: not a polygonal enclosure by straight bisectors. None of these: Not applicable; rectangle is correct. Common Pitfalls: Thinking the bisectors themselves must be at 45°; they depend on α but remain mutually perpendicular for supplementary pairs. Final Answer: Rectangle

Question 18

In right-angled triangle △PQR with ∠QPR = 90°, the hypotenuse QR = 26 cm. A point M lies on PR such that ∠PMR = 90° with PM = 6 cm and MR = 8 cm (so △PMR is right-angled at M). Find the area of △PQR (in cm^2).

Accepted Answer

Introduction / Context: This problem combines two right triangles. △PQR is right-angled at P, and a smaller right triangle △PMR sits along PR with legs PM = 6 and MR = 8. We use these to determine the legs of △PQR and hence its area. Given Data / Assumptions: In △PQR, ∠QPR = 90° (right at P). QR = 26 cm (hypotenuse of △PQR). △PMR is right at M with PM = 6 cm and MR = 8 cm. M lies on PR, so PR is the hypotenuse of △PMR. Concept / Approach: For △PMR: hypotenuse PR = sqrt(6^2 + 8^2) = 10 cm. In △PQR, legs are PQ and PR (right at P). With PR = 10 and hypotenuse QR = 26, use Pythagoras to get PQ. Area(△PQR) = (1/2) * PQ * PR. Step-by-Step Solution: PR = √(6^2 + 8^2) = √(36 + 64) = √100 = 10 cmQR^2 = PQ^2 + PR^2 ⇒ 26^2 = PQ^2 + 10^2PQ^2 = 676 − 100 = 576 ⇒ PQ = 24 cmArea = (1/2) * PQ * PR = (1/2) * 24 * 10 = 120 cm^2 Verification / Alternative check: Triple (10,24,26) is a scaled 5-12-13 triple (×2), confirming consistency. Area from legs is therefore 120 cm^2. Why Other Options Are Wrong: 180, 150, 240 cm^2: Each implies incorrect leg lengths inconsistent with 10-24-26. None of these: Incorrect since we get an exact 120 cm^2. Common Pitfalls: Misreading PR as 14 by adding 6 and 8; PR is hypotenuse √(6^2+8^2)=10. Using QR with wrong right-angle assumption; right angle is at P. Final Answer: 120 cm2

Question 19

In triangle △ABC, points D and E are the midpoints of AB and AC respectively. What is the ratio of areas ar(△ADE) : ar(△ABC)?

Accepted Answer

Introduction / Context: This is a standard midpoint/similarity fact in triangles. Joining a vertex to the midpoints of the two adjacent sides creates a triangle similar to the original with a linear scale factor 1/2, so its area is 1/4 of the original. Given Data / Assumptions: D is the midpoint of AB and E is the midpoint of AC. Segment DE is parallel to BC (Midpoint Theorem). △ADE is formed by joining A to D and E. Concept / Approach: By the Midpoint Theorem, DE ∥ BC and AD/AB = AE/AC = 1/2. Thus △ADE ~ △ABC with similarity ratio (linear) k = 1/2. Area scales as k^2, hence area(△ADE) = (1/2)^2 * area(△ABC) = 1/4 area(△ABC). Step-by-Step Solution: AD = AB/2, AE = AC/2 ⇒ similarity ratio = 1/2Area ratio = (1/2)^2 = 1/4 Verification / Alternative check: Pick coordinates: A(0,0), B(2,0), C(0,2). Then D(1,0), E(0,1). Area(△ABC)=2, Area(△ADE)=0.5 ⇒ ratio 1:4. Why Other Options Are Wrong: 1:2 and 3:4 overestimate the smaller triangle. 1:8 underestimates; that would correspond to a linear scale of 1/√8, not 1/2. None of these: Incorrect because 1:4 is exact. Common Pitfalls: Confusing length ratios (1/2) with area ratios (1/4). Forgetting that DE ∥ BC guarantees similarity. Final Answer: 1 : 4

Question 20

A 12 cm vertical stick casts an 8 cm shadow at the same time a tower casts a 40 m shadow. Assuming the same sun elevation (similar triangles), find the height of the tower (in metres).

Accepted Answer

Introduction / Context: When two vertical objects cast shadows simultaneously, the triangles formed by height and shadow length are similar (same sun elevation). Therefore, height is proportional to shadow length; we scale from the stick to the tower. Given Data / Assumptions: Stick height = 12 cm; its shadow = 8 cm. Tower shadow = 40 m. Both are vertical; ground is level; same sunlight angle (similarity). Concept / Approach: For similar triangles: height/shadow is constant. Convert units consistently (use metres). Step-by-Step Solution: Convert: 12 cm = 0.12 m, 8 cm = 0.08 mHeight ratio = 0.12 / 0.08 = 3/2 = 1.5Tower height H = 1.5 * (tower shadow) = 1.5 * 40 = 60 m Verification / Alternative check: Proportion in centimetres also works: 12/8 = H_cm / 4000 ⇒ H_cm = 6000 cm ⇒ 60 m. Why Other Options Are Wrong: 600 m and 160 m: Incorrect scaling or unit mistakes. 52 m: Random deviation; not in 3:2 ratio. None of these: Not applicable; 60 m is exact. Common Pitfalls: Mismatched units (cm with m) causing wrong scale. Assuming non-parallel sun rays; here standard assumption holds. Final Answer: 60 m

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