CBSE Class 10 Mathematics Triangles Assignment Set 14

Previous Year Problems (CBSE Board)

Teacher's Note:
1. Match the common angle B and the two right angles.
2. Write the ratios with B as the first vertex.

Teacher's Note:
1. A median to a leg joins a vertex to the mid-point of the other leg.
2. AC² = AB² + BC² in the right triangle.

Question 3. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio. OR Prove that in a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. [4 marks, CBSE 2013, 2014, 2015]
Answer:
Basic proportionality theorem. In \( \triangle ABC \), let \( DE \parallel BC \) with D on AB and E on AC. Join BE and CD.
\( \dfrac{ar(ADE)}{ar(DBE)} = \dfrac{AD}{DB} \) and \( \dfrac{ar(ADE)}{ar(DEC)} = \dfrac{AE}{EC} \) (same height from E and D respectively).
\( \triangle DBE \) and \( \triangle DEC \) have the same base DE and lie between the same parallels DE and BC, so \( ar(DBE) = ar(DEC) \). Hence \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \).
Converse of the Pythagoras theorem. In \( \triangle ABC \), let \( AC^2 = AB^2 + BC^2 \). Draw \( \triangle PQR \) with \( \angle Q = 90^\circ \), PQ = AB and QR = BC. Then \( PR^2 = PQ^2 + QR^2 = AB^2 + BC^2 = AC^2 \), so PR = AC.
By SSS, \( \triangle ABC \cong \triangle PQR \), so \( \angle B = \angle Q = 90^\circ \).
In simple words: Equal heights make area ratios equal base ratios.

Teacher's Note:
1. Triangles on the same base between the same parallels have equal area.
2. For the converse, build a right triangle with the same two short sides.

Teacher's Note:
1. The ratio AE : EC equals AD : DB.

Teacher's Note:
1. Vertically opposite angles are equal.
2. Write the proportion with E as the common vertex.

Question 6. A triangle has sides 5 cm, 12 cm and 13 cm. Find the length to one decimal place of the perpendicular from the opposite vertex to the side whose length is 13 cm. [CBSE 2014]
Answer:
\( 5^2 + 12^2 = 13^2 \), so it is a right triangle with hypotenuse 13 cm. Its area is \( \dfrac{1}{2} \times 5 \times 12 = 30 \) cm².
\( \dfrac{1}{2} \times 13 \times h = 30 \Rightarrow h = \dfrac{60}{13} = \) 4.6 cm.
In simple words: Find the area two ways.

Teacher's Note:
1. Spot the 5-12-13 right triangle.
2. Area = half base × height with either side as base.

Teacher's Note:
1. The 6-8-10 triangle gives PR.
2. Then the 10-24-26 triangle gives PQ.

Question 8. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides. OR Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides. [CBSE 2014]
Answer:
Areas. Let \( \triangle ABC \sim \triangle PQR \). Draw \( AM \perp BC \) and \( PN \perp QR \). Then \( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{\frac{1}{2}BC \cdot AM}{\frac{1}{2}QR \cdot PN} = \dfrac{BC}{QR} \cdot \dfrac{AM}{PN} \).
\( \triangle ABM \sim \triangle PQN \) (AA), so \( \dfrac{AM}{PN} = \dfrac{AB}{PQ} = \dfrac{BC}{QR} \). Therefore \( \dfrac{ar(ABC)}{ar(PQR)} = \dfrac{BC^2}{QR^2} = \dfrac{AB^2}{PQ^2} = \dfrac{AC^2}{PR^2} \).
Pythagoras theorem. In \( \triangle ABC \), \( \angle B = 90^\circ \), draw \( BD \perp AC \). \( \triangle ADB \sim \triangle ABC \) gives \( AB^2 = AD \cdot AC \); \( \triangle BDC \sim \triangle ABC \) gives \( BC^2 = CD \cdot AC \).
Adding: \( AB^2 + BC^2 = AC(AD + CD) = AC^2 \).
In simple words: Similar triangles have heights in the same ratio as their bases.

Teacher's Note:
1. The altitude from the right angle splits the triangle into two similar ones.
2. Add the two relations to get the Pythagoras theorem.

Question 9. \( \triangle ABC \) and \( \triangle PQR \) are similar triangles such that \( \angle A = 32^\circ \) and \( \angle R = 65^\circ \). Then \( \angle B \) is [CBSE 2015]
(A) 83°   (B) 33°   (C) 63°   (D) 93°
Answer:
Ans. (A) 83°
\( \angle C = \angle R = 65^\circ \), so \( \angle B = 180^\circ - 32^\circ - 65^\circ = 83^\circ \).
In simple words: C corresponds to R.

Teacher's Note:
1. Use the order of the letters to match angles.
2. Then the angle sum.

Teacher's Note:
1. Use the basic proportionality theorem first.
2. Then add the equal segments.

Question 11. In a \( \triangle ABC \), P and Q are points on sides AB and AC respectively such that \( PQ \parallel BC \). If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ. [CBSE 2015]
Answer:
\( \dfrac{AP}{PB} = \dfrac{AQ}{QC} = \dfrac{2}{3} \Rightarrow PB = 3.6 \) cm, so \( AB = 2.4 + 3.6 = \) 6 cm.
\( \triangle APQ \sim \triangle ABC \): \( \dfrac{PQ}{BC} = \dfrac{AQ}{AC} = \dfrac{2}{5} \Rightarrow PQ = \dfrac{2}{5} \times 6 = \) 2.4 cm.
In simple words: AQ is 2 of the 5 parts of AC.

Teacher's Note:
1. AC = AQ + QC = 5 cm.
2. PQ : BC = AQ : AC.

Question 12. Find the length of an altitude of an equilateral triangle of side 2 cm. [CBSE 2015]
Answer:
\( h = \sqrt{2^2 - 1^2} = \) \( \sqrt{3} \) cm.
In simple words: The altitude bisects the base.

Teacher's Note:
1. Half the base is 1 cm.

Question 13. If ABC is an equilateral triangle with \( AD \perp BC \), then prove that \( AD^2 = 3DC^2 \). [CBSE 2016]
Answer:
In an equilateral triangle the altitude bisects the base, so \( BD = DC \) and \( AB = BC = 2DC \).
\( AD^2 = AB^2 - BD^2 = 4DC^2 - DC^2 = 3DC^2 \).
In simple words: Use the Pythagoras theorem in triangle ABD.

Teacher's Note:
1. The altitude of an equilateral triangle is also a median.

Teacher's Note:
1. Alternate angles between the parallel lines are equal.

Question 15. If \( \triangle ABC \cong \triangle RQP \), \( \angle A = 80^\circ \) and \( \angle B = 60^\circ \), the value of \( \angle P \) is [CBSE 2017]
(A) 80°   (B) 30°   (C) 40°   (D) 50°
Answer:
Ans. (C) 40°
A, B, C match R, Q, P. So \( \angle P = \angle C = 180^\circ - 80^\circ - 60^\circ = 40^\circ \).
In simple words: P corresponds to C.

Teacher's Note:
1. Match the letters in order.

Teacher's Note:
1. Both angles 1 and 3 equal 90° − ∠C.

Question 17. In an isosceles triangle ABC, AB = AC = 25 cm and BC = 14 cm. Calculate the altitude from A on BC. [CBSE 2017]
Answer:
The altitude bisects BC, so half the base is 7 cm. \( h = \sqrt{25^2 - 7^2} = \sqrt{576} = \) 24 cm.
In simple words: 7-24-25 right triangle.

Teacher's Note:
1. In an isosceles triangle the altitude from the apex bisects the base.

Teacher's Note:
1. Use PQ = PX + XQ to get PX.