CBSE Class 9 Mathematics Plane Geometry I MCQs Set 18

Answer: (a) 110°

Explanation:
1. Angle B = 180° − 60° − 50° = 70°.
2. D and E are mid-points, so DE ∥ BC (mid-point theorem).
3. ∠ADE = ∠B = 70° (corresponding angles).
4. ∠EDB = 180° − 70° = 110°.

Teacher's Note:
1. The mid-point theorem gives a line parallel to the third side.
2. Corresponding angles on parallel lines are equal; angles on a straight line add to 180°.

Answer: (d) 105°

Explanation:
1. Join A to P and extend it to meet BC. The exterior angle at P equals the sum of the two opposite interior angles in each small triangle.
2. ∠BPC = ∠BAP + ∠ABP + ∠CAP + ∠ACP = ∠BAC + ∠ABP + ∠ACP.
3. ∠BPC = 50° + 20° + 35° = 105°.

Teacher's Note:
1. For a point inside a triangle, the angle it subtends at a side equals the vertex angle plus the two base angles made by the joins.
2. An extra line through P is the standard trick.

Answer: (c) 15

Explanation:
1. Let the polygon have n sides; the angle sum is (n − 2) × 180°.
2. The missing angle lies between 0° and 180°, so 2180° < (n − 2) × 180° < 2360°.
3. This gives 12.1 < n − 2 < 13.1, so n − 2 = 13 and n = 15.
4. Check: 13 × 180° = 2340°, the missing angle is 160°, which is less than 180°.

Teacher's Note:
1. An interior angle of a convex polygon is always less than 180°.
2. Bracket the angle sum between two multiples of 180°.

Answer: (a) \( 10\frac{10}{11} \) minutes past 5

Explanation:
1. At 5 o'clock the minute hand is 150° behind the hour hand (measured the long way round).
2. The minute hand gains 5.5° per minute on the hour hand.
3. For a right angle, 150° − 5.5t = 90°, so t = 60 ÷ 5.5 = 120/11 = 10 10/11 minutes.
4. So the hands are at right angles at \( 10\frac{10}{11} \) minutes past 5.

Teacher's Note:
1. Use relative speed: the minute hand moves 6° per minute and the hour hand 0.5° per minute.
2. Always set up the starting angle first, then the required angle.

Answer: (c) 37.5°

Explanation:
1. At 4:15 the minute hand is at 15 × 6° = 90° from 12.
2. The hour hand is at 4 × 30° + 15 × 0.5° = 127.5°.
3. Angle between them = 127.5° − 90° = 37.5°.

Teacher's Note:
1. The hour hand also moves during the extra minutes.
2. Hour hand: 30° per hour; minute hand: 6° per minute.

Answer: (d) All of these

Explanation:
1. The mid-point of the hypotenuse is the centre of the circumcircle of a right triangle.
2. So PA = PB = PC, each equal to the circumradius AC/2.
3. Hence all three statements are true.

Teacher's Note:
1. The hypotenuse is a diameter of the circle through the three vertices (angle in a semicircle).
2. Hence the median to the hypotenuse is half the hypotenuse.

Answer: (d) impossible

Explanation:
1. Take sides 4k, 6k, 11k.
2. The sum of two sides must exceed the third: 4k + 6k = 10k, which is less than 11k.
3. The triangle inequality fails, so no such triangle exists.

Teacher's Note:
1. Always test the triangle inequality first when ratios are given.
2. The two shorter sides together must be longer than the longest side.

Answer: (b) QR > PR > PQ

Explanation:
1. ∠P = 180° − 70° − 30° = 80°.
2. The larger the angle, the longer the opposite side: ∠P = 80° (opposite QR) > ∠Q = 70° (opposite PR) > ∠R = 30° (opposite PQ).
3. Hence QR > PR > PQ.

Teacher's Note:
1. The side opposite the greatest angle is the longest.
2. The bisector PS is extra information that is not needed.

Answer: (b) I and III are correct

Explanation:
1. In a right triangle the altitude to the hypotenuse divides it into two triangles similar to each other and to the whole triangle.
2. So I and III are true.
3. The two small triangles are congruent only if PQ = PR, which is not given, so II is false.

Teacher's Note:
1. Similar by AA: each small triangle shares an acute angle with the big one and has a right angle.
2. Similarity does not imply congruence.

Answer: (b) 3

Explanation:
1. Take BL = 1, then CL = AL = 2. Place L at the origin: B(1, 0), C(−2, 0), A(0, 2).
2. AB = √5. Area of ABC = ½ × BC × AL = ½ × 3 × 2 = 3, so CM = 2 × 3 ÷ √5 = 6/√5.
3. AM² = AC² − CM² = 8 − 36/5 = 4/5, so AM = 2/√5.
4. MC/AM = (6/√5) ÷ (2/√5) = 3.

Teacher's Note:
1. Coordinates make perpendiculars easy.
2. Use area to find an altitude, then Pythagoras for the rest.

Answer: (b) 18 cm

Explanation:
1. AE = 2 parts and AB = 3 parts.
2. Triangle ADE ~ triangle ACB (DE ∥ BC), so DE/BC = AE/AB = 2/3.
3. BC = 12 × 3/2 = 18 cm.

Teacher's Note:
1. Equal division of a side gives ratios of 1/3 and 2/3.
2. Parallel lines give similar triangles with a common vertex.

Answer: (a) 1/6

Explanation:
1. Let AB = 3a and BC = h. Then EF = a.
2. Area of triangle CEF = ½ × EF × h = ah/2.
3. Area of rectangle = 3a × h = 3ah.
4. Ratio = (ah/2) ÷ 3ah = 1/6.

Teacher's Note:
1. The triangle CEF has base EF on AB and height equal to the rectangle's width.
2. Express everything in terms of the same two lengths.

Answer: (c) 2 : 1

Explanation:
1. DE ∥ AB gives CD/CA = DE/AB = 10/15 = 2/3, so CA = 12 and x = AD = 4.
2. DF ∥ AG gives CF/CG = CD/CA = 2/3, so CF = 4 and y = FG = 6 − 4 = 2.
3. x : y = 4 : 2 = 2 : 1.

Teacher's Note:
1. Apply the basic proportionality theorem twice.
2. Keep track of which segment is the 'whole' (CA, CG).

Answer: (a) 13 cm

Explanation:
1. Triangles CED and CAB share ∠C and have ∠CED = ∠CAB, so they are similar with C ↔ C, E ↔ A, D ↔ B.
2. CE/CA = CD/CB = ED/AB. CB = CE + EB = 12, so CD/CB = 8/12 = 2/3.
3. CA = 10 × 3/2 = 15, so b = AD = 15 − 8 = 7. DE = 9 × 2/3 = 6, so a = 6.
4. a + b = 13 cm.

Teacher's Note:
1. Match the equal angles first, then the sides.
2. The ratio 2/3 is common to all three pairs of corresponding sides.

Answer: (a) 20 : 15 : 12

Explanation:
1. Area = ½ × BC × AP = ½ × AC × BQ = ½ × AB × CR.
2. So BC × 3 = AC × 4 = AB × 5, which makes the sides inversely proportional to the altitudes.
3. BC : AC : AB = 1/3 : 1/4 : 1/5 = 20 : 15 : 12.

Teacher's Note:
1. Equal area means side × altitude is constant.
2. Multiply by the LCM (60) to clear fractions.

Answer: (b) 337.5

Explanation:
1. For any triangle, the sum of the squares of the medians = ¾ × (sum of the squares of the sides).
2. Sum of squares of sides = 81 + 144 + 225 = 450.
3. Required sum = ¾ × 450 = 337.5.

Teacher's Note:
1. This follows from Apollonius' theorem applied to each median.
2. You can verify by finding the three medians: 7.5, √(…), √(…).

Answer: (a) 1.5 cm

Explanation:
1. The altitude from A bisects BC, so the half-base is 2 cm.
2. h² = 2.5² − 2² = 6.25 − 4 = 2.25.
3. h = 1.5 cm.

Teacher's Note:
1. In an isosceles triangle the altitude bisects the base.
2. Use Pythagoras in one half.

Answer: (c) 1/y

Explanation:
1. Let AB = m and BC = n. In triangles CQB and CPA: y/x = n/(m + n).
2. In triangles AQB and ARC: y/z = m/(m + n).
3. Adding: y/x + y/z = 1.
4. So 1/x + 1/z = 1/y.

Teacher's Note:
1. This is the classic 'crossed ladders' result.
2. Use similar triangles twice and add.

Answer: (b) 23.2

Explanation:
1. Area of the triangle by Heron's formula: s = 24, area = √(24 × 14 × 7 × 3) = 84.
2. Height on the side 21 is 2 × 84 ÷ 21 = 8.
3. Side of inscribed square = (base × height)/(base + height) = 21 × 8/29 = 168/29 ≈ 5.79.
4. Perimeter = 4 × 168/29 ≈ 23.2.

Teacher's Note:
1. An inscribed square cuts the triangle into similar pieces; that gives the standard formula.
2. Heron's formula gives the height easily.

Answer: (c) 25 : 81

Explanation:
1. AD/AB = 5/9, so DE/BC = 5/9.
2. Triangles DFE and CFB are similar (DE ∥ BC, vertically opposite angles at F).
3. The ratio of their areas = (5/9)² = 25/81.

Teacher's Note:
1. The areas of similar triangles are in the ratio of the squares of corresponding sides.
2. DE/BC = AD/AB from the parallel line.

Answer: (b) \( \frac{12\sqrt{3}}{7} \)

Explanation:
1. Area of ABC = ½ × 4 × 3 × sin 60° = 3√3.
2. The area equals the sum of the areas of ABD and ACD: ½ × 4 × AD × sin 30° + ½ × 3 × AD × sin 30° = (7/4) AD.
3. So (7/4) AD = 3√3 and AD = 12√3/7.

Teacher's Note:
1. Splitting an area along an angle bisector is a very handy method.
2. sin 30° = ½.

Answer: (b) AC² = AD² + 5CD²

Explanation:
1. BC = BD + DC = 3 CD.
2. AC² = AB² + BC² and AD² = AB² + BD².
3. AC² − AD² = BC² − BD² = 9CD² − 4CD² = 5CD².
4. So AC² = AD² + 5CD².

Teacher's Note:
1. Subtract the two Pythagoras relations to remove AB.
2. Express everything in terms of CD.

Answer: (b) 26

Explanation:
1. Let OX = a, OY = b. XN² = a² + b²/4 = 361 and YM² = b² + a²/4 = 484.
2. Adding: (5/4)(a² + b²) = 845, so a² + b² = 676.
3. XY² = a² + b² = 676, so XY = 26.

Teacher's Note:
1. Write each median in terms of the legs and add.
2. XY is the hypotenuse, so XY² = a² + b².

Answer: (b) 18

Explanation:
1. By the angle bisector theorem, AC/AB = CD/DB = 15/9 = 5/3.
2. Let AB = 3k and AC = 5k; then BC = 4k (3-4-5 triangle). BC = CD + DB = 24, so k = 6.
3. AB = 3k = 18.

Teacher's Note:
1. The angle bisector divides the opposite side in the ratio of the other two sides.
2. The ratio 3 : 5 gives the 3-4-5 right triangle.