CBSE Class 10 · Maths 
In the above figure, $\triangle ABC \sim \triangle QPR$. If $AC = 6$ cm, $BC = 5$ cm, $QR = 3$ cm and $PR = x$, then the value of $x$ is
CBSE Class 10 Maths Triangles Previous Year Questions
Help your child master CBSE Class 10 Maths Triangles Previous Year Questions with this curated collection sourced from real board papers spanning 2018–2023. Every question comes with a detailed step-by-step solution, helping your child confidently tackle the Basic Proportionality Theorem, AA/SAS/SSS similarity criteria, and properties of similar triangles — topics that consistently carry significant marks in the board exam.
CBSE Class 10 Maths Triangles — Questions with Solutions
$\triangle PQR$ is shown below. $ST$ is drawn such that $\angle PRQ = \angle STQ$. If $ST$ divides $QR$ in a ratio of $2:3$, then what is the length of $ST$? 
Solution
Answer: Option (B) is correct.
Explanation: In $\triangle PQR$ and $\triangle STQ$:
• $\angle Q = \angle Q$ (Common)
• $\angle PRQ = \angle STQ$ (Given)
By AA rule, $\triangle PQR \sim \triangle SQT$. Therefore: $$\frac{ST}{PR} = \frac{QT}{QR}$$ $$\frac{ST}{20} = \frac{2x}{5x} \Rightarrow ST = \frac{2 \times 20}{5} = 8 \text{ cm}$$
Explanation: In $\triangle PQR$ and $\triangle STQ$:
• $\angle Q = \angle Q$ (Common)
• $\angle PRQ = \angle STQ$ (Given)
By AA rule, $\triangle PQR \sim \triangle SQT$. Therefore: $$\frac{ST}{PR} = \frac{QT}{QR}$$ $$\frac{ST}{20} = \frac{2x}{5x} \Rightarrow ST = \frac{2 \times 20}{5} = 8 \text{ cm}$$
Two scalene triangles are given below: 
Anas and Rishi observed them and said the following:
Anas: $\triangle PQR$ is similar to $\triangle CBA$
Rishu: $\triangle PQR$ is congruent to $\triangle CBA$
Which of them is/are correct?
Anas and Rishi observed them and said the following:Anas: $\triangle PQR$ is similar to $\triangle CBA$
Rishu: $\triangle PQR$ is congruent to $\triangle CBA$
Which of them is/are correct?
Solution
Answer: Option (A) is correct.
Explanation: Only Anas is correct. In $\triangle PQR$ and $\triangle CBA$:
$\angle P = \angle C$ and $\angle R = \angle A$
Therefore, both triangles are similar by the AA rule. They are not congruent because the corresponding sides are not equal.
Explanation: Only Anas is correct. In $\triangle PQR$ and $\triangle CBA$:
$\angle P = \angle C$ and $\angle R = \angle A$
Therefore, both triangles are similar by the AA rule. They are not congruent because the corresponding sides are not equal.
In the above figure, $\triangle ABC \sim \triangle QPR$. If $AC = 6$ cm, $BC = 5$ cm, $QR = 3$ cm and $PR = x$, then the value of $x$ isSolution
Answer: Option (B) is correct.
Explanation: Since $\triangle ABC \sim \triangle QPR$: $$\frac{AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR}$$ $$\Rightarrow \frac{AB}{QP} = \frac{5}{x} = \frac{6}{3}$$ Equating the last two: $$x = \frac{5 \times 3}{6} = \frac{5}{2} = 2.5 \text{ cm}$$
Explanation: Since $\triangle ABC \sim \triangle QPR$: $$\frac{AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR}$$ $$\Rightarrow \frac{AB}{QP} = \frac{5}{x} = \frac{6}{3}$$ Equating the last two: $$x = \frac{5 \times 3}{6} = \frac{5}{2} = 2.5 \text{ cm}$$
If $\triangle PQR \sim \triangle ABC$, $PQ = 6$ cm, $AB = 8$ cm and the perimeter of $\triangle ABC$ is 36 cm, then the perimeter of $\triangle PQR$ is
Solution
Answer: Option (B) is correct.
Explanation: If two triangles are similar, the ratio of their corresponding sides equals the ratio of their respective perimeters. Therefore: $$\frac{PQ}{AB} = \frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle ABC}$$ $$\frac{6}{8} = \frac{\text{Perimeter of } \triangle PQR}{36}$$ $$\text{Perimeter of } \triangle PQR = \frac{6 \times 36}{8} = 27 \text{ cm}$$
Explanation: If two triangles are similar, the ratio of their corresponding sides equals the ratio of their respective perimeters. Therefore: $$\frac{PQ}{AB} = \frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle ABC}$$ $$\frac{6}{8} = \frac{\text{Perimeter of } \triangle PQR}{36}$$ $$\text{Perimeter of } \triangle PQR = \frac{6 \times 36}{8} = 27 \text{ cm}$$
If in two triangles $DEF$ and $PQR$, $\angle D = \angle Q$ and $\angle R = \angle E$, then which of the following is not true?
Solution
Answer: Option (B) is correct (i.e., Option B is the statement that is NOT true).
Explanation: In triangles $DEF$ and $PQR$, since $\angle D = \angle Q$ and $\angle R = \angle E$, by AA similarity:
$\triangle DEF \sim \triangle QRP$
Therefore: $\dfrac{EF}{PR} = \dfrac{DF}{PQ} = \dfrac{DE}{QR}$
Option (B) states $\dfrac{EF}{RP} = \dfrac{DE}{PQ}$, which does not follow from the above correspondence and is hence not true.
Explanation: In triangles $DEF$ and $PQR$, since $\angle D = \angle Q$ and $\angle R = \angle E$, by AA similarity:
$\triangle DEF \sim \triangle QRP$
Therefore: $\dfrac{EF}{PR} = \dfrac{DF}{PQ} = \dfrac{DE}{QR}$
Option (B) states $\dfrac{EF}{RP} = \dfrac{DE}{PQ}$, which does not follow from the above correspondence and is hence not true.
In the given figure, $PQRS$ is a parallelogram. If $AT = AQ = 6$ cm, $AS = 3$ cm and $TS = 4$ cm, then find $x$ and $y$. 
Solution
Answer: Option (D) is correct.
Explanation: Since $PQRS$ is a parallelogram, in $\triangle ATS$ and $\triangle APQ$:
• $\angle PQA = \angle ATS$ (Alternate angles)
• $\angle PAQ = \angle TAS$ (Vertically opposite angles)
By AA similarity, $\triangle ATS \sim \triangle AQP$. Therefore: $$\frac{PA}{AS} = \frac{PQ}{TS} = \frac{AQ}{AT}$$ $$\frac{x}{3} = \frac{y}{4} = \frac{6}{6} = 1$$ Clearly, $x = 3$ and $y = 4$.
Explanation: Since $PQRS$ is a parallelogram, in $\triangle ATS$ and $\triangle APQ$:
• $\angle PQA = \angle ATS$ (Alternate angles)
• $\angle PAQ = \angle TAS$ (Vertically opposite angles)
By AA similarity, $\triangle ATS \sim \triangle AQP$. Therefore: $$\frac{PA}{AS} = \frac{PQ}{TS} = \frac{AQ}{AT}$$ $$\frac{x}{3} = \frac{y}{4} = \frac{6}{6} = 1$$ Clearly, $x = 3$ and $y = 4$.
In the given figure, $DE \| BC$. If $AD = 3$ cm, $AB = 7$ cm and $EC = 3$ cm, then the length of $AE$ is 
Solution
Answer: Option (B) is correct.
Explanation: In $\triangle ABC$, since $DE \| BC$, by the Basic Proportionality Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Since $DB = AB – AD = 7 – 3 = 4$ cm: $$\frac{3}{4} = \frac{AE}{3} \Rightarrow AE = \frac{9}{4} = 2.25 \text{ cm}$$
Explanation: In $\triangle ABC$, since $DE \| BC$, by the Basic Proportionality Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Since $DB = AB – AD = 7 – 3 = 4$ cm: $$\frac{3}{4} = \frac{AE}{3} \Rightarrow AE = \frac{9}{4} = 2.25 \text{ cm}$$
In $\triangle ABC$ and $\triangle DEF$, $\angle F = \angle C$, $\angle B = \angle E$ and $AB = \dfrac{1}{2}DE$. Then the two triangles are
Solution
Answer: Option (B) is correct.
Explanation: Since $\angle F = \angle C$ and $\angle B = \angle E$, the two triangles are similar by the AA criterion. However, since $AB = \dfrac{1}{2}DE$, the corresponding sides are not equal, so the triangles are similar but not congruent.
Explanation: Since $\angle F = \angle C$ and $\angle B = \angle E$, the two triangles are similar by the AA criterion. However, since $AB = \dfrac{1}{2}DE$, the corresponding sides are not equal, so the triangles are similar but not congruent.
Shown below are three triangles. The measures of two adjacent sides and included angle are given for each triangle. 
(Note: The figure is not to scale.)
Which of these triangles are similar?
(Note: The figure is not to scale.)Which of these triangles are similar?
Solution
Answer: Option (A) is correct.
Explanation: In $\triangle RPQ$ and $\triangle XZY$: $$\frac{RP}{XZ} = \frac{6 \text{ cm}}{9 \text{ cm}} = \frac{2}{3}, \quad \frac{PQ}{ZY} = \frac{4 \text{ cm}}{6 \text{ cm}} = \frac{2}{3}$$ Since $\dfrac{RP}{XZ} = \dfrac{PQ}{ZY}$ and $\angle RPQ = \angle XZY = 60°$, by SAS similarity, $\triangle RPQ \sim \triangle XZY$.
Explanation: In $\triangle RPQ$ and $\triangle XZY$: $$\frac{RP}{XZ} = \frac{6 \text{ cm}}{9 \text{ cm}} = \frac{2}{3}, \quad \frac{PQ}{ZY} = \frac{4 \text{ cm}}{6 \text{ cm}} = \frac{2}{3}$$ Since $\dfrac{RP}{XZ} = \dfrac{PQ}{ZY}$ and $\angle RPQ = \angle XZY = 60°$, by SAS similarity, $\triangle RPQ \sim \triangle XZY$.
In the figure below, $DE \| AC$ and $DF \| AE$. Which of these is equal to $\dfrac{BF}{FE}$? 
Solution
Answer: Option (B) is correct.
Explanation: In $\triangle ABC$, since $DE \| AC$ (given), by BPT: $$\frac{BD}{DA} = \frac{BE}{EC} \quad \cdots (i)$$ In $\triangle ABE$, since $DF \| AE$ (given), by BPT: $$\frac{BD}{DA} = \frac{FB}{FE} \quad \cdots (ii)$$ From (i) and (ii): $$\frac{BF}{FE} = \frac{BE}{EC}$$
Explanation: In $\triangle ABC$, since $DE \| AC$ (given), by BPT: $$\frac{BD}{DA} = \frac{BE}{EC} \quad \cdots (i)$$ In $\triangle ABE$, since $DF \| AE$ (given), by BPT: $$\frac{BD}{DA} = \frac{FB}{FE} \quad \cdots (ii)$$ From (i) and (ii): $$\frac{BF}{FE} = \frac{BE}{EC}$$
$\triangle ABC \sim \triangle PQR$. If $AM$ and $PN$ are altitudes of $\triangle ABC$ and $\triangle PQR$ respectively and $AB^2 : PQ^2 = 4 : 9$, then $AM : PN =$
Solution
Answer: Option (D) is correct.
Explanation: Since $\triangle ABC \sim \triangle PQR$, the ratio of corresponding sides equals the ratio of corresponding altitudes: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{AM}{PN}$$ Given $\dfrac{AB^2}{PQ^2} = \dfrac{4}{9}$, so $\left(\dfrac{AB}{PQ}\right)^2 = \left(\dfrac{2}{3}\right)^2$, giving $\dfrac{AB}{PQ} = \dfrac{2}{3}$.
Therefore, $AM : PN = 2 : 3$.
Explanation: Since $\triangle ABC \sim \triangle PQR$, the ratio of corresponding sides equals the ratio of corresponding altitudes: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{AM}{PN}$$ Given $\dfrac{AB^2}{PQ^2} = \dfrac{4}{9}$, so $\left(\dfrac{AB}{PQ}\right)^2 = \left(\dfrac{2}{3}\right)^2$, giving $\dfrac{AB}{PQ} = \dfrac{2}{3}$.
Therefore, $AM : PN = 2 : 3$.
$ABCD$ is a trapezium with $AD \| BC$ and $AD = 4$ cm. If the diagonals $AC$ and $BD$ intersect each other at $O$ such that $\dfrac{AO}{OC} = \dfrac{DO}{OB} = \dfrac{1}{2}$, then $BC =$
Solution
Answer: Option (C) is correct.
Explanation: Since $AD \| BC$ and the diagonals intersect at $O$ with $\dfrac{AO}{OC} = \dfrac{DO}{OB} = \dfrac{1}{2}$, by the property of similar triangles in a trapezium: $$\frac{AD}{BC} = \frac{AO}{OC} = \frac{1}{2}$$ $$\frac{4}{BC} = \frac{1}{2} \Rightarrow BC = 8 \text{ cm}$$
Explanation: Since $AD \| BC$ and the diagonals intersect at $O$ with $\dfrac{AO}{OC} = \dfrac{DO}{OB} = \dfrac{1}{2}$, by the property of similar triangles in a trapezium: $$\frac{AD}{BC} = \frac{AO}{OC} = \frac{1}{2}$$ $$\frac{4}{BC} = \frac{1}{2} \Rightarrow BC = 8 \text{ cm}$$
In $\triangle ABC$, $DE \| AB$. If $AB = a$, $DE = x$, $BE = b$ and $EC = c$, express $x$ in terms of $a$, $b$ and $c$. 
Solution
Answer: Option (B) is correct.
Explanation: By the Basic Proportionality Theorem and properties of similar triangles: $$\frac{DE}{AB} = \frac{CE}{BC}$$ Here $CE = c$ and $BC = BE + EC = b + c$: $$\frac{x}{a} = \frac{c}{b+c} \Rightarrow x = \frac{ac}{b+c}$$
Explanation: By the Basic Proportionality Theorem and properties of similar triangles: $$\frac{DE}{AB} = \frac{CE}{BC}$$ Here $CE = c$ and $BC = BE + EC = b + c$: $$\frac{x}{a} = \frac{c}{b+c} \Rightarrow x = \frac{ac}{b+c}$$
Leela has a triangular cabinet that fits under his staircase. There are four parallel shelves as shown below. 
(Note: The figure is not to scale.)
The total height of the cabinet is 144 cm. What is the maximum height of a book that can stand upright on the bottom-most shelf?
(Note: The figure is not to scale.)The total height of the cabinet is 144 cm. What is the maximum height of a book that can stand upright on the bottom-most shelf?
Solution
Answer: Option (C) is correct.
Step 1 — Set up the equation (shelf heights in ratio $y : 2y : 2y : 3y$): $$y + 2y + 2y + 3y = 144$$ Step 2 — Solve for $y$: $$8y = 144 \Rightarrow y = 18$$ Step 3 — Bottom-most shelf height is $3y$: $$3y = 3 \times 18 = 54 \text{ cm}$$ The maximum height of a book on the bottom-most shelf is 54 cm.
Step 1 — Set up the equation (shelf heights in ratio $y : 2y : 2y : 3y$): $$y + 2y + 2y + 3y = 144$$ Step 2 — Solve for $y$: $$8y = 144 \Rightarrow y = 18$$ Step 3 — Bottom-most shelf height is $3y$: $$3y = 3 \times 18 = 54 \text{ cm}$$ The maximum height of a book on the bottom-most shelf is 54 cm.
In the following figure, $Q$ is a point on $PR$ and $S$ is a point on $TR$. $QS$ is drawn and $\angle RPT = \angle RQS$. 
Which of these criteria can be used to prove that $\triangle RSQ$ is similar to $\triangle RTP$?
Which of these criteria can be used to prove that $\triangle RSQ$ is similar to $\triangle RTP$?Solution
Answer: Option (A) is correct.
Explanation: In $\triangle RQS$ and $\triangle RPT$:
• $\angle RQS = \angle RPT$ (Given)
• $\angle R = \angle R$ (Common)
Since two pairs of angles are equal, by the AAA similarity criterion (reduces to AA), $\triangle RQS \sim \triangle RPT$.
Explanation: In $\triangle RQS$ and $\triangle RPT$:
• $\angle RQS = \angle RPT$ (Given)
• $\angle R = \angle R$ (Common)
Since two pairs of angles are equal, by the AAA similarity criterion (reduces to AA), $\triangle RQS \sim \triangle RPT$.
If two triangles are similar, then find the relation of their corresponding sides.
Answer
If two triangles are similar, then their corresponding sides are in the same ratio (i.e., they are proportional).
In the figure, if $\angle ACB = \angle CDA$, $AC = 6$ cm and $AD = 3$ cm, then find the length of $AB$. 
Answer
Since $\angle CDA = \angle ACB$ (given) and $\angle CAD = \angle CAB$ (common angle),
by AA Similarity: $\triangle ADC \sim \triangle ACB$.
Therefore: $$\frac{AC}{AB} = \frac{AD}{AC}$$ $$\frac{6}{AB} = \frac{3}{6} \Rightarrow AB = \frac{6 \times 6}{3} = 12 \text{ cm}$$
by AA Similarity: $\triangle ADC \sim \triangle ACB$.
Therefore: $$\frac{AC}{AB} = \frac{AD}{AC}$$ $$\frac{6}{AB} = \frac{3}{6} \Rightarrow AB = \frac{6 \times 6}{3} = 12 \text{ cm}$$
In the given figure, $DB \perp BC$, $DE \perp AB$ and $AC \perp BC$. Prove that $\dfrac{BE}{DE} = \dfrac{AC}{BC}$. 
Answer
$\angle DEB = \angle ACB = 90°$
Since $DB \perp BC$ and $DE \perp AB$:
$\angle ABC = 90° – \angle DBE$
$\angle BDE = 90° – \angle DBE$
Therefore, $\angle ABC = \angle BDE$
In $\triangle BDE$ and $\triangle ABC$:
• $\angle DEB = \angle ACB = 90°$
• $\angle ABС = \angle BDE$
By AA similarity, $\triangle BDE \sim \triangle ABC$.
Therefore, $\dfrac{BE}{DE} = \dfrac{AC}{BC}$. Hence Proved.
Since $DB \perp BC$ and $DE \perp AB$:
$\angle ABC = 90° – \angle DBE$
$\angle BDE = 90° – \angle DBE$
Therefore, $\angle ABC = \angle BDE$
In $\triangle BDE$ and $\triangle ABC$:
• $\angle DEB = \angle ACB = 90°$
• $\angle ABС = \angle BDE$
By AA similarity, $\triangle BDE \sim \triangle ABC$.
Therefore, $\dfrac{BE}{DE} = \dfrac{AC}{BC}$. Hence Proved.
In the given figure, $ABC$ is a triangle in which $DE \| BC$. If $AD = x$, $DB = x – 2$, $AE = x + 2$ and $EC = x – 1$, then find the value of $x$. 
Answer
In $\triangle ABC$, since $DE \| BC$, by the Basic Proportionality Theorem:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{x}{x-2} = \frac{x+2}{x-1}$$
$$x(x-1) = (x-2)(x+2)$$
$$x^2 – x = x^2 – 4 \quad \left[\because (a-b)(a+b) = a^2 – b^2\right]$$
$$-x = -4$$
$$\therefore x = 4$$
In the given figure, $E$ is a point on the side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, then prove that $\triangle ABD \sim \triangle ECF$. 
Answer
Since $AB = AC$ (isosceles $\triangle ABC$): $\angle B = \angle C \quad \cdots (i)$
Since $AD \perp BC$: $\angle ADB = 90° \quad \cdots (ii)$
Since $EF \perp AC$: $\angle EFC = 90° \quad \cdots (iii)$
In $\triangle ABD$ and $\triangle ECF$:
• $\angle B = \angle C$ [from (i)]
• $\angle ADB = \angle EFC = 90°$ [from (ii) and (iii)]
By AA similarity, $\triangle ABD \sim \triangle ECF$. Hence Proved.
Since $AD \perp BC$: $\angle ADB = 90° \quad \cdots (ii)$
Since $EF \perp AC$: $\angle EFC = 90° \quad \cdots (iii)$
In $\triangle ABD$ and $\triangle ECF$:
• $\angle B = \angle C$ [from (i)]
• $\angle ADB = \angle EFC = 90°$ [from (ii) and (iii)]
By AA similarity, $\triangle ABD \sim \triangle ECF$. Hence Proved.
The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm, find the length of the corresponding side of the second triangle.
Answer
Let the triangles be $\triangle ABC$ and $\triangle DEF$.
$AB + BC + CA = 25$ cm and $DE + EF + FD = 15$ cm.
Given $CA = 9$ cm. Let $FD = x$ cm.
Since the triangles are similar, the ratio of perimeters equals the ratio of corresponding sides: $$\frac{25}{15} = \frac{CA}{FD} = \frac{9}{x}$$ $$25x = 135 \Rightarrow x = \frac{135}{25} = 5.4 \text{ cm}$$ The length of the corresponding side of the second triangle is 5.4 cm.
$AB + BC + CA = 25$ cm and $DE + EF + FD = 15$ cm.
Given $CA = 9$ cm. Let $FD = x$ cm.
Since the triangles are similar, the ratio of perimeters equals the ratio of corresponding sides: $$\frac{25}{15} = \frac{CA}{FD} = \frac{9}{x}$$ $$25x = 135 \Rightarrow x = \frac{135}{25} = 5.4 \text{ cm}$$ The length of the corresponding side of the second triangle is 5.4 cm.
In the given figure, if $\triangle ABC \sim \triangle DEF$ and the lengths (in cm) of their sides are marked along them, then find the lengths of sides of each triangle. 
Answer
Since $\triangle ABC \sim \triangle DEF$, by the corresponding sides property:
$$\frac{AB}{BC} = \frac{DE}{EF} \Rightarrow \frac{2x-1}{2x+2} = \frac{18}{3x+9}$$
$$(2x-1)(3x+9) = 18(2x+2)$$
$$(2x-1)(x+3) = 6(2x+2)$$
$$2x^2 + 5x – 3 = 12x + 12$$
$$2x^2 – 7x – 15 = 0$$
$$(x-5)(2x+3) = 0$$
Either $x = 5$ or $x = -\dfrac{3}{2}$ (not valid, as lengths must be positive).
So $x = 5$.
In $\triangle ABC$: $AB = 2(5)-1 = 9$ cm, $BC = 2(5)+2 = 12$ cm, $AC = 3(5) = 15$ cm
In $\triangle DEF$: $DE = 18$ cm, $EF = 3(5)+9 = 24$ cm, $DF = 6(5) = 30$ cm
So $x = 5$.
In $\triangle ABC$: $AB = 2(5)-1 = 9$ cm, $BC = 2(5)+2 = 12$ cm, $AC = 3(5) = 15$ cm
In $\triangle DEF$: $DE = 18$ cm, $EF = 3(5)+9 = 24$ cm, $DF = 6(5) = 30$ cm
Shown below is a circle with centre $O$. $YX$ is the tangent to the circle at $Y$. 
(Note: The figure is not to scale.)
(i) Prove that $\triangle ZWY \sim \triangle ZYX$.
(ii) Using part (i), find the length of $ZY$.
(Note: The figure is not to scale.)(i) Prove that $\triangle ZWY \sim \triangle ZYX$.
(ii) Using part (i), find the length of $ZY$.
Answer
(i) In $\triangle XYZ$: $\angle ZYX = 90°$ (tangent at any point is perpendicular to the radius at the point of contact).
In $\triangle WYZ$: $\angle ZWY = 90°$ (angle in a semicircle is a right angle).
In $\triangle XYZ$ and $\triangle WYZ$:
• $\angle ZYX = \angle ZWY = 90°$ (proved above)
• $\angle Z = \angle Z$ (common)
By AA similarity, $\triangle XYZ \sim \triangle YWZ$, i.e., $\triangle ZWY \sim \triangle ZYX$. Hence Proved.
(ii) From the similarity, corresponding sides are proportional: $$\frac{ZX}{ZY} = \frac{ZY}{ZW}$$ $$\frac{25 + 11}{ZY} = \frac{ZY}{25}$$ $$ZY^2 = 36 \times 25 \Rightarrow ZY = 6 \times 5 = 30 \text{ cm}$$
In $\triangle WYZ$: $\angle ZWY = 90°$ (angle in a semicircle is a right angle).
In $\triangle XYZ$ and $\triangle WYZ$:
• $\angle ZYX = \angle ZWY = 90°$ (proved above)
• $\angle Z = \angle Z$ (common)
By AA similarity, $\triangle XYZ \sim \triangle YWZ$, i.e., $\triangle ZWY \sim \triangle ZYX$. Hence Proved.
(ii) From the similarity, corresponding sides are proportional: $$\frac{ZX}{ZY} = \frac{ZY}{ZW}$$ $$\frac{25 + 11}{ZY} = \frac{ZY}{25}$$ $$ZY^2 = 36 \times 25 \Rightarrow ZY = 6 \times 5 = 30 \text{ cm}$$
In the below figure, $QR = 4$ cm, $RP = 8$ cm and $ST = 6$ cm. 
(Note: The figure is not to scale.)
If the perimeter of $\triangle STU$ is 27 cm, find the length of $PQ$. Show your steps.
(Note: The figure is not to scale.)If the perimeter of $\triangle STU$ is 27 cm, find the length of $PQ$. Show your steps.
Answer
In $\triangle PQR$ and $\triangle UTS$: $\angle P = \angle U$ and $\angle Q = \angle T$ (given).
By AA similarity, $\triangle PQR \sim \triangle UTS$.
Since the ratio of perimeters of two similar triangles equals the ratio of their corresponding sides: $$\frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle STU} = \frac{QR}{ST}$$ $$\frac{\text{Perimeter of } \triangle PQR}{27} = \frac{4}{6} \Rightarrow \text{Perimeter of } \triangle PQR = \frac{4 \times 27}{6} = 18 \text{ cm}$$ Now, Perimeter of $\triangle PQR = PQ + QR + PR$: $$18 = PQ + 4 + 8 \Rightarrow PQ = 18 – 12 = 6 \text{ cm}$$ The length of $PQ$ is 6 cm.
By AA similarity, $\triangle PQR \sim \triangle UTS$.
Since the ratio of perimeters of two similar triangles equals the ratio of their corresponding sides: $$\frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle STU} = \frac{QR}{ST}$$ $$\frac{\text{Perimeter of } \triangle PQR}{27} = \frac{4}{6} \Rightarrow \text{Perimeter of } \triangle PQR = \frac{4 \times 27}{6} = 18 \text{ cm}$$ Now, Perimeter of $\triangle PQR = PQ + QR + PR$: $$18 = PQ + 4 + 8 \Rightarrow PQ = 18 – 12 = 6 \text{ cm}$$ The length of $PQ$ is 6 cm.
Are the two quadrilaterals shown below similar? Give a reason for your answer. 
(Note: The figure is not to scale.)
(Note: The figure is not to scale.)Answer
No, the two quadrilaterals are not similar.
From the figure, the corresponding angles are equal, but the corresponding sides are not in proportion. For two polygons to be similar, both conditions must hold simultaneously — equal corresponding angles and proportional corresponding sides. Since the sides fail the proportionality condition, the quadrilaterals are not similar.
From the figure, the corresponding angles are equal, but the corresponding sides are not in proportion. For two polygons to be similar, both conditions must hold simultaneously — equal corresponding angles and proportional corresponding sides. Since the sides fail the proportionality condition, the quadrilaterals are not similar.
In figure, if $AD = 6$ cm, $DB = 9$ cm, $AE = 8$ cm and $EC = 12$ cm and $\angle ADE = 48°$, find $\angle ABC$. 
Answer
$AB = AD + DB = 6 + 9 = 15$ cm and $AC = AE + EC = 8 + 12 = 20$ cm.
$$\frac{AD}{AB} = \frac{6}{15} = \frac{2}{5} \quad \text{and} \quad \frac{AE}{AC} = \frac{8}{20} = \frac{2}{5}$$ Since $\dfrac{AD}{AB} = \dfrac{AE}{AC}$, by the converse of BPT, $DE \| BC$.
Therefore, $\angle ABC = \angle ADE$ (corresponding angles).
Hence, $\angle ABC = 48°$.
$$\frac{AD}{AB} = \frac{6}{15} = \frac{2}{5} \quad \text{and} \quad \frac{AE}{AC} = \frac{8}{20} = \frac{2}{5}$$ Since $\dfrac{AD}{AB} = \dfrac{AE}{AC}$, by the converse of BPT, $DE \| BC$.
Therefore, $\angle ABC = \angle ADE$ (corresponding angles).
Hence, $\angle ABC = 48°$.
In the adjoining figure, $DE \| AC$ and $DC \| AP$. Prove that $\dfrac{BE}{EC} = \dfrac{BC}{CP}$. 
Answer
In $\triangle ABP$, since $DC \| AP$ (given), by BPT:
$$\frac{BD}{DA} = \frac{BC}{CP} \quad \cdots (i)$$
In $\triangle ABC$, since $DE \| AC$ (given), by BPT:
$$\frac{BD}{DA} = \frac{BE}{EC} \quad \cdots (ii)$$
From equations (i) and (ii):
$$\frac{BE}{EC} = \frac{BC}{CP}$$ Hence Proved.
In the given figure, $DEFG$ is a square and $\angle BAC = 90°$. Show that $FG^2 = BG \times FC$. 
Answer
Since $DEFG$ is a square: $DE = EF = FG = GD$.
In $\triangle ADE$ and $\triangle GDB$:
• $\angle A = \angle DGB = 90°$
• $\angle ADE = \angle GBD$ (corresponding angles)
By AA similarity, $\triangle EAD \sim \triangle GDB$ $\quad \cdots (i)$
In $\triangle AED$ and $\triangle FCE$:
• $\angle A = \angle EFC = 90°$
• $\angle AED = \angle FCE$ (corresponding angles)
By AA similarity, $\triangle AED \sim \triangle FCE$ $\quad \cdots (ii)$
From (i) and (ii): $\triangle GDB \sim \triangle FCE$. Since corresponding sides are proportional: $$\frac{GD}{FC} = \frac{BG}{EF} \Rightarrow GD \times EF = BG \times FC$$ Since $GD = EF = FG$: $$FG^2 = BG \times FC$$ Hence Proved.
In $\triangle ADE$ and $\triangle GDB$:
• $\angle A = \angle DGB = 90°$
• $\angle ADE = \angle GBD$ (corresponding angles)
By AA similarity, $\triangle EAD \sim \triangle GDB$ $\quad \cdots (i)$
In $\triangle AED$ and $\triangle FCE$:
• $\angle A = \angle EFC = 90°$
• $\angle AED = \angle FCE$ (corresponding angles)
By AA similarity, $\triangle AED \sim \triangle FCE$ $\quad \cdots (ii)$
From (i) and (ii): $\triangle GDB \sim \triangle FCE$. Since corresponding sides are proportional: $$\frac{GD}{FC} = \frac{BG}{EF} \Rightarrow GD \times EF = BG \times FC$$ Since $GD = EF = FG$: $$FG^2 = BG \times FC$$ Hence Proved.
In the given figure below, $\dfrac{AD}{AE} = \dfrac{AC}{BD}$ and $\angle 1 = \angle 2$. Show that $\triangle BAE \sim \triangle CAD$. 
Answer
In $\triangle ABD$: $\angle 1 = \angle 2$ (given), so $AB = BD$ (sides opposite equal angles). $\quad \cdots (i)$
Given: $\dfrac{AD}{AE} = \dfrac{AC}{BD}$. Substituting (i): $$\frac{AD}{AE} = \frac{AC}{AB} \quad \cdots (ii)$$ In $\triangle BAE$ and $\triangle CAD$, from equation (ii): $$\frac{AC}{AB} = \frac{AD}{AE}, \quad \text{and } \angle A = \angle A \text{ (common)}$$ By SAS similarity, $\triangle BAE \sim \triangle CAD$. Hence Proved.
Given: $\dfrac{AD}{AE} = \dfrac{AC}{BD}$. Substituting (i): $$\frac{AD}{AE} = \frac{AC}{AB} \quad \cdots (ii)$$ In $\triangle BAE$ and $\triangle CAD$, from equation (ii): $$\frac{AC}{AB} = \frac{AD}{AE}, \quad \text{and } \angle A = \angle A \text{ (common)}$$ By SAS similarity, $\triangle BAE \sim \triangle CAD$. Hence Proved.
In similar triangles $\triangle ABC$ and $\triangle PQR$, $AD$ and $PM$ are the medians respectively. Prove that $\dfrac{AD}{PM} = \dfrac{AB}{PQ}$.
Answer
Since $AD$ is a median of $\triangle ABC$: $BD = \dfrac{BC}{2}$.
Since $PM$ is a median of $\triangle PQR$: $QM = \dfrac{QR}{2}$.
Since $\triangle ABC \sim \triangle PQR$: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$$ In $\triangle ABD$ and $\triangle PQM$:
• $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ (shown above)
• $\angle B = \angle Q$ (since $\triangle ABC \sim \triangle PQR$)
By SAS similarity, $\triangle ABD \sim \triangle PQM$. Therefore: $$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$ Hence, $\dfrac{AD}{PM} = \dfrac{AB}{PQ}$. Hence Proved.
Since $PM$ is a median of $\triangle PQR$: $QM = \dfrac{QR}{2}$.
Since $\triangle ABC \sim \triangle PQR$: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$$ In $\triangle ABD$ and $\triangle PQM$:
• $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ (shown above)
• $\angle B = \angle Q$ (since $\triangle ABC \sim \triangle PQR$)
By SAS similarity, $\triangle ABD \sim \triangle PQM$. Therefore: $$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$ Hence, $\dfrac{AD}{PM} = \dfrac{AB}{PQ}$. Hence Proved.
Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then the other two sides are divided in the same ratio.
Answer
Given: $\triangle ABC$ where $PQ \| BC$, intersecting $AB$ at $P$ and $AC$ at $Q$.
To Prove: $\dfrac{AP}{BP} = \dfrac{AQ}{CQ}$
Construction: Join $BQ$ and $PC$. Draw $PR \perp AQ$ and $QS \perp AP$.
Proof: $\triangle PQB$ and $\triangle PQC$ are on the same base $PQ$ and lie between the same parallels $PQ$ and $BC$.
So, $\text{ar}(\triangle PQB) = \text{ar}(\triangle PCQ)$.
$$\frac{\text{ar}(\triangle APQ)}{\text{ar}(\triangle PBQ)} = \frac{\frac{1}{2} \times AP \times SQ}{\frac{1}{2} \times PB \times SQ} = \frac{AP}{PB} \quad \cdots (ii)$$ $$\frac{\text{ar}(\triangle APQ)}{\text{ar}(\triangle PCQ)} = \frac{\frac{1}{2} \times AQ \times PR}{\frac{1}{2} \times QC \times PR} = \frac{AQ}{QC} \quad \cdots (iii)$$ Since $\text{ar}(\triangle PBQ) = \text{ar}(\triangle PCQ)$, from (ii) and (iii): $$\frac{AP}{BP} = \frac{AQ}{QC}$$ Hence Proved.
To Prove: $\dfrac{AP}{BP} = \dfrac{AQ}{CQ}$
Construction: Join $BQ$ and $PC$. Draw $PR \perp AQ$ and $QS \perp AP$.
Proof: $\triangle PQB$ and $\triangle PQC$ are on the same base $PQ$ and lie between the same parallels $PQ$ and $BC$.
So, $\text{ar}(\triangle PQB) = \text{ar}(\triangle PCQ)$.
$$\frac{\text{ar}(\triangle APQ)}{\text{ar}(\triangle PBQ)} = \frac{\frac{1}{2} \times AP \times SQ}{\frac{1}{2} \times PB \times SQ} = \frac{AP}{PB} \quad \cdots (ii)$$ $$\frac{\text{ar}(\triangle APQ)}{\text{ar}(\triangle PCQ)} = \frac{\frac{1}{2} \times AQ \times PR}{\frac{1}{2} \times QC \times PR} = \frac{AQ}{QC} \quad \cdots (iii)$$ Since $\text{ar}(\triangle PBQ) = \text{ar}(\triangle PCQ)$, from (ii) and (iii): $$\frac{AP}{BP} = \frac{AQ}{QC}$$ Hence Proved.
(a) State and prove the Basic Proportionality Theorem.
(b) In the given figure, $\angle CEF = \angle CFE$. $F$ is the midpoint of $DC$. Prove that $\dfrac{AB}{BD} = \dfrac{AE}{FD}$.
(b) In the given figure, $\angle CEF = \angle CFE$. $F$ is the midpoint of $DC$. Prove that $\dfrac{AB}{BD} = \dfrac{AE}{FD}$.
Answer
(a) Basic Proportionality Theorem (BPT):
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
(Proof: as in Question 31 above.)
(b) Proof:
Draw $DG \| BE$. In $\triangle ABE$, by BPT: $$\frac{AB}{BD} = \frac{AE}{GE} \quad \cdots (i)$$ Since $F$ is the midpoint of $DC$: $CF = FD \quad \cdots (ii)$
In $\triangle CDG$, by the Midpoint Theorem: $\dfrac{DF}{CF} = \dfrac{GE}{CE} = 1$, so $GE = CE \quad \cdots (iii)$
Given $\angle CEF = \angle CFE$, therefore $CF = CE$ (sides opposite equal angles). $\quad \cdots (iv)$
From (iii) and (iv): $CF = GE$. Combined with (ii): $GE = FD \quad \cdots (v)$
Substituting (v) into (i): $$\frac{AB}{BD} = \frac{AE}{FD}$$ Hence Proved.
If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
(Proof: as in Question 31 above.)
(b) Proof:
Draw $DG \| BE$. In $\triangle ABE$, by BPT: $$\frac{AB}{BD} = \frac{AE}{GE} \quad \cdots (i)$$ Since $F$ is the midpoint of $DC$: $CF = FD \quad \cdots (ii)$
In $\triangle CDG$, by the Midpoint Theorem: $\dfrac{DF}{CF} = \dfrac{GE}{CE} = 1$, so $GE = CE \quad \cdots (iii)$
Given $\angle CEF = \angle CFE$, therefore $CF = CE$ (sides opposite equal angles). $\quad \cdots (iv)$
From (iii) and (iv): $CF = GE$. Combined with (ii): $GE = FD \quad \cdots (v)$
Substituting (v) into (i): $$\frac{AB}{BD} = \frac{AE}{FD}$$ Hence Proved.
State and prove the Basic Proportionality Theorem.
In $\triangle ABC$, if $DE \| BC$, $AD = x$, $DB = x – 2$, $AE = x + 2$ and $EC = x – 1$, then using the above result find the value of $x$.
In $\triangle ABC$, if $DE \| BC$, $AD = x$, $DB = x – 2$, $AE = x + 2$ and $EC = x – 1$, then using the above result find the value of $x$.
Answer
Basic Proportionality Theorem: If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.
(Proof: as in Question 31 above.)
In $\triangle ABC$, since $DE \| BC$, by BPT: $$\frac{AD}{DB} = \frac{AE}{EC}$$ $$\frac{x}{x-2} = \frac{x+2}{x-1}$$ $$x(x-1) = (x-2)(x+2) = x^2 – 4$$ $$x^2 – x = x^2 – 4$$ $$-x = -4 \Rightarrow \boxed{x = 4}$$
(Proof: as in Question 31 above.)
In $\triangle ABC$, since $DE \| BC$, by BPT: $$\frac{AD}{DB} = \frac{AE}{EC}$$ $$\frac{x}{x-2} = \frac{x+2}{x-1}$$ $$x(x-1) = (x-2)(x+2) = x^2 – 4$$ $$x^2 – x = x^2 – 4$$ $$-x = -4 \Rightarrow \boxed{x = 4}$$
$D$ is a point on the side $BC$ of a triangle $ABC$ such that $\angle ADC = \angle BAC$. Prove that $CA^2 = CB \cdot CD$.
OR
In the given figure, $\angle ADC = \angle BCA$. Prove that $\triangle ACB \sim \triangle ADC$. Hence find $BD$, if $AC = 8$ cm and $AD = 3$ cm.
OR
In the given figure, $\angle ADC = \angle BCA$. Prove that $\triangle ACB \sim \triangle ADC$. Hence find $BD$, if $AC = 8$ cm and $AD = 3$ cm.
Answer
In $\triangle ACB$ and $\triangle ADC$:
• $\angle ADC = \angle BCA$ (given)
• $\angle A = \angle A$ (common)
By AA similarity, $\triangle ACB \sim \triangle ADC$. Hence Proved.
Since corresponding sides are proportional: $$\frac{AC}{AD} = \frac{BC}{CD} = \frac{AB}{AC}$$ Using $\dfrac{AC}{AD} = \dfrac{AB}{AC}$: $$AC^2 = AD \times AB$$ With $AC = 8$ cm and $AD = 3$ cm: $$64 = 3 \times AB \Rightarrow AB = \frac{64}{3} \text{ cm}$$ $$BD = AB – AD = \frac{64}{3} – 3 = \frac{64 – 9}{3} = \frac{55}{3} \approx 18.3 \text{ cm}$$
• $\angle ADC = \angle BCA$ (given)
• $\angle A = \angle A$ (common)
By AA similarity, $\triangle ACB \sim \triangle ADC$. Hence Proved.
Since corresponding sides are proportional: $$\frac{AC}{AD} = \frac{BC}{CD} = \frac{AB}{AC}$$ Using $\dfrac{AC}{AD} = \dfrac{AB}{AC}$: $$AC^2 = AD \times AB$$ With $AC = 8$ cm and $AD = 3$ cm: $$64 = 3 \times AB \Rightarrow AB = \frac{64}{3} \text{ cm}$$ $$BD = AB – AD = \frac{64}{3} – 3 = \frac{64 – 9}{3} = \frac{55}{3} \approx 18.3 \text{ cm}$$
In $\triangle PQR$, $N$ is a point on $PR$ such that $QN \perp PR$. If $PN \times NR = QN^2$, prove that $\angle PQR = 90°$.
Answer
Given: In $\triangle PQR$, $QN \perp PR$ and $PN \cdot NR = QN^2$.
To Prove: $\angle PQR = 90°$.
Proof: From $PN \cdot NR = QN \cdot QN$: $$\frac{PN}{QN} = \frac{QN}{NR}$$ In $\triangle QNP$ and $\triangle RNQ$:
• $\dfrac{PN}{QN} = \dfrac{QN}{NR}$
• $\angle PNQ = \angle RNQ = 90°$
By SAS similarity, $\triangle QNP \sim \triangle RNQ$. Therefore the triangles are equiangular: $$\angle PQN = \angle QRN \quad \text{and} \quad \angle RQN = \angle QPN$$ Adding both sides: $$\angle PQN + \angle RQN = \angle QRN + \angle QPN$$ $$\angle PQR = \angle QRP + \angle QPR$$ In $\triangle PQR$: $\angle PQR + \angle QPR + \angle QRP = 180°$, so: $$\angle PQR + \angle PQR = 180° \Rightarrow \angle PQR = 90°$$ Hence Proved.
To Prove: $\angle PQR = 90°$.
Proof: From $PN \cdot NR = QN \cdot QN$: $$\frac{PN}{QN} = \frac{QN}{NR}$$ In $\triangle QNP$ and $\triangle RNQ$:
• $\dfrac{PN}{QN} = \dfrac{QN}{NR}$
• $\angle PNQ = \angle RNQ = 90°$
By SAS similarity, $\triangle QNP \sim \triangle RNQ$. Therefore the triangles are equiangular: $$\angle PQN = \angle QRN \quad \text{and} \quad \angle RQN = \angle QPN$$ Adding both sides: $$\angle PQN + \angle RQN = \angle QRN + \angle QPN$$ $$\angle PQR = \angle QRP + \angle QPR$$ In $\triangle PQR$: $\angle PQR + \angle QPR + \angle QRP = 180°$, so: $$\angle PQR + \angle PQR = 180° \Rightarrow \angle PQR = 90°$$ Hence Proved.
Read the following text and answer the questions that follow:
A scale drawing of an object is the same shape as the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents, written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor. $$\text{Scale factor} = \frac{\text{length in image}}{\text{corresponding length in object}}$$ In the photograph below showing the side view of a train engine, the scale factor is 1:200. This means a length of 1 cm on the photograph corresponds to 200 cm (or 2 m) of the actual engine.
(i) If the length of the model is 11 cm, find the overall length of the engine, including the couplings.
(ii) What will affect the similarity of any two polygons?
OR
What is the actual width of the door if the width of the door in the photograph is 0.35 cm?
(iii) Find the length of $AB$ in the given figure.
A scale drawing of an object is the same shape as the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents, written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor. $$\text{Scale factor} = \frac{\text{length in image}}{\text{corresponding length in object}}$$ In the photograph below showing the side view of a train engine, the scale factor is 1:200. This means a length of 1 cm on the photograph corresponds to 200 cm (or 2 m) of the actual engine.
(i) If the length of the model is 11 cm, find the overall length of the engine, including the couplings.
(ii) What will affect the similarity of any two polygons?
OR
What is the actual width of the door if the width of the door in the photograph is 0.35 cm?
(iii) Find the length of $AB$ in the given figure.
Answer
(i) Length of model = 11 cm.
Total length of engine $= 200 \times 11 = 2200$ cm $= 22$ m.
(ii) The similarity of two polygons is affected when the corresponding sides are no longer proportional or the corresponding angles are no longer equal — for example, if one polygon is a mirror image of the other (and mirror images are not considered similar in that context).
OR
$$\text{Actual width of door} = \frac{0.35}{1/200} = 0.35 \times 200 = 70 \text{ cm} = 0.7 \text{ m}$$ (iii) Since $\triangle ABC \sim \triangle ADE$, corresponding sides are proportional. Let $AB = x$: $$\frac{AB}{BC} = \frac{AB + BD}{DE}$$ $$\frac{x}{3} = \frac{x+4}{6}$$ $$6x = 3(x+4) \Rightarrow 6x = 3x + 12 \Rightarrow 3x = 12 \Rightarrow x = 4$$ Therefore, $AB = 4$ cm.
Total length of engine $= 200 \times 11 = 2200$ cm $= 22$ m.
(ii) The similarity of two polygons is affected when the corresponding sides are no longer proportional or the corresponding angles are no longer equal — for example, if one polygon is a mirror image of the other (and mirror images are not considered similar in that context).
OR
$$\text{Actual width of door} = \frac{0.35}{1/200} = 0.35 \times 200 = 70 \text{ cm} = 0.7 \text{ m}$$ (iii) Since $\triangle ABC \sim \triangle ADE$, corresponding sides are proportional. Let $AB = x$: $$\frac{AB}{BC} = \frac{AB + BD}{DE}$$ $$\frac{x}{3} = \frac{x+4}{6}$$ $$6x = 3(x+4) \Rightarrow 6x = 3x + 12 \Rightarrow 3x = 12 \Rightarrow x = 4$$ Therefore, $AB = 4$ cm.
Read the following text and answer the questions that follow:
The triangle proportionality theorem states that when you draw a line parallel to one side of a triangle, it intersects the other two sides and divides them proportionally. In the given $\triangle PQR$, $ST \| QR$, $\dfrac{PS}{SQ} = \dfrac{3}{5}$ and $PR = 28$ cm.
(i) What is the length of $PQ$ and $PT$?
OR
If $QR = 32$ cm, then find $ST$.
(ii) Which property is used in the given case?
(iii) What is the length of $TR$?
The triangle proportionality theorem states that when you draw a line parallel to one side of a triangle, it intersects the other two sides and divides them proportionally. In the given $\triangle PQR$, $ST \| QR$, $\dfrac{PS}{SQ} = \dfrac{3}{5}$ and $PR = 28$ cm.
(i) What is the length of $PQ$ and $PT$?
OR
If $QR = 32$ cm, then find $ST$.
(ii) Which property is used in the given case?
(iii) What is the length of $TR$?
Answer
(i) $PQ = PS + SQ = 3 + 5 = 8$ cm.
Since $ST \| QR$, by BPT: $\dfrac{PS}{PQ} = \dfrac{PT}{PR}$ $$\frac{3}{8} = \frac{PT}{28} \Rightarrow PT = \frac{3 \times 28}{8} = 10.5 \text{ cm}$$ OR
Since $\triangle PST \sim \triangle PQR$: $$\frac{PS}{PQ} = \frac{ST}{QR} \Rightarrow \frac{3}{8} = \frac{ST}{32} \Rightarrow ST = \frac{3 \times 32}{8} = 12 \text{ cm}$$
(ii) The Basic Proportionality Theorem (BPT) is used.
(iii) $$TR = PR – PT = 28 – 10.5 = 17.5 \text{ cm}$$
Since $ST \| QR$, by BPT: $\dfrac{PS}{PQ} = \dfrac{PT}{PR}$ $$\frac{3}{8} = \frac{PT}{28} \Rightarrow PT = \frac{3 \times 28}{8} = 10.5 \text{ cm}$$ OR
Since $\triangle PST \sim \triangle PQR$: $$\frac{PS}{PQ} = \frac{ST}{QR} \Rightarrow \frac{3}{8} = \frac{ST}{32} \Rightarrow ST = \frac{3 \times 32}{8} = 12 \text{ cm}$$
(ii) The Basic Proportionality Theorem (BPT) is used.
(iii) $$TR = PR – PT = 28 – 10.5 = 17.5 \text{ cm}$$
Frequently Asked Questions
What does the Triangles chapter cover in CBSE Class 10 Maths? ⌄
The Triangles chapter in CBSE Class 10 Maths covers the Basic Proportionality Theorem (Thales’ Theorem) and its converse, the three criteria for similarity of triangles (AA, SAS, and SSS), the relationship between the areas and corresponding sides of similar triangles, and the Pythagoras Theorem with its converse. It is one of the most geometry-intensive chapters in the syllabus and requires both conceptual clarity and neat proof-writing skills.
How many marks does the Triangles chapter carry in the CBSE Class 10 board exam? ⌄
Triangles falls under the Geometry unit in CBSE Class 10 Mathematics, which carries approximately 15 marks in the board paper. Questions from this chapter appear across multiple question types — 1-mark MCQs, 2-mark short answers, 3-mark proofs, and 5-mark long-answer or case-study questions — making it one of the highest-weightage individual chapters in the exam.
What are the most important topics to focus on in Class 10 Triangles? ⌄
The most frequently tested topics are: the Basic Proportionality Theorem (statement, proof, and application), AA, SAS, and SSS similarity criteria, the property that the ratio of areas of similar triangles equals the square of the ratio of their corresponding sides, and the Pythagoras Theorem. Board papers consistently include at least one proof and one application question from this chapter every year.
What common mistakes do students make in the Triangles chapter? ⌄
The most common error is writing the correspondence of similar triangles in the wrong order — for example, writing $\triangle ABC \sim \triangle PQR$ when the correct correspondence is $\triangle ABC \sim \triangle QPR$. Students also lose marks in proofs by skipping logical steps or not citing the theorem used (BPT, AA, etc.) at each stage. Consistent practice with step-by-step solutions helps your child build the structured proof-writing habit that earns full marks.
How does Angle Belearn help students score well in the Triangles chapter? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn from real board papers, each paired with clear, step-by-step solutions. For Triangles — a chapter that demands both conceptual understanding and structured proof-writing — students practising on Angle Belearn develop the logical, mark-by-mark approach that examiners reward. Regular practice with these verified questions builds speed, accuracy, and exam-day confidence.
