Triangles: 4 marks Questions (CBSE Class 10)

4 different 4 marks questions on Triangles from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (70)2 marks (56)3 marks (21)4 marks (4)5 marks (59)

Q364 marksCase StudyTriangles CBSE 2026 · Standard 30/4/1

Carom board is a very popular game. The board is a square of side length 65 cm. It has circular pockets in each corner.
Ansh strikes a disc, kept at position P with a striker. The disc, hits the boundary of the board at R and goes straight to pocket at corner C. It is given that PS = 9 cm, PQ = 35 cm, BR = $x$ , $∠ P R Q = α$ and $∠ C R B = θ$ .
Based on the above information, answer the following questions :
(i) Using law of reflection i.e. $∠ P R T = ∠ C R T$ , prove that $θ = α$ . (1)
(ii) Prove that $△ P QR \sim △ C B R$ given that PQ is perpendicular to AB. (1)
(iii) (a) Find the value of $x$ using similarity of triangles. (2)
OR (b) If $\frac{Area △ P QR}{Area △ C B R} = \frac{P Q ^{2}}{C B ^{2}}$ , then find the value of $x$ .

Diagram for CBSE 2026 Class 10 Maths question 36

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Answer: (i) Proved. (ii) Proved. (iii) (a)

x = 36.4

cm OR (b)

x = 36.4

(i) RT is perpendicular to AB at R, so $θ = 9 0^{\circ} - ∠ C R T$ and $α = 9 0^{\circ} - ∠ P R T$ . As $∠ P R T = ∠ C R T$ , $θ = α$ .
(ii) In $△ P QR$ and $△ C B R$ : $∠ P QR = ∠ C B R = 9 0^{\circ}$ and $∠ P R Q = ∠ C R B$ (part (i)). So $△ P QR \sim △ C B R$ (AA).
(iii)(a) AQ = PS = 9 cm, so $QR = 65 - 9 - x = 56 - x$ .
$\frac{P Q}{C B} = \frac{QR}{B R} \Rightarrow \frac{35}{65} = \frac{56 - x}{x}$
$35 x = 3640 - 65 x \Rightarrow 100 x = 3640 \Rightarrow x = 36.4$ cm
(iii)(b) For similar triangles the area ratio equals the square of the ratio of corresponding sides, so $\frac{P Q ^{2}}{C B ^{2}} = \frac{Q R ^{2}}{B R ^{2}}$ , i.e. $\frac{35}{65} = \frac{56 - x}{x}$ , giving $x = 36.4$ cm.

Also asked in: 2026 Standard 30/4/2, 2026 Standard 30/4/3

Q364 marksCase StudyTriangles CBSE 2025 · Basic 430/6/1

A triangular window of a building is shown above. Its diagram represents a $△ A B C$ with $∠ A = 9 0^{\circ}$ and AB = AC. Points P and R trisect AB and PQ $∥$ RS $∥$ AC.
Based on the above, answer the following questions :
(i) Show that $△ B P Q \sim △ B A C$ . (1)
(ii) Prove that PQ $= \frac{1}{3}$ AC. (1)
(iii) (a) If AB = 3 m, find length BQ and BS. Verify that BQ $= \frac{1}{2}$ BS. (2)
OR
(iii) (b) Prove that BR $^{2}$ + RS $^{2}$ $= \frac{4}{9}$ BC $^{2}$ . (2)

Diagram for CBSE 2025 Class 10 Maths question 36

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Answer: (i) Proved (AA similarity) (ii) Proved (iii) (a) BQ

= 2

m, BS

= 22

m, so BQ

= \frac{1}{2}

BS OR (b) Proved

(i) In $△ B P Q$ and $△ B A C$ : $∠ B$ is common and $∠ B P Q = ∠ B A C$ (corresponding angles, PQ $∥$ AC). So $△ B P Q \sim △ B A C$ (AA).
(ii) From (i), $\frac{P Q}{A C} = \frac{B P}{B A}$ . P and R trisect AB with P nearer B, so BP $= \frac{1}{3}$ AB. Hence PQ $= \frac{1}{3}$ AC.
(iii) (a) AC = AB = 3 m, so BC $= 9 + 9 = 32$ m. From the similarity, BQ $= \frac{1}{3}$ BC $= 2$ m.
Similarly $△ B R S \sim △ B A C$ with BR $= \frac{2}{3}$ AB, so BS $= \frac{2}{3}$ BC $= 22$ m. Thus BQ $= \frac{1}{2}$ BS.
(iii) (b) $△ B R S \sim △ B A C$ with ratio $\frac{B R}{B A} = \frac{2}{3}$ , so BR $= \frac{2}{3}$ AB and RS $= \frac{2}{3}$ AC.
BR $^{2}$ + RS $^{2}$ $= \frac{4}{9}$ (AB $^{2}$ + AC $^{2}$ ) $= \frac{4}{9}$ BC $^{2}$ (Pythagoras in $△ A B C$ , right-angled at A).

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q384 marksCase StudyTriangles CBSE 2024 · Basic 430/3/1

In the figure given below, a folding table is shown :
The legs of the table are represented by line segments AB and CD intersecting at O. Join AC and BD.
Considering table top is parallel to the ground, and $O B = x$ , $O D = x + 3$ , $O C = 3 x + 19$ and $O A = 3 x + 4$ , answer the following questions :
(i) Prove that $△ O A C$ is similar to $△ O B D$ . (1)
(ii) Prove that $\frac{O A}{A C} = \frac{O B}{B D}$ . (1)
(iii) Observe the figure and find the value of $x$ . Hence, find the length of OC. (2)
OR Observe the figure and find $\frac{B D}{A C}$ . (2)

Diagram for CBSE 2024 Class 10 Maths question 38

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Answer: (i) Proved (AA) (ii) Proved (iii)

x = 2

O C = 25

; OR

\frac{B D}{A C} = \frac{1}{5}

(i) $D B ∥ A C$ , so $∠ O B D = ∠ O A C$ and $∠ O D B = ∠ O C A$ (alternate angles); also $∠ A O C = ∠ B O D$ (vertically opposite). Hence $△ O A C \sim △ O B D$ (AA).
(ii) From the similarity, $\frac{O A}{O B} = \frac{A C}{B D}$ , so $\frac{O A}{A C} = \frac{O B}{B D}$ .
(iii) $\frac{O A}{O B} = \frac{O C}{O D}$ : $\frac{3 x + 4}{x} = \frac{3 x + 19}{x + 3}$ .
$(3 x + 4) (x + 3) = x (3 x + 19)$ gives $3 x^{2} + 13 x + 12 = 3 x^{2} + 19 x$ , so $x = 2$ .
$O C = 3 (2) + 19 = 25$ .
OR: $\frac{B D}{A C} = \frac{O B}{O A} = \frac{2}{3 ( 2 ) + 4} = \frac{2}{10} = \frac{1}{5}$ .

Also asked in: 2024 Basic 430/3/2, 2024 Basic 430/3/3

Q384 marksCase StudyTriangles CBSE 2023 · Basic 430/4/1

Observe the figures given below carefully and answer the questions :
(i) Name the figure(s) wherein two figures are similar. (1)
(ii) Name the figure(s) wherein the figures are congruent. (1)
(iii) (a) Prove that congruent triangles are also similar but not the converse. (2)
OR (b) What more is least needed for two similar triangles to be congruent ? (2)

Diagram for CBSE 2023 Class 10 Maths question 38

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Answer: (i) Figure A (and Figure C, since congruent figures are also similar) (ii) Figure C (iii) (a) Proved. OR (b) At least one pair of corresponding sides must be equal.

(i) In Figure A, corresponding angles are equal ( $8 5^{\circ}$ , $7 0^{\circ}$ , $10 5^{\circ}$ , $10 0^{\circ}$ ) and sides are in the same ratio: $\frac{2.4}{4.8} = \frac{2}{4} = \frac{2.5}{5} = \frac{1.2}{2.4} = \frac{1}{2}$ , so A(i) and A(ii) are similar. In Figure B the sides are not proportional ( $\frac{3.5}{3} \neq = \frac{3}{3}$ ). Figure C triangles are congruent, hence also similar.
(ii) In Figure C, AB = PQ = 2 cm, BC = QR = 3 cm, AC = PR = 4.5 cm, so $△ A B C ≅ △ P QR$ (SSS).
(iii) (a) If $△ A B C ≅ △ P QR$ , then corresponding angles are equal and $\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} = 1$ , so the triangles are similar. Converse: an equilateral triangle of side 1 cm and one of side 2 cm are similar (equal angles) but not congruent, since their sides are unequal.
(iii) (b) Similar triangles already have equal angles and proportional sides; if one pair of corresponding sides is equal, the ratio is 1 and all sides are equal, so they are congruent.

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

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