15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
Q1
·1 mark·Multiple choiceSimilar figures
Two polygons with the same number of sides are similar when
(a)their corresponding angles are equal, whatever their sides
(b)their corresponding sides are proportional, whatever their angles
(c)their corresponding angles are equal and their corresponding sides are proportional
(d)their perimeters are equal
Show answer
Answer: (c) their corresponding angles are equal and their corresponding sides are proportional
Why: Polygons in general need both conditions; only for triangles does one imply the other.
A square and a non-square rectangle have equal angles but are not similar; a square and a non-square rhombus have proportional sides but are not similar. Both conditions are needed.
Q2
·1 mark·Multiple choiceSimilar figures
Which of these pairs of figures need not be similar?
(a)Two circles
(b)Two squares
(c)Two equilateral triangles
(d)Two rhombuses
Show answer
Answer: (d) Two rhombuses
Why: Two rhombuses can have different angles (e.g. a square and a 60°–120° rhombus).
All circles, all squares and all equilateral triangles are similar. Two rhombuses with equal sides can have different angles, so they need not be similar.
Q3
·1 mark·Multiple choiceBasic Proportionality Theorem and its converse
In \(\triangle ABC\), \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\). If \(AD = 5\) cm, \(DB = 8\) cm and \(AE = 10\) cm, then \(EC\) is
\(\triangle ABC \sim \triangle DEF\) with \(AB = 8\) cm, \(DE = 12\) cm and \(BC = 10\) cm. Then \(EF\) is
(a)\(12\) cm
(b)\(20\) cm
(c)\(15\) cm
(d)\(18\) cm
Show answer
Answer: (c) \(15\) cm
Why: Corresponding sides are in the ratio DE/AB = 3/2.
\(\dfrac{EF}{BC} = \dfrac{DE}{AB} = \dfrac{12}{8} = \dfrac32\), so \(EF = \dfrac32 \times 10 = 15\) cm.
Q6
·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)
A triangle with sides \(6\) cm, \(8\) cm and \(10\) cm is similar to a triangle whose shortest side is \(9\) cm. The longest side of the second triangle is
(a)\(12\) cm
(b)\(13\) cm
(c)\(18\) cm
(d)\(15\) cm
Show answer
Answer: (d) \(15\) cm
Why: Scale factor 9/6 = 3/2 applies to every side.
Scale factor \(= \dfrac96 = \dfrac32\). Longest side \(= \dfrac32 \times 10 = 15\) cm.
Q7
·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)
At the same moment, a girl \(1.5\) m tall casts a shadow \(2\) m long and a water tank casts a shadow \(24\) m long. The height of the tank is
(a)\(18\) m
(b)\(32\) m
(c)\(16\) m
(d)\(24\) m
Show answer
Answer: (a) \(18\) m
Why: The sun's rays make equal angles, so the two triangles are similar (AA).
\(\dfrac{h}{24} = \dfrac{1.5}{2} \Rightarrow h = 24 \times \dfrac34 = 18\) m.
Q8
·1 mark·Multiple choiceBasic Proportionality Theorem and its converse
In \(\triangle ABC\), \(DE \parallel BC\) with \(D\) on \(AB\), \(E\) on \(AC\) and \(\dfrac{AD}{DB} = \dfrac23\). Then \(\dfrac{AE}{AC}\) equals
(a)\(\dfrac23\)
(b)\(\dfrac35\)
(c)\(\dfrac25\)
(d)\(\dfrac32\)
Show answer
Answer: (c) \(\dfrac25\)
Why: AE/EC = 2/3, so AE is 2 parts out of 2 + 3 = 5.
By BPT \(\dfrac{AE}{EC} = \dfrac23\). So \(\dfrac{AE}{AC} = \dfrac{2}{2 + 3} = \dfrac25\).
Q9
·1 mark·Multiple choiceBasic Proportionality Theorem and its converse
In \(\triangle PQR\), \(D\) is on \(PQ\) and \(E\) is on \(PR\) with \(PD = 3\) cm, \(DQ = 4.5\) cm, \(PE = 4\) cm and \(ER = 6\) cm. Then
(a)\(DE \parallel QR\)
(b)\(DE \perp QR\)
(c)\(DE = QR\)
(d)nothing can be said about \(DE\) and \(QR\)
Show answer
Answer: (a) \(DE \parallel QR\)
Why: PD/DQ = 3/4.5 = 2/3 = PE/ER, so the converse of BPT applies.
\(\dfrac{PD}{DQ} = \dfrac{3}{4.5} = \dfrac23\) and \(\dfrac{PE}{ER} = \dfrac46 = \dfrac23\). A line dividing two sides in the same ratio is parallel to the third side: \(DE \parallel QR\).
Q10
·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)
Which of the following is not a criterion for the similarity of two triangles?
(a)AA
(b)SSS
(c)SAS
(d)SSA
Show answer
Answer: (d) SSA
Why: Two sides in proportion with a non-included equal angle does not force similarity.
AA, SSS and SAS are similarity criteria. SSA is not: two proportional sides and an equal angle that is not between them can give triangles of different shapes.
Q11
·1 mark·Multiple choiceSimilar figures
If \(\triangle ABC \sim \triangle QRP\), which of the following is correct?
(a)\(\dfrac{AB}{PQ} = \dfrac{BC}{QR}\)
(b)\(\dfrac{AB}{QR} = \dfrac{BC}{RP}\)
(c)\(\dfrac{AB}{RP} = \dfrac{BC}{PQ}\)
(d)\(\dfrac{AB}{QR} = \dfrac{BC}{PQ}\)
Show answer
Answer: (b) \(\dfrac{AB}{QR} = \dfrac{BC}{RP}\)
Why: The order of letters gives A↔Q, B↔R, C↔P, so AB↔QR and BC↔RP.
\(\dfrac{x}{x - 2} = \dfrac{x + 2}{x - 1} \Rightarrow x^2 - x = x^2 - 4 \Rightarrow x = 4\). (Then \(DB = 2\), \(EC = 3\), both positive.)
Q13
·1 mark·Multiple choiceCriteria for similarity (AA, SSS, SAS)
In \(\triangle ABC\), \(AB = 3\) cm, \(AC = 5\) cm and \(\angle A = 70^\circ\). In \(\triangle PQR\), \(PQ = 6\) cm, \(PR = 10\) cm and \(\angle P = 70^\circ\). Which statement is true?
(a)\(\triangle ABC \sim \triangle PRQ\) by SAS
(b)\(\triangle ABC \sim \triangle PQR\) by SAS
(c)\(\triangle ABC \cong \triangle PQR\)
(d)The triangles are not similar
Show answer
Answer: (b) \(\triangle ABC \sim \triangle PQR\) by SAS
Why: AB/PQ = AC/PR = 1/2 and the included angles are equal.
\(\dfrac{AB}{PQ} = \dfrac36 = \dfrac12\), \(\dfrac{AC}{PR} = \dfrac{5}{10} = \dfrac12\) and the included angles \(\angle A = \angle P\). So \(\triangle ABC \sim \triangle PQR\) (SAS), with \(A \leftrightarrow P\), \(B \leftrightarrow Q\), \(C \leftrightarrow R\). They are not congruent (ratio \(\tfrac12\)).
Q14
·1 mark·Multiple choiceBasic Proportionality Theorem and its converse
In \(\triangle PQR\), \(S\) and \(T\) lie on \(PQ\) and \(PR\) with \(ST \parallel QR\). If \(PS = 2\) cm, \(SQ = 3\) cm and \(ST = 4\) cm, then \(QR\) is
(a)\(6\) cm
(b)\(10\) cm
(c)\(8\) cm
(d)\(12\) cm
Show answer
Answer: (b) \(10\) cm
Why: △PST ∼ △PQR (AA), so ST/QR = PS/PQ = 2/5.
\(\triangle PST \sim \triangle PQR\), so \(\dfrac{ST}{QR} = \dfrac{PS}{PQ} = \dfrac25\). \(QR = 4 \times \dfrac52 = 10\) cm. (Not \(\tfrac{PS}{SQ}\): that is the BPT ratio of the side pieces.)
Q15
·1 mark·Multiple choiceSimilar figures
\(\triangle ABC \sim \triangle DEF\) with \(AB = 4\) cm, \(DE = 6\) cm, \(EF = 9\) cm and \(FD = 12\) cm. The perimeter of \(\triangle ABC\) is
(a)\(27\) cm
(b)\(20\) cm
(c)\(18\) cm
(d)\(16\) cm
Show answer
Answer: (c) \(18\) cm
Why: Scale factor AB/DE = 2/3, so BC = 6 and CA = 8.
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
Case study 1: Pinhole camera (4 marks)
In a science project, Kabir builds a pinhole camera. Light from the top \(A\) and bottom \(B\) of a tree passes through the pinhole \(O\) and forms an inverted image \(A'B'\) on a screen \(20\) cm behind the pinhole. The tree is \(9\) m tall and stands \(30\) m in front of the pinhole, and the tree and the screen are both vertical, so \(AB \parallel A'B'\).
(i) Which similarity criterion shows that \(\triangle OAB \sim \triangle OA'B'\)? [1 mark]
(ii) Find the ratio of the height of the image to the height of the tree. [1 mark]
Show answer
Answer: \(1 : 150\)
\(\triangle OAB \sim \triangle OA'B'\), so their corresponding altitudes (the perpendicular distances of \(O\) from \(AB\) and from \(A'B'\)) are in the same ratio as corresponding sides: \(\dfrac{0.2}{30} = \dfrac{1}{150}\). A1
(iii) Find the height of the image in centimetres. [2 marks]
Show answer
Answer: \(6\) cm
\(\dfrac{A'B'}{9} = \dfrac{1}{150}\) M1, so \(A'B' = \dfrac{9}{150}\) m \(= 0.06\) m \(= 6\) cm. A1
OR Kabir moves back so that the tree is \(45\) m from the pinhole (the screen stays \(20\) cm behind it). Find the new height of the image. [2 marks]
Show answer
Answer: \(4\) cm
\(\dfrac{A'B'}{900 \text{ cm}} = \dfrac{20}{4500}\) M1, so \(A'B' = 4\) cm. A1
Case study 2: Width of a river (4 marks)
To find the width of a river without crossing it, Ravi stands at \(Q\) on one bank, exactly opposite a tree \(P\) on the far bank. He walks \(12\) m along the bank to a point \(R\), puts a stick there, walks a further \(4\) m to \(S\), and then walks \(6\) m away from the river at right angles to the bank to a point \(T\) from which the stick \(R\) and the tree \(P\) are in a straight line. \(PQ\) and \(TS\) are both perpendicular to the bank \(QS\).
(i) Name the criterion by which \(\triangle PQR \sim \triangle TSR\). [1 mark]
Show answer
Answer: AA
\(\angle PQR = \angle TSR = 90^\circ\) and \(\angle PRQ = \angle TRS\) (vertically opposite), so AA. A1
(ii) Write the equal ratios of corresponding sides. [1 mark]
(iii) Find the width \(PQ\) of the river. [2 marks]
Show answer
Answer: \(18\) m
\(\dfrac{PQ}{6} = \dfrac{12}{4}\) M1, so \(PQ = 18\) m. A1
OR If Ravi had walked \(5\) m (instead of \(4\) m) from \(R\) to \(S\), how far from the bank would he have had to walk for \(T\), \(R\), \(P\) to be in line? [2 marks]
Show answer
Answer: \(7.5\) m
\(\dfrac{PQ}{TS} = \dfrac{QR}{SR} \Rightarrow \dfrac{18}{TS} = \dfrac{12}{5}\) M1, so \(TS = \dfrac{90}{12} = 7.5\) m. A1
Next steps for Triangles
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Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.