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 
A triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here, the largest triangle is represented by $\triangle ABC$ and the smallest one with shelf is represented by $\triangle DEF$. $PQ$ is parallel to $EF$.
(i) Show that $\triangle DPQ \sim \triangle DEF$
(ii) If $DP = 50$ cm and $PE = 70$ cm, then find $\dfrac{PQ}{EF}$
(iii) If $2AB = 5DE$ and $\triangle ABC \sim \triangle DEF$, then show that $\dfrac{\text{perimeter of } \triangle ABC}{\text{perimeter of } \triangle DEF}$ is constant. 
In the figure given above, 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 the table top is parallel to the ground, and $OB = x$, $OD = x + 3$, $OC = 3x + 19$, $OA = 3x + 4$, answer the following questions:
(i) Prove that $\triangle OAC$ is similar to $\triangle OBD$.
(ii) Prove that $\dfrac{OA}{AC} = \dfrac{OB}{BD}$.
(iii) Find the value of $x$. Hence, find the length of $OC$.
CBSE Class 10 Maths Triangles Competency Based Questions
Help your child build deep conceptual understanding of CBSE Class 10 Maths Triangles with these competency based questions, covering similarity criteria, the Basic Proportionality Theorem, and real-life applications. Each question comes with a detailed, step-by-step solution prepared and verified by Angle Belearn’s CBSE specialists — so your child always knows exactly where they went right or wrong.
CBSE Class 10 Maths Triangles — Multiple Choice Questions
$\triangle PQR$ 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
Option (B) is correct.
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, $$\frac{ST}{PR} = \frac{QT}{QR}$$ $$\frac{ST}{20} = \frac{2x}{5x}$$ $$ST = \frac{2 \times 20}{5} = 8 \text{ cm}$$
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, $$\frac{ST}{PR} = \frac{QT}{QR}$$ $$\frac{ST}{20} = \frac{2x}{5x}$$ $$ST = \frac{2 \times 20}{5} = 8 \text{ cm}$$
Two scalene triangles are given below:
Anas and Rishu 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 Rishu 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
Option (A) is correct.
Only Anas is correct. In $\triangle PQR$ and $\triangle CBA$: $$\angle P = \angle C \quad \text{and} \quad \angle R = \angle A$$ Therefore, both triangles are similar according to the AA rule. The triangles are not congruent because their side lengths are different (the triangles are scalene with unequal scales).
Only Anas is correct. In $\triangle PQR$ and $\triangle CBA$: $$\angle P = \angle C \quad \text{and} \quad \angle R = \angle A$$ Therefore, both triangles are similar according to the AA rule. The triangles are not congruent because their side lengths are different (the triangles are scalene with unequal scales).
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
Solution
Option (B) is correct.
Given $\triangle ABC \sim \triangle QPR$, the corresponding sides are proportional: $$\frac{AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR}$$ $$\Rightarrow \frac{AB}{QP} = \frac{5}{x} = \frac{6}{3}$$ Equating the last two terms: $$x = \frac{5 \times 3}{6} = \frac{5}{2} = 2.5 \text{ cm}$$
Given $\triangle ABC \sim \triangle QPR$, the corresponding sides are proportional: $$\frac{AB}{QP} = \frac{BC}{PR} = \frac{AC}{QR}$$ $$\Rightarrow \frac{AB}{QP} = \frac{5}{x} = \frac{6}{3}$$ Equating the last two terms: $$x = \frac{5 \times 3}{6} = \frac{5}{2} = 2.5 \text{ cm}$$
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
Option (B) is correct.
If two triangles are similar, the ratio of their corresponding sides equals the ratio of their perimeters. Therefore: $$\frac{PQ}{AB} = \frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle ABC}$$ $$\Rightarrow \frac{6}{8} = \frac{\text{Perimeter of } \triangle PQR}{36}$$ $$\Rightarrow \text{Perimeter of } \triangle PQR = \frac{6 \times 36}{8} = 27 \text{ cm}$$
If two triangles are similar, the ratio of their corresponding sides equals the ratio of their perimeters. Therefore: $$\frac{PQ}{AB} = \frac{\text{Perimeter of } \triangle PQR}{\text{Perimeter of } \triangle ABC}$$ $$\Rightarrow \frac{6}{8} = \frac{\text{Perimeter of } \triangle PQR}{36}$$ $$\Rightarrow \text{Perimeter of } \triangle PQR = \frac{6 \times 36}{8} = 27 \text{ cm}$$
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
Option (B) is correct (it is the one that is not true).
Given $\angle D = \angle Q$ and $\angle R = \angle E$, by AA similarity: $$\triangle DEF \sim \triangle QRP$$ Therefore the correct ratios are: $$\frac{EF}{PR} = \frac{DF}{PQ} = \frac{DE}{QR}$$ Option (B) states $\dfrac{EF}{RP} = \dfrac{DE}{PQ}$, which does not follow from this correspondence, so it is not true.
Given $\angle D = \angle Q$ and $\angle R = \angle E$, by AA similarity: $$\triangle DEF \sim \triangle QRP$$ Therefore the correct ratios are: $$\frac{EF}{PR} = \frac{DF}{PQ} = \frac{DE}{QR}$$ Option (B) states $\dfrac{EF}{RP} = \dfrac{DE}{PQ}$, which does not follow from this correspondence, so it is 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
Option (D) is correct.
In $\triangle ATS$ and $\triangle AQP$:
$\angle PQA = \angle ATS$ (Alternate angles)
$\angle PAQ = \angle TAS$ (Vertically opposite angles)
By AA similarity rule, $\triangle ATS \sim \triangle AQP$. Therefore: $$\frac{PA}{AS} = \frac{PQ}{TS} = \frac{AQ}{AT}$$ $$\frac{x}{3} = \frac{y}{4} = \frac{6}{6} = 1$$ Therefore $x = 3$ cm and $y = 4$ cm.
In $\triangle ATS$ and $\triangle AQP$:
$\angle PQA = \angle ATS$ (Alternate angles)
$\angle PAQ = \angle TAS$ (Vertically opposite angles)
By AA similarity rule, $\triangle ATS \sim \triangle AQP$. Therefore: $$\frac{PA}{AS} = \frac{PQ}{TS} = \frac{AQ}{AT}$$ $$\frac{x}{3} = \frac{y}{4} = \frac{6}{6} = 1$$ Therefore $x = 3$ cm and $y = 4$ cm.
In the given figure, $DE \| BC$. If $AD = 3$ cm, $AB = 7$ cm and $EC = 3$ cm, then the length of $AE$ is
Solution
Option (B) is correct.
In $\triangle ABC$, since $DE \| BC$, by the Basic Proportionality Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Here $DB = AB – AD = 7 – 3 = 4$ cm. Substituting: $$\frac{3}{4} = \frac{AE}{3}$$ $$AE = \frac{9}{4} = 2.25 \text{ cm}$$
In $\triangle ABC$, since $DE \| BC$, by the Basic Proportionality Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Here $DB = AB – AD = 7 – 3 = 4$ cm. Substituting: $$\frac{3}{4} = \frac{AE}{3}$$ $$AE = \frac{9}{4} = 2.25 \text{ cm}$$
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
Option (B) is correct.
Since $\angle F = \angle C$ and $\angle B = \angle E$, the two triangles satisfy the AA similarity criterion, so they are similar.
However, since $AB = \dfrac{1}{2}DE$, the corresponding sides are in ratio $1:2$ — they are not equal, so the triangles are not congruent.
Since $\angle F = \angle C$ and $\angle B = \angle E$, the two triangles satisfy the AA similarity criterion, so they are similar.
However, since $AB = \dfrac{1}{2}DE$, the corresponding sides are in ratio $1:2$ — they are not equal, so the triangles are not congruent.
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
Option (A) is correct.
In $\triangle RQS$ and $\triangle RPT$:
$\angle RQS = \angle RPT$ (Given)
$\angle R = \angle R$ (Common angle)
Since two pairs of corresponding angles are equal, by the AAA (AA) similarity criterion, $\triangle RSQ \sim \triangle RTP$.
In $\triangle RQS$ and $\triangle RPT$:
$\angle RQS = \angle RPT$ (Given)
$\angle R = \angle R$ (Common angle)
Since two pairs of corresponding angles are equal, by the AAA (AA) similarity criterion, $\triangle RSQ \sim \triangle RTP$.
Shown below are three triangles. The measures of two adjacent sides and the included angle are given for each triangle.
Which of these triangles are similar?
Which of these triangles are similar?
Solution
Option (A) is correct.
In $\triangle RPQ$ and $\triangle XZY$: $$\frac{RP}{XZ} = \frac{6 \text{ cm}}{9 \text{ cm}} = \frac{2}{3}, \quad \frac{PQ}{ZY} = \frac{4 \text{ cm}}{6 \text{ cm}} = \frac{2}{3}$$ $$\therefore \frac{RP}{XZ} = \frac{PQ}{ZY}$$ And $\angle RPQ = \angle XZY = 60°$ (included angle).
By SAS similarity criterion, $\triangle RPQ \sim \triangle XZY$.
In $\triangle RPQ$ and $\triangle XZY$: $$\frac{RP}{XZ} = \frac{6 \text{ cm}}{9 \text{ cm}} = \frac{2}{3}, \quad \frac{PQ}{ZY} = \frac{4 \text{ cm}}{6 \text{ cm}} = \frac{2}{3}$$ $$\therefore \frac{RP}{XZ} = \frac{PQ}{ZY}$$ And $\angle RPQ = \angle XZY = 60°$ (included angle).
By SAS similarity criterion, $\triangle RPQ \sim \triangle XZY$.
In the figure below, $DE \| AC$ and $DF \| AE$. Which of these is equal to $\dfrac{BF}{FE}$?
Solution
Option (B) is correct.
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): $$\frac{BF}{FE} = \frac{BE}{EC}$$
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
Option (D) is correct.
Since $\triangle ABC \sim \triangle PQR$, the ratio of corresponding altitudes equals the ratio of corresponding sides: $$\frac{AM}{PN} = \frac{AB}{PQ}$$ Given $\dfrac{AB^2}{PQ^2} = \dfrac{4}{9}$, we get: $$\left(\frac{AB}{PQ}\right)^2 = \left(\frac{2}{3}\right)^2 \implies \frac{AB}{PQ} = \frac{2}{3}$$ Therefore $AM : PN = 2 : 3$.
Since $\triangle ABC \sim \triangle PQR$, the ratio of corresponding altitudes equals the ratio of corresponding sides: $$\frac{AM}{PN} = \frac{AB}{PQ}$$ Given $\dfrac{AB^2}{PQ^2} = \dfrac{4}{9}$, we get: $$\left(\frac{AB}{PQ}\right)^2 = \left(\frac{2}{3}\right)^2 \implies \frac{AB}{PQ} = \frac{2}{3}$$ 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
Option (C) is correct.
Since $AD \| BC$ and the diagonals intersect at $O$, triangles $AOD$ and $COB$ are similar. Therefore: $$\frac{AD}{BC} = \frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}$$ $$\Rightarrow \frac{4}{BC} = \frac{1}{2} \Rightarrow BC = 8 \text{ cm}$$
Since $AD \| BC$ and the diagonals intersect at $O$, triangles $AOD$ and $COB$ are similar. Therefore: $$\frac{AD}{BC} = \frac{AO}{OC} = \frac{DO}{OB} = \frac{1}{2}$$ $$\Rightarrow \frac{4}{BC} = \frac{1}{2} \Rightarrow BC = 8 \text{ cm}$$
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
Option (B) is correct.
Using the Basic Proportionality Theorem, since $DE \| AB$: $$\frac{DE}{AB} = \frac{CE}{BC}$$ Here $BC = BE + EC = b + c$. Substituting: $$\frac{x}{a} = \frac{c}{b+c}$$ $$\Rightarrow x = \frac{ac}{b+c}$$
Using the Basic Proportionality Theorem, since $DE \| AB$: $$\frac{DE}{AB} = \frac{CE}{BC}$$ Here $BC = BE + EC = b + c$. Substituting: $$\frac{x}{a} = \frac{c}{b+c}$$ $$\Rightarrow x = \frac{ac}{b+c}$$
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
Option (C) is correct.
In $\triangle ABC$, since $HI \| BC$, by the Basic Proportionality Theorem: $$\frac{AB}{HB} = \frac{AC}{IC}$$ The shelf divides the height in ratio $8y : 3y$ (from the figure). Substituting: $$\frac{144}{HB} = \frac{8y}{3y} = \frac{8}{3}$$ $$HB = \frac{144 \times 3}{8} = 54 \text{ cm}$$ Thus, the maximum height of a book on the bottom-most shelf is 54 cm.
In $\triangle ABC$, since $HI \| BC$, by the Basic Proportionality Theorem: $$\frac{AB}{HB} = \frac{AC}{IC}$$ The shelf divides the height in ratio $8y : 3y$ (from the figure). Substituting: $$\frac{144}{HB} = \frac{8y}{3y} = \frac{8}{3}$$ $$HB = \frac{144 \times 3}{8} = 54 \text{ cm}$$ Thus, the maximum height of a book on the bottom-most shelf is 54 cm.
CBSE Class 10 Maths Triangles — Case Study & Subjective Questions
A triangle is a very popular shape used in interior designing. The picture given above shows a cabinet designed by a famous interior designer. Here, the largest triangle is represented by $\triangle ABC$ and the smallest one with shelf is represented by $\triangle DEF$. $PQ$ is parallel to $EF$.
(i) Show that $\triangle DPQ \sim \triangle DEF$
(ii) If $DP = 50$ cm and $PE = 70$ cm, then find $\dfrac{PQ}{EF}$
(iii) If $2AB = 5DE$ and $\triangle ABC \sim \triangle DEF$, then show that $\dfrac{\text{perimeter of } \triangle ABC}{\text{perimeter of } \triangle DEF}$ is constant.
Answer
(i) Proof: Since $PQ \parallel EF$,
$\angle DPQ = \angle DEF$ (corresponding angles)
$\angle DQP = \angle DFE$ (corresponding angles)
Two corresponding angles of $\triangle DPQ$ and $\triangle DEF$ are equal. $$\therefore \triangle DPQ \sim \triangle DEF \quad \text{(by AA similarity)}$$
$\angle DPQ = \angle DEF$ (corresponding angles)
$\angle DQP = \angle DFE$ (corresponding angles)
Two corresponding angles of $\triangle DPQ$ and $\triangle DEF$ are equal. $$\therefore \triangle DPQ \sim \triangle DEF \quad \text{(by AA similarity)}$$
(ii) Finding PQ/EF: $DE = DP + PE = 50 + 70 = 120$ cm
Since $\triangle DPQ \sim \triangle DEF$: $$\frac{DP}{DE} = \frac{PQ}{EF}$$ $$\therefore \frac{PQ}{EF} = \frac{50}{120} = \frac{5}{12}$$
Since $\triangle DPQ \sim \triangle DEF$: $$\frac{DP}{DE} = \frac{PQ}{EF}$$ $$\therefore \frac{PQ}{EF} = \frac{50}{120} = \frac{5}{12}$$
(iii) Proof that ratio of perimeters is constant: Given $2AB = 5DE \Rightarrow \dfrac{AB}{DE} = \dfrac{5}{2}$
Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{5}{2}$$ $$\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF} = \frac{\frac{5}{2}(DE + EF + FD)}{DE + EF + FD} = \frac{5}{2}$$ This is a constant value. Hence proved.
Since $\triangle ABC \sim \triangle DEF$: $$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{5}{2}$$ $$\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle DEF} = \frac{\frac{5}{2}(DE + EF + FD)}{DE + EF + FD} = \frac{5}{2}$$ This is a constant value. Hence proved.
A student is trying to find the height of a tower near his house using the properties of similar triangles. The height of the building is 20 m when it casts a shadow 10 m long on the ground. At the same time, the tower casts a shadow 50 m long on the ground. At the same time, his house casts a shadow 20 m on the ground.
On the basis of the above situation, answer the following questions:
(i) What is the height of the tower?
(ii) What is the height of the student’s house?
(iii) What will be the length of the shadow of the tower when the building casts a shadow of 12 m at the same time?
On the basis of the above situation, answer the following questions:
(i) What is the height of the tower?
(ii) What is the height of the student’s house?
(iii) What will be the length of the shadow of the tower when the building casts a shadow of 12 m at the same time?
Answer
(i) Height of the tower: Using similarity of triangles (objects and their shadows form similar triangles at the same time):
$$\frac{\text{Height of building}}{\text{Shadow of building}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}$$
$$\frac{20}{10} = \frac{h}{50} \implies h = 100 \text{ m}$$
(ii) Height of the student’s house: $$\frac{20}{10} = \frac{h_1}{20} \implies h_1 = 40 \text{ m}$$
(iii) Shadow of the tower when building’s shadow = 12 m: The new ratio of shadow to height for the building: $\dfrac{12}{20} = \dfrac{3}{5}$
Applying the same ratio to the tower (height 100 m): $$\text{New tower shadow} = \frac{100}{20} \times 12 = 60 \text{ m}$$
Applying the same ratio to the tower (height 100 m): $$\text{New tower shadow} = \frac{100}{20} \times 12 = 60 \text{ m}$$
In the figure given above, 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 the table top is parallel to the ground, and $OB = x$, $OD = x + 3$, $OC = 3x + 19$, $OA = 3x + 4$, answer the following questions:
(i) Prove that $\triangle OAC$ is similar to $\triangle OBD$.
(ii) Prove that $\dfrac{OA}{AC} = \dfrac{OB}{BD}$.
(iii) Find the value of $x$. Hence, find the length of $OC$.
Answer
(i) Proof that △OAC ~ △OBD: Since the table top is parallel to the ground (i.e., $AC \parallel BD$):
$\angle D = \angle C$ and $\angle B = \angle A$ (alternate interior angles)
By AA similarity: $$\therefore \triangle OAC \sim \triangle OBD$$
$\angle D = \angle C$ and $\angle B = \angle A$ (alternate interior angles)
By AA similarity: $$\therefore \triangle OAC \sim \triangle OBD$$
(ii) Proof that OA/AC = OB/BD: From $\triangle OAC \sim \triangle OBD$, corresponding sides are proportional:
$$\frac{OA}{OB} = \frac{AC}{BD}$$
Rearranging:
$$\frac{OA}{AC} = \frac{OB}{BD} \quad \text{(Hence proved)}$$
(iii) Finding x and OC: From the similarity $\triangle OAC \sim \triangle OBD$:
$$\frac{OA}{OB} = \frac{OC}{OD}$$
$$\frac{3x+4}{x} = \frac{3x+19}{x+3}$$
Cross-multiplying:
$$(3x+4)(x+3) = x(3x+19)$$
$$3x^2 + 13x + 12 = 3x^2 + 19x$$
$$6x = 12 \implies x = 2$$
$$\therefore OC = 3(2) + 19 = 25 \text{ cm}$$
Digvijay is trying to find the average height of a tower near his house using the properties of similar triangles. The height of Digvijay’s house is 20 m. Digvijay’s house casts a shadow of 10 m on the ground. At the same time, the tower casts a shadow of 50 m on the ground. At the same time, the house of Anshul casts a shadow of 20 m on the ground.
(i) The height of the tower is:
(A) 10 m (B) 20 m (C) 50 m (D) 100 m
(ii) When Digvijay’s house casts a shadow of 18 m, the length of the shadow of the tower is:
(A) 18 m (B) 20 m (C) 90 m (D) 100 m
(iii) The height of Anshul’s house is:
(A) 20 m (B) 40 m (C) 50 m (D) 100 m
(iv) When the tower casts a shadow of 40 m, at the same time the length of the shadow of Anshul’s house is:
(A) 16 m (B) 40 m (C) 100 m (D) None of these
(i) The height of the tower is:
(A) 10 m (B) 20 m (C) 50 m (D) 100 m
(ii) When Digvijay’s house casts a shadow of 18 m, the length of the shadow of the tower is:
(A) 18 m (B) 20 m (C) 90 m (D) 100 m
(iii) The height of Anshul’s house is:
(A) 20 m (B) 40 m (C) 50 m (D) 100 m
(iv) When the tower casts a shadow of 40 m, at the same time the length of the shadow of Anshul’s house is:
(A) 16 m (B) 40 m (C) 100 m (D) None of these
Answer
(i) Option (D) — Height = 100 m (Note: derivation gives 100 m from the given shadow ratio) Using similar triangles ($\triangle ACD \sim \triangle ABE$):
$$\frac{\text{Shadow of tower}}{\text{Shadow of house}} = \frac{\text{Height of tower}}{\text{Height of house}}$$
$$\frac{50}{10} = \frac{h}{20} \implies h = 100 \text{ m}$$
(ii) Option (C) — Shadow of tower = 90 m When Digvijay’s house shadow $= 18$ m:
$$\frac{18}{x} = \frac{20}{100} \implies x = \frac{18 \times 100}{20} = 90 \text{ m}$$
(iii) Option (B) — Height of Anshul’s house = 40 m $$\frac{20}{50} = \frac{h_1}{100} \implies h_1 = \frac{20 \times 100}{50} = 40 \text{ m}$$
(iv) Option (A) — Shadow of Anshul’s house = 16 m Tower shadow $= 40$ m, Tower height $= 100$ m.
Since $\triangle HFG \sim \triangle HCD$: $$\frac{HF}{40} = \frac{40}{100} \implies HF = \frac{40 \times 40}{100} = 16 \text{ m}$$
Since $\triangle HFG \sim \triangle HCD$: $$\frac{HF}{40} = \frac{40}{100} \implies HF = \frac{40 \times 40}{100} = 16 \text{ m}$$
Gaurav placed a light bulb at point $O$ on the ceiling and directly below it placed a table. He cuts a quadrilateral $PQRS$ from cardboard and places it parallel to the ground between the lighted bulb and the table. Then the shadow of $PQRS$ is cast on the table as $P’Q’R’S’$. Quadrilateral $P’Q’R’S’$ is an enlargement of $PQRS$ with scale factor $1:3$.
Given: $PQ = 2.5$ cm, $QR = 3.5$ cm, $RS = 3.4$ cm, $PS = 3.1$ cm, $\angle P = 115°$, $\angle Q = 95°$, $\angle R = 65°$, $\angle S = 85°$
(i) The length of $R’S’$ is: (A) 3.4 cm (B) 10.2 cm (C) 6.8 cm (D) 9.5 cm
(ii) The ratio of sides $P’Q’$ and $Q’R’$ is: (A) 5:7 (B) 7:5 (C) 7:2 (D) 2:7
(iii) The measurement of $\angle Q’$ is: (A) 115° (B) 95° (C) 65° (D) 85°
(iv) The sum of lengths $Q’R’$ and $P’S’$ is: (A) 12.3 cm (B) 6.7 cm (C) 19.8 cm (D) 9 cm
(v) The sum of angles of quadrilateral $P’Q’R’S’$ is: (A) 180° (B) 270° (C) 300° (D) 360°
Given: $PQ = 2.5$ cm, $QR = 3.5$ cm, $RS = 3.4$ cm, $PS = 3.1$ cm, $\angle P = 115°$, $\angle Q = 95°$, $\angle R = 65°$, $\angle S = 85°$
(i) The length of $R’S’$ is: (A) 3.4 cm (B) 10.2 cm (C) 6.8 cm (D) 9.5 cm
(ii) The ratio of sides $P’Q’$ and $Q’R’$ is: (A) 5:7 (B) 7:5 (C) 7:2 (D) 2:7
(iii) The measurement of $\angle Q’$ is: (A) 115° (B) 95° (C) 65° (D) 85°
(iv) The sum of lengths $Q’R’$ and $P’S’$ is: (A) 12.3 cm (B) 6.7 cm (C) 19.8 cm (D) 9 cm
(v) The sum of angles of quadrilateral $P’Q’R’S’$ is: (A) 180° (B) 270° (C) 300° (D) 360°
Answer
(i) Option (B) — R’S’ = 10.2 cm Scale factor $= 1:3$, so $R’S’ = 3 \times RS = 3 \times 3.4 = 10.2$ cm
(ii) Option (A) — P’Q’ : Q’R’ = 5 : 7 $P’Q’ = 3 \times 2.5 = 7.5$ cm; $Q’R’ = 3 \times 3.5 = 10.5$ cm
$$\frac{P’Q’}{Q’R’} = \frac{7.5}{10.5} = \frac{5}{7}$$
(iii) Option (B) — ∠Q’ = 95° Since $P’Q’R’S’$ is similar to $PQRS$, all corresponding angles are equal. $\angle Q’ = \angle Q = 95°$.
(iv) Option (C) — Q’R’ + P’S’ = 19.8 cm $Q’R’ = 3 \times 3.5 = 10.5$ cm; $P’S’ = 3 \times 3.1 = 9.3$ cm
$Q’R’ + P’S’ = 10.5 + 9.3 = 19.8$ cm
$Q’R’ + P’S’ = 10.5 + 9.3 = 19.8$ cm
(v) Option (D) — Sum of angles = 360° $\angle P’ + \angle Q’ + \angle R’ + \angle S’ = 115° + 95° + 65° + 85° = 360°$
(The angle sum of any quadrilateral is always 360°.)
(The angle sum of any quadrilateral is always 360°.)
A scale drawing of an object is of the same shape as the object but of a different size. The scale factor is:
$$\text{Scale factor} = \frac{\text{Length in image}}{\text{Corresponding length in object}}$$ 
(i) A model of a boat is made on the scale of $1:4$. The model is 120 cm long. The full-size boat has a width of 60 cm. What is the width of the scale model?
(A) 20 cm (B) 25 cm (C) 15 cm (D) 240 cm
(ii) What will affect the similarity of any two polygons?
(A) They are flipped horizontally. (B) They are dilated by a scale factor. (C) They are translated down. (D) They are not the mirror image of one another.
(iii) If two similar triangles have a scale factor of $a:b$, which statement is true?
(A) The ratio of their perimeters is $3a:b$ (B) Their altitudes have a ratio $a:b$ (C) Their medians have a ratio $\frac{a}{2}:b$ (D) Their angle bisectors have a ratio $a^2:b^2$
(iv) The shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is:
(A) 3 m (B) 3.5 m (C) 4.5 m (D) 5 m
(v) In a model of a farmhouse roof, all edges of the pyramid have length 12 m. $E$ is the middle of $AT$, $F$ is the middle of $BT$. What is the length of $EF$?
(A) 24 m (B) 3 m (C) 6 m (D) 10 m
(i) A model of a boat is made on the scale of $1:4$. The model is 120 cm long. The full-size boat has a width of 60 cm. What is the width of the scale model?
(A) 20 cm (B) 25 cm (C) 15 cm (D) 240 cm
(ii) What will affect the similarity of any two polygons?
(A) They are flipped horizontally. (B) They are dilated by a scale factor. (C) They are translated down. (D) They are not the mirror image of one another.
(iii) If two similar triangles have a scale factor of $a:b$, which statement is true?
(A) The ratio of their perimeters is $3a:b$ (B) Their altitudes have a ratio $a:b$ (C) Their medians have a ratio $\frac{a}{2}:b$ (D) Their angle bisectors have a ratio $a^2:b^2$
(iv) The shadow of a stick 5 m long is 2 m. At the same time the shadow of a tree 12.5 m high is:
(A) 3 m (B) 3.5 m (C) 4.5 m (D) 5 m
(v) In a model of a farmhouse roof, all edges of the pyramid have length 12 m. $E$ is the middle of $AT$, $F$ is the middle of $BT$. What is the length of $EF$?
(A) 24 m (B) 3 m (C) 6 m (D) 10 m
Answer
(i) Option (C) — Width of scale model = 15 cm Width of scale model $= \dfrac{60}{4} = 15$ cm
(ii) Option (D) — They are not the mirror image of one another. Flipping, translating, rotating, and dilating all preserve similarity. Two polygons that are not mirror images of each other may still be similar in most cases; however, the option that describes a condition that can affect similarity correspondence is (D).
(iii) Option (B) — Altitudes have ratio a : b For similar triangles with scale factor $a:b$, all corresponding linear measurements (sides, altitudes, medians, angle bisectors, perimeters) are in the ratio $a:b$.
$$\frac{AD}{PE} = \frac{AB}{PQ} = \frac{a}{b}$$
(iv) Option (D) — Shadow of tree = 5 m Let shadow of tree $= x$. By similar triangles:
$$\frac{5}{2} = \frac{12.5}{x} \implies x = \frac{12.5 \times 2}{5} = 5 \text{ m}$$
(v) Option (C) — EF = 6 m Since $E$ is the midpoint of $AT$ and $F$ is the midpoint of $BT$, by the midpoint theorem, $EF$ is parallel to $AB$ and:
$$EF = \frac{1}{2} \times AB = \frac{12}{2} = 6 \text{ m}$$
Some concrete water towers have been built to supply water to localities nearby. They are usually mounted with a cylindrical tank. A water tower for a locality is 40 m high.
(i) The water tower casts a shadow of 25 m. At the same time, a tree casts a shadow of 5 m. What is the height of the tree?
(A) 3.12 m (B) 8 m (C) 20 m (D) 25 m
(ii) A scale model of the water tower of 100 cm height is created. The height of its pillars is 75 cm each. What is the height of a pillar (in m) in the actual water tower?
(A) 7.5 m (B) 25 m (C) 30 m (D) 53.4 m
(iii) Dharmendra made a scale model of a water tower. The radius of the reservoir in the model is 6 cm and its volume is 216 cm³. The radius of the actual water reservoir is 2.5 m. What is its volume?
(A) $15.625 \text{ m}^3$ (B) $15625 \text{ m}^3$ (C) $15625 \text{ cm}^3$ (D) $15.625 \text{ cm}^3$
(iv) In the figure, $\angle ABC$ is equal to:
(A) 60° (B) 55° (C) 75° (D) 65°
(v) From the figure of (iv), $\dfrac{DE}{BC}$ is equal to:
(A) $\dfrac{3}{4}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{2}{3}$
(i) The water tower casts a shadow of 25 m. At the same time, a tree casts a shadow of 5 m. What is the height of the tree?
(A) 3.12 m (B) 8 m (C) 20 m (D) 25 m
(ii) A scale model of the water tower of 100 cm height is created. The height of its pillars is 75 cm each. What is the height of a pillar (in m) in the actual water tower?
(A) 7.5 m (B) 25 m (C) 30 m (D) 53.4 m
(iii) Dharmendra made a scale model of a water tower. The radius of the reservoir in the model is 6 cm and its volume is 216 cm³. The radius of the actual water reservoir is 2.5 m. What is its volume?
(A) $15.625 \text{ m}^3$ (B) $15625 \text{ m}^3$ (C) $15625 \text{ cm}^3$ (D) $15.625 \text{ cm}^3$
(iv) In the figure, $\angle ABC$ is equal to:
(A) 60° (B) 55° (C) 75° (D) 65°
(v) From the figure of (iv), $\dfrac{DE}{BC}$ is equal to:
(A) $\dfrac{3}{4}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{2}{3}$
Answer
(i) Option (B) — Height of tree = 8 m $$\frac{\text{Height of water tower}}{\text{Height of tree}} = \frac{\text{Shadow of water tower}}{\text{Shadow of tree}}$$
$$\frac{40}{x} = \frac{25}{5} \implies 25x = 200 \implies x = 8 \text{ m}$$
(ii) Option (C) — Height of pillar = 30 m Height of pillar in actual tower $= \dfrac{40}{100} \times 75 = 30$ m
(iii) Option (A) — Volume = 15.625 m³ Volume of actual reservoir $= 2.5 \times 2.5 \times 2.5 = 15.625 \text{ m}^3$
(iv) Option (B) — ∠ABC = 55° From the figure: $\dfrac{AD}{DB} = \dfrac{10}{15} = \dfrac{2}{3}$ and $\dfrac{AE}{EC} = \dfrac{14}{21} = \dfrac{2}{3}$
Since $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, by converse of BPT: $DE \| BC$
$\therefore \angle ABC = \angle ADE = 55°$ (corresponding angles)
Since $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, by converse of BPT: $DE \| BC$
$\therefore \angle ABC = \angle ADE = 55°$ (corresponding angles)
(v) Option (D) — DE/BC = 2/3 Since $DE \| BC$ (proved above):
$$\frac{DE}{BC} = \frac{AD}{DB} = \frac{10}{15} = \frac{2}{3}$$
A scale drawing of an object is the same shape as the object but a different size. The scale factor $= \dfrac{\text{length in image}}{\text{Corresponding length in object}}$
In the photograph of a train engine, the scale factor is $1:200$. This means a length of 1 cm on the photograph corresponds to 200 cm (2 m) on the actual engine.
(i) If the length of the model is 11 cm, find the overall length of the engine.
(ii) What will affect the similarity of any two polygons? OR What is the actual width of the door if the width in the photograph is 0.35 cm?
(iii) Find the length of $AB$ in the given figure:
In the photograph of a train engine, the scale factor is $1:200$. This means a length of 1 cm on the photograph corresponds to 200 cm (2 m) on the actual engine.
(i) If the length of the model is 11 cm, find the overall length of the engine.
(ii) What will affect the similarity of any two polygons? OR What is the actual width of the door if the width in the photograph is 0.35 cm?
(iii) Find the length of $AB$ in the given figure:
Answer
(i) Overall length of the engine: Total length $= 200 \times 11 = 2200$ cm $= 22$ m
(ii) Similarity affected when polygons are not mirror images of one another. OR Actual width of door: Actual width $= 0.35 \times 200 = 70$ cm $= 0.7$ m
(iii) Finding length AB: Since $\triangle ABC \sim \triangle ADE$, corresponding sides are proportional:
$$\frac{AB}{BC} = \frac{AB + BD}{DE}$$
Let $AB = x$:
$$\frac{x}{3} = \frac{x + 4}{6}$$
$$6x = 3(x + 4) \implies 6x = 3x + 12 \implies 3x = 12 \implies x = 4$$
$$\therefore AB = 4 \text{ cm}$$
The triangle proportionality theorem states that when a line is drawn parallel to one side of a triangle, it intersects the other two sides and divides them proportionally. In the given $\triangle PQR$, $ST \| QR$ and $\dfrac{PS}{SQ} = \dfrac{3}{5}$ and $PR = 28$ cm.
(i) What is the length of $PQ$ and $PT$? OR If $QR = 32$ cm, find $ST$.
(ii) Which property is used in the given case?
(iii) What is the length of $TR$?
(i) What is the length of $PQ$ and $PT$? OR If $QR = 32$ cm, find $ST$.
(ii) Which property is used in the given case?
(iii) What is the length of $TR$?
Answer
(i) Length of PQ and PT (or ST): $PQ = PS + SQ = 3 + 5 = 8$ cm
Since $ST \| QR$: $\dfrac{PS}{PQ} = \dfrac{PT}{PR}$ $$\frac{3}{8} = \frac{PT}{28} \implies PT = \frac{3 \times 28}{8} = 10.5 \text{ cm}$$ OR (If $QR = 32$ cm): $\triangle PST \sim \triangle PQR$, so: $$\frac{ST}{QR} = \frac{PS}{PQ} = \frac{3}{8} \implies ST = \frac{3 \times 32}{8} = 12 \text{ cm}$$
Since $ST \| QR$: $\dfrac{PS}{PQ} = \dfrac{PT}{PR}$ $$\frac{3}{8} = \frac{PT}{28} \implies PT = \frac{3 \times 28}{8} = 10.5 \text{ cm}$$ OR (If $QR = 32$ cm): $\triangle PST \sim \triangle PQR$, so: $$\frac{ST}{QR} = \frac{PS}{PQ} = \frac{3}{8} \implies ST = \frac{3 \times 32}{8} = 12 \text{ cm}$$
(ii) Property used: Basic Proportionality Theorem (BPT), also known as Thales’ Theorem.
(iii) Length of TR: $$TR = PR – PT = 28 – 10.5 = 17.5 \text{ cm}$$
Observe the figures given below carefully and answer the questions:
(i) Name the figure(s) wherein two figures are similar.
(ii) Name the figure(s) wherein the figures are congruent.
(iii) Prove that congruent triangles are also similar, but not the converse.
(i) Name the figure(s) wherein two figures are similar.
(ii) Name the figure(s) wherein the figures are congruent.
(iii) Prove that congruent triangles are also similar, but not the converse.
Answer
(i) Similar figures: Figure A and Figure C
(ii) Congruent figures: Figure C
(iii) Proof: If two triangles are congruent, then their corresponding angles are equal and their corresponding sides are equal. Since equal sides are trivially in ratio $1:1$, the condition for similarity (equal angles + proportional sides) is satisfied. Therefore, congruent triangles are always similar.
Conversely, if two triangles are similar, their corresponding angles are equal and sides are proportional — but proportional does not imply equal (the ratio could be $k \neq 1$). Therefore, similar triangles may not be congruent. Hence the converse does not hold.
Conversely, if two triangles are similar, their corresponding angles are equal and sides are proportional — but proportional does not imply equal (the ratio could be $k \neq 1$). Therefore, similar triangles may not be congruent. Hence the converse does not hold.
CBSE Class 10 Maths Triangles — Assertion-Reason Questions
Directions: Mark the correct choice:
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, $DE \| BC$, such that $AD = (7x-4)$ cm, $AE = (5x-2)$ cm, $DB = (3x+4)$ cm, $EC = 3x$ cm, then the value of $x$ is 5.
Reason (R): If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, $DE \| BC$, such that $AD = (7x-4)$ cm, $AE = (5x-2)$ cm, $DB = (3x+4)$ cm, $EC = 3x$ cm, then the value of $x$ is 5.
Reason (R): If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.
Answer
Option (D) is correct. — A is false but R is True.
By Thales’ Theorem (which is the true Reason R): $$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{7x-4}{3x+4} = \frac{5x-2}{3x}$$ Cross-multiplying: $(7x-4)(3x) = (5x-2)(3x+4)$ $$21x^2 – 12x = 15x^2 + 14x – 8$$ $$6x^2 – 26x + 8 = 0 \implies 3x^2 – 13x + 4 = 0$$ $$(3x-1)(x-4) = 0 \implies x = \frac{1}{3} \text{ or } x = 4$$ The correct value is $x = 4$, not 5. So Assertion is false; Reason is true.
By Thales’ Theorem (which is the true Reason R): $$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{7x-4}{3x+4} = \frac{5x-2}{3x}$$ Cross-multiplying: $(7x-4)(3x) = (5x-2)(3x+4)$ $$21x^2 – 12x = 15x^2 + 14x – 8$$ $$6x^2 – 26x + 8 = 0 \implies 3x^2 – 13x + 4 = 0$$ $$(3x-1)(x-4) = 0 \implies x = \frac{1}{3} \text{ or } x = 4$$ The correct value is $x = 4$, not 5. So Assertion is false; Reason is true.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $\triangle ABC$ and $\triangle PQR$ are congruent triangles, then they are also similar triangles.
Reason (R): All congruent triangles are similar, but similar triangles need not be congruent.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If $\triangle ABC$ and $\triangle PQR$ are congruent triangles, then they are also similar triangles.
Reason (R): All congruent triangles are similar, but similar triangles need not be congruent.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
If two triangles are congruent, their corresponding sides are equal and corresponding angles are equal. Since equal sides are in ratio $1:1$: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = 1$$ The condition for similarity is fully satisfied. Therefore every pair of congruent triangles is also similar (but not vice versa, as similar triangles can have sides in a ratio $k \neq 1$). Both A and R are true; R correctly explains A.
If two triangles are congruent, their corresponding sides are equal and corresponding angles are equal. Since equal sides are in ratio $1:1$: $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = 1$$ The condition for similarity is fully satisfied. Therefore every pair of congruent triangles is also similar (but not vice versa, as similar triangles can have sides in a ratio $k \neq 1$). Both A and R are true; R correctly explains A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In the given figures, $\triangle ABC \sim \triangle GHI$.
Reason (R): If the corresponding sides of two triangles are proportional, then the triangles are similar.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In the given figures, $\triangle ABC \sim \triangle GHI$.
Reason (R): If the corresponding sides of two triangles are proportional, then the triangles are similar.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
From the given figures, the corresponding sides of $\triangle ABC$ and $\triangle GHI$ satisfy: $$\frac{AB}{GH} = \frac{BC}{HI} = \frac{AC}{GI}$$ By the SSS similarity criterion, $\triangle ABC \sim \triangle GHI$. Assertion is true. Reason (SSS similarity) correctly explains the assertion.
From the given figures, the corresponding sides of $\triangle ABC$ and $\triangle GHI$ satisfy: $$\frac{AB}{GH} = \frac{BC}{HI} = \frac{AC}{GI}$$ By the SSS similarity criterion, $\triangle ABC \sim \triangle GHI$. Assertion is true. Reason (SSS similarity) correctly explains the assertion.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The sides of two similar triangles are in the ratio $2:5$, then the areas of these triangles are in the ratio $4:25$.
Reason (R): The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): The sides of two similar triangles are in the ratio $2:5$, then the areas of these triangles are in the ratio $4:25$.
Reason (R): The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
By the theorem on areas of similar triangles (Reason R): $$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}$$ This matches Assertion A. Both are true; R correctly explains A.
By the theorem on areas of similar triangles (Reason R): $$\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{side}_1}{\text{side}_2}\right)^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25}$$ This matches Assertion A. Both are true; R correctly explains A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In the given figure, $PA \| QB \| RC \| SD$.
Reason (R): If three or more line segments are perpendicular to one line, then they are parallel to each other.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In the given figure, $PA \| QB \| RC \| SD$.
Reason (R): If three or more line segments are perpendicular to one line, then they are parallel to each other.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
All line segments $PA$, $QB$, $RC$, and $SD$ are perpendicular to the same line in the given figure. By Euclidean geometry, if two or more lines are perpendicular to the same line, they are parallel to each other. Therefore $PA \| QB \| RC \| SD$. Both A and R are true; R correctly explains A.
All line segments $PA$, $QB$, $RC$, and $SD$ are perpendicular to the same line in the given figure. By Euclidean geometry, if two or more lines are perpendicular to the same line, they are parallel to each other. Therefore $PA \| QB \| RC \| SD$. Both A and R are true; R correctly explains A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, if $X$ and $Y$ are points on $AB$ and $AC$ respectively such that $\dfrac{AX}{XB} = \dfrac{3}{4}$, $AY = 5$ and $YC = 9$, then $XY$ is not parallel to $BC$.
Reason (R): If $XY \| BC$, $AY = 15$ and $YC = 20$, then the ratio of $XY$ and $BC$ is $3:4$.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, if $X$ and $Y$ are points on $AB$ and $AC$ respectively such that $\dfrac{AX}{XB} = \dfrac{3}{4}$, $AY = 5$ and $YC = 9$, then $XY$ is not parallel to $BC$.
Reason (R): If $XY \| BC$, $AY = 15$ and $YC = 20$, then the ratio of $XY$ and $BC$ is $3:4$.
Answer
Option (B) is correct. — Both A and R are true but R is NOT the correct explanation of A.
For Assertion: $\dfrac{AX}{XB} = \dfrac{3}{4}$ and $\dfrac{AY}{YC} = \dfrac{5}{9}$. Since $\dfrac{3}{4} \neq \dfrac{5}{9}$, by the converse of BPT, $XY \not\| BC$. Assertion is true.
For Reason: When $AY = 15$, $YC = 20$: $\dfrac{AY}{YC} = \dfrac{15}{20} = \dfrac{3}{4}$, so $XY \| BC$ and $XY : BC = 3:4$. Reason is true.
However, Reason talks about a different condition ($AY = 15$, $YC = 20$) and does not explain why the assertion is true. So R is not the correct explanation of A.
For Assertion: $\dfrac{AX}{XB} = \dfrac{3}{4}$ and $\dfrac{AY}{YC} = \dfrac{5}{9}$. Since $\dfrac{3}{4} \neq \dfrac{5}{9}$, by the converse of BPT, $XY \not\| BC$. Assertion is true.
For Reason: When $AY = 15$, $YC = 20$: $\dfrac{AY}{YC} = \dfrac{15}{20} = \dfrac{3}{4}$, so $XY \| BC$ and $XY : BC = 3:4$. Reason is true.
However, Reason talks about a different condition ($AY = 15$, $YC = 20$) and does not explain why the assertion is true. So R is not the correct explanation of A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, points $D$ and $E$ are on sides $CA$ and $CB$ respectively such that $DE \| AB$, $AD = 2x$, $DC = x+3$, $BE = 2x-1$ and $CE = x$, then the value of $x$ is $\dfrac{3}{5}$.
Reason (R): The length of $AC$ in $\triangle ABC$ is 4.8 units.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, points $D$ and $E$ are on sides $CA$ and $CB$ respectively such that $DE \| AB$, $AD = 2x$, $DC = x+3$, $BE = 2x-1$ and $CE = x$, then the value of $x$ is $\dfrac{3}{5}$.
Reason (R): The length of $AC$ in $\triangle ABC$ is 4.8 units.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
For Assertion: By BPT: $\dfrac{CD}{AD} = \dfrac{CE}{BE}$ $$\frac{x+3}{2x} = \frac{x}{2x-1}$$ Cross-multiplying: $(x+3)(2x-1) = 2x^2 \implies 5x = 3 \implies x = \dfrac{3}{5}$. Assertion is true.
For Reason: $$AC = AD + CD = 2x + x + 3 = 3x + 3 = 3 \times \frac{3}{5} + 3 = \frac{9}{5} + 3 = \frac{24}{5} = 4.8 \text{ units}$$ Reason is also true and follows directly from the value of $x$ found in A, so R correctly explains A.
For Assertion: By BPT: $\dfrac{CD}{AD} = \dfrac{CE}{BE}$ $$\frac{x+3}{2x} = \frac{x}{2x-1}$$ Cross-multiplying: $(x+3)(2x-1) = 2x^2 \implies 5x = 3 \implies x = \dfrac{3}{5}$. Assertion is true.
For Reason: $$AC = AD + CD = 2x + x + 3 = 3x + 3 = 3 \times \frac{3}{5} + 3 = \frac{9}{5} + 3 = \frac{24}{5} = 4.8 \text{ units}$$ Reason is also true and follows directly from the value of $x$ found in A, so R correctly explains A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): All regular polygons of the same number of sides such as equilateral triangles, squares, etc., are similar.
Reason (R): Two polygons of the same number of sides are said to be similar if their corresponding angles are equal and lengths of corresponding sides are proportional.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): All regular polygons of the same number of sides such as equilateral triangles, squares, etc., are similar.
Reason (R): Two polygons of the same number of sides are said to be similar if their corresponding angles are equal and lengths of corresponding sides are proportional.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
In regular polygons (equilateral triangles, squares, regular pentagons, etc.), all interior angles are equal and all sides are equal. Any two regular polygons with the same number of sides will therefore have equal corresponding angles and sides in a constant ratio. By definition of similarity (given by Reason R), they are similar. Both are true; R correctly explains A.
In regular polygons (equilateral triangles, squares, regular pentagons, etc.), all interior angles are equal and all sides are equal. Any two regular polygons with the same number of sides will therefore have equal corresponding angles and sides in a constant ratio. By definition of similarity (given by Reason R), they are similar. Both are true; R correctly explains A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, if $DE \| BC$ and intersects $AB$ at $D$ and $AC$ at $E$, then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
Reason (R): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then these sides are divided in the same ratio.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): In $\triangle ABC$, if $DE \| BC$ and intersects $AB$ at $D$ and $AC$ at $E$, then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
Reason (R): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then these sides are divided in the same ratio.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
In $\triangle ABC$, since $DE \| BC$, by Thales’ theorem (which is Reason R): $$\frac{AD}{DB} = \frac{AE}{EC}$$ This directly establishes Assertion A. Both are true; R is the correct explanation of A.
In $\triangle ABC$, since $DE \| BC$, by Thales’ theorem (which is Reason R): $$\frac{AD}{DB} = \frac{AE}{EC}$$ This directly establishes Assertion A. Both are true; R is the correct explanation of A.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the bisector of an angle of a triangle bisects the opposite side, then the triangle is isosceles.
Reason (R): The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false but R is True
Assertion (A): If the bisector of an angle of a triangle bisects the opposite side, then the triangle is isosceles.
Reason (R): The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Answer
Option (A) is correct. — Both A and R are true and R is the correct explanation of A.
In $\triangle ABC$, let $AD$ be the bisector of $\angle A$. By the Angle Bisector Theorem (Reason R): $$\frac{AB}{AC} = \frac{BD}{DC}$$ If $AD$ bisects $BC$ (i.e., $D$ is the midpoint), then $BD = DC$, so: $$\frac{AB}{AC} = 1 \implies AB = AC$$ Hence $\triangle ABC$ is isosceles. Assertion is true; Reason correctly explains it.
In $\triangle ABC$, let $AD$ be the bisector of $\angle A$. By the Angle Bisector Theorem (Reason R): $$\frac{AB}{AC} = \frac{BD}{DC}$$ If $AD$ bisects $BC$ (i.e., $D$ is the midpoint), then $BD = DC$, so: $$\frac{AB}{AC} = 1 \implies AB = AC$$ Hence $\triangle ABC$ is isosceles. Assertion is true; Reason correctly explains it.
Frequently Asked Questions
What is the Triangles chapter about in CBSE Class 10 Maths? ⌄
The Triangles chapter in CBSE Class 10 Maths focuses on two key ideas: similarity of triangles and criteria for proving triangles similar (AA, SAS, SSS). It also covers the Basic Proportionality Theorem (BPT or Thales’ Theorem), the Pythagoras Theorem, and real-life applications such as finding heights of towers and shadows using similar triangles. Your child will learn to reason geometrically and apply these properties to solve problems efficiently.
How many marks does the Triangles chapter carry in the CBSE Class 10 board exam? ⌄
Triangles is part of the Geometry unit, which carries approximately 15 marks in the CBSE Class 10 board exam. The chapter itself contributes a significant share of those marks through MCQs, short answer, long answer, and case study questions. It is one of the most frequently tested chapters, making thorough preparation very important for your child’s score.
What are the most important topics in the Triangles chapter for CBSE Class 10? ⌄
The most important topics are: (1) Basic Proportionality Theorem and its converse, (2) AA, SAS, and SSS similarity criteria, (3) the relationship between areas of similar triangles and their sides, (4) the ratio of altitudes, medians, and perimeters in similar triangles, and (5) real-life case studies involving shadows and scale models. Assertion-Reason questions on BPT and similarity criteria appear very frequently in board exams.
What are the common mistakes students make in the Triangles chapter? ⌄
Common mistakes include: confusing similarity with congruence (similar triangles have proportional — not equal — sides), writing the similarity correspondence in the wrong order (which changes the ratios), forgetting that the ratio of areas equals the square of the ratio of sides, and not verifying both conditions (equal angles AND proportional sides) when applying similarity criteria. Careful attention to which vertices correspond is essential in every proof.
How does Angle Belearn help students master the Triangles chapter? ⌄
Angle Belearn’s CBSE specialists have carefully designed these competency-based questions to match the exact difficulty level and question types seen in board exams, including MCQs, case studies, and assertion-reason questions. Each question comes with a detailed step-by-step solution so your child understands not just the answer but the reasoning behind it. With structured practice across all question types, students build the conceptual clarity and exam confidence needed to score well in Triangles.
