IMO Maths Olympiad Class 10 Chapter 06 Triangles MCQs with Answers

Answer: (d) DC × AC

Explanation:
1. △ABC ~ △DCB matches A ↔ D, B ↔ C and C ↔ B.
2. So the corresponding sides are in proportion: \( \dfrac{AB}{DC} = \dfrac{AC}{DB} \).
3. Cross-multiplying, AB × DB = DC × AC.

Teacher's Note:
1. Read the vertex order in the similarity statement to pair the sides correctly.
2. Write the ratio first, then cross-multiply.

Answer: (d) AE² + BF² + CD²

Explanation:
1. By Pythagoras in △ODB and △ODC: OB² − OC² = BD² − CD² …(i)
2. Similarly OC² − OA² = CE² − AE² …(ii) and OA² − OB² = AF² − BF² …(iii)
3. Adding: 0 = BD² + CE² + AF² − CD² − AE² − BF², so AF² + BD² + CE² = AE² + BF² + CD².

Teacher's Note:
1. Write each length from O using Pythagoras on both sides of a foot of a perpendicular.
2. Adding the three differences makes the OA², OB², OC² terms cancel.

Answer: (a) \( \dfrac{1}{p^2} \)

Explanation:
1. Area of △ABC = \( \dfrac{1}{2} \times c \times p = \dfrac{1}{2} \times a \times b \), so cp = ab and \( \dfrac{1}{p^2} = \dfrac{c^2}{a^2b^2} \).
2. By Pythagoras, c² = a² + b², so \( \dfrac{1}{p^2} = \dfrac{a^2+b^2}{a^2b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} \).
3. Hence the value is \( \dfrac{1}{p^2} \).

Teacher's Note:
1. Writing the area in two ways (using either leg or the hypotenuse as base) is a powerful trick.
2. Then substitute c² = a² + b².

Answer: (b) BD × CD

Explanation:
1. ∠DAC = 90° − ∠BAD = ∠ABD.
2. In △BDA and △ADC: ∠ABD = ∠CAD and ∠BDA = ∠ADC = 90°, so △BDA ~ △ADC (AA).
3. \( \dfrac{BD}{AD} = \dfrac{AD}{DC} \), so AD² = BD × CD.

Teacher's Note:
1. The altitude to the hypotenuse splits a right triangle into two triangles similar to each other and to the whole.
2. The altitude is the geometric mean of the two segments of the hypotenuse.

Answer: (c) 21 sq.cm

Explanation:
1. ∠AOB = ∠COD (vertically opposite) and ∠OAB = ∠OCD (alternate angles), so △AOB ~ △COD.
2. \( \dfrac{\text{ar}(\triangle AOB)}{\text{ar}(\triangle COD)} = \dfrac{AB^2}{CD^2} = \dfrac{(2CD)^2}{CD^2} = 4 \).
3. ar(△COD) = 84 ÷ 4 = 21 sq.cm.

Teacher's Note:
1. The ratio of areas of similar triangles is the square of the ratio of sides.
2. A common slip is to divide by 2 instead of 4.

Answer: (c) 48 cm

Explanation:
1. The light is 15 cm from the rod and 15 + 45 = 60 cm from the wall.
2. The rod and its shadow form similar triangles with the light as the common vertex, so \( \dfrac{\text{shadow}}{12} = \dfrac{60}{15} = 4 \).
3. Length of the shadow = 4 × 12 = 48 cm.

Teacher's Note:
1. Measure both distances from the light source, not from the rod.
2. The shadow is enlarged by the ratio of the distances.

Answer: (c) (i) 8.75 m, (ii) 12 m

Explanation:
1. Since CD ∥ AB, △RCD ~ △RBA (alternate angles), with RC ↔ RB and CD ↔ BA.
2. (i) \( \dfrac{AB}{CD} = \dfrac{RB}{RC} \Rightarrow AB = \dfrac{7.5 \times 1.4}{1.2} \) = 8.75 m.
3. (ii) In right △CRD, RD² = 1.4² − 1.2² = 0.52, so RD ≈ 0.72 m and \( AR = \dfrac{7.5 \times 0.72}{1.2} \) ≈ 4.5 m.
4. Distance through R = AR + RB ≈ 4.5 + 7.5 = 12 m.

Teacher's Note:
1. Use the parallel lines to get equal alternate angles and hence similar triangles.
2. The right angle at R lets you find RD by Pythagoras.

Answer: (b) \( 2\sqrt{5} \) cm

Explanation:
1. In right △ABD: \( AD^2 = AB^2 + \dfrac{BC^2}{4} \); in right △EBC: \( CE^2 = BC^2 + \dfrac{AB^2}{4} \).
2. Adding: \( AD^2 + CE^2 = \dfrac{5}{4}\left(AB^2 + BC^2\right) = \dfrac{5}{4}AC^2 = \dfrac{5}{4} \times 25 \).
3. \( \dfrac{45}{4} + CE^2 = \dfrac{125}{4} \Rightarrow CE^2 = 20 \), so CE = \( 2\sqrt{5} \) cm.

Teacher's Note:
1. A median to a leg of a right triangle meets it at the mid-point, so half that leg appears in Pythagoras.
2. Adding the two equations avoids finding AB and BC separately.

Answer: (b) (P) → (2), (Q) → (1), (R) → (3), (S) → (4)

Explanation:
1. (P) Two sides in proportion and the included angle equal: SAS similarity → (2).
2. (Q) Two angles equal: AA similarity → (1).
3. (R) All three sides in proportion: SSS similarity → (3).
4. (S) A line parallel to one side divides the other two sides in the same ratio: BPT → (4).

Teacher's Note:
1. In SAS similarity the equal angle must lie between the two proportional sides.
2. BPT is about one triangle cut by a parallel line, not about two triangles.

Answer: (b) (i) \( \left(2-\sqrt{2}\right):2 \), (ii) \( \sqrt{2} \)

Explanation:
1. XY ∥ AC gives △XBY ~ △ABC (AA), and ar(△ABC) = 2 × ar(△XBY).
2. So \( \dfrac{AB^2}{XB^2} = 2 \Rightarrow AB = \sqrt{2}\,XB \), hence \( \dfrac{AC}{XY} = \dfrac{AB}{XB} = \sqrt{2} \).
3. \( \dfrac{AX}{AB} = \dfrac{AB - XB}{AB} = 1 - \dfrac{1}{\sqrt{2}} = \dfrac{2-\sqrt{2}}{2} \), so AX : AB = \( \left(2-\sqrt{2}\right):2 \).

Teacher's Note:
1. Equal halves by area means the area ratio of the small and whole triangles is 1 : 2, so the side ratio is 1 : √2.
2. AC/XY is the ratio of corresponding sides, so it equals AB/XB.