Solution: (B) Similar Reason: All circles have the same shape (round) but can have different sizes (radii). Thus, they are similar but not necessarily congruent.
Question 2: In $\Delta ABC$, $DE || BC$ such that $D$ is on $AB$ and $E$ is on $AC$. If $AD = 1.5$ cm, $DB = 3$ cm and $AE = 1$ cm, then $EC$ is:
(A) 2 cm
(B) 1.5 cm
(C) 3 cm
(D) 2.5 cm
Solution: (A) 2 cm Step 1: By Basic Proportionality Theorem (BPT), $\frac{AD}{DB} = \frac{AE}{EC}$. Step 2: $\frac{1.5}{3} = \frac{1}{EC} \Rightarrow \frac{1}{2} = \frac{1}{EC} \Rightarrow EC = 2$ cm.
Question 3: If $\Delta ABC \sim \Delta DEF$, $\angle A = 47^\circ$ and $\angle E = 83^\circ$, then $\angle C$ is:
(A) $47^\circ$
(B) $50^\circ$
(C) $83^\circ$
(D) $60^\circ$
Solution: (B) $50^\circ$ Step 1: Since $\Delta ABC \sim \Delta DEF$, corresponding angles are equal. $\angle A = \angle D = 47^\circ$ and $\angle B = \angle E = 83^\circ$. Step 2: In $\Delta ABC$, $\angle A + \angle B + \angle C = 180^\circ$. Step 3: $47^\circ + 83^\circ + \angle C = 180^\circ \Rightarrow 130^\circ + \angle C = 180^\circ \Rightarrow \angle C = 50^\circ$.
Question 4: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. The height of the tower is:
(A) 40 m
(B) 42 m
(C) 38 m
(D) 44 m
Solution: (B) 42 m Step 1: The triangles formed by the pole/tower and their shadows are similar (sun's elevation is same). Step 2: $\frac{\text{Height of Pole}}{\text{Shadow of Pole}} = \frac{\text{Height of Tower}}{\text{Shadow of Tower}}$. Step 3: $\frac{6}{4} = \frac{h}{28} \Rightarrow h = \frac{6 \times 28}{4} = 6 \times 7 = 42$ m.
Question 5: In $\Delta ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE || BC$. If $AD = x, DB = x-2, AE = x+2$ and $EC = x-1$, then the value of $x$ is:
Question 6: The diagonals of a trapezium $ABCD$ with $AB || DC$ intersect each other at the point $O$. If $AB = 2CD$, then the ratio of the areas of triangles $AOB$ and $COD$ is:
(A) 1:2
(B) 1:4
(C) 4:1
(D) 2:1
Solution: (C) 4:1 Step 1: $\Delta AOB \sim \Delta COD$ (AA similarity using alternate interior angles). Step 2: Ratio of areas = Ratio of squares of corresponding sides. Step 3: $\frac{\text{Area}(AOB)}{\text{Area}(COD)} = \left(\frac{AB}{CD}\right)^2 = \left(\frac{2CD}{CD}\right)^2 = (2)^2 = 4:1$.
Question 7: If $\Delta ABC \sim \Delta QRP$, $\frac{\text{Area}(ABC)}{\text{Area}(QRP)} = \frac{9}{4}$, $AB = 18$ cm and $BC = 15$ cm, then $PR$ is equal to:
Question 8: It is given that $\Delta ABC \sim \Delta DFE$, $\angle A = 30^\circ$, $\angle C = 50^\circ$, $AB = 5$ cm, $AC = 8$ cm and $DF = 7.5$ cm. Then which of the following is true?
(A) $DE = 12$ cm, $\angle F = 50^\circ$
(B) $DE = 12$ cm, $\angle F = 100^\circ$
(C) $EF = 12$ cm, $\angle D = 100^\circ$
(D) $EF = 12$ cm, $\angle D = 30^\circ$
Solution: (B) $DE = 12$ cm, $\angle F = 100^\circ$ Step 1: $\Delta ABC \sim \Delta DFE \Rightarrow \angle A = \angle D = 30^\circ$, $\angle C = \angle E = 50^\circ$. Step 2: $\angle B = \angle F = 180 - (30+50) = 100^\circ$. Step 3: $\frac{AB}{DF} = \frac{AC}{DE} \Rightarrow \frac{5}{7.5} = \frac{8}{DE} \Rightarrow DE = \frac{8 \times 7.5}{5} = 12$ cm.
Question 9: In $\Delta ABC$, $AB = 6\sqrt{3}$ cm, $AC = 12$ cm and $BC = 6$ cm. The angle $B$ is:
(A) $120^\circ$
(B) $60^\circ$
(C) $90^\circ$
(D) $45^\circ$
Solution: (C) $90^\circ$ Step 1: Check Pythagoras theorem. $AB^2 = (6\sqrt{3})^2 = 108$, $BC^2 = 6^2 = 36$, $AC^2 = 12^2 = 144$. Step 2: $AB^2 + BC^2 = 108 + 36 = 144 = AC^2$. Step 3: Since square of one side is sum of squares of other two, angle opposite to longest side ($AC$) is $90^\circ$. So $\angle B = 90^\circ$.
Question 10: If in two triangles $DEF$ and $PQR$, $\angle D = \angle Q$ and $\angle R = \angle E$, then which of the following is not true?
(A) $\frac{EF}{PR} = \frac{DF}{PQ}$
(B) $\frac{DE}{PQ} = \frac{EF}{RP}$
(C) $\frac{DE}{QR} = \frac{DF}{PQ}$
(D) $\frac{EF}{RP} = \frac{DE}{QR}$
Solution: (B) $\frac{DE}{PQ} = \frac{EF}{RP}$ Step 1: Match correspondence: $\angle D = \angle Q$, $\angle E = \angle R$, $\angle F = \angle P$. Step 2: So $\Delta DEF \sim \Delta QRP$. Step 3: Ratios: $\frac{DE}{QR} = \frac{EF}{RP} = \frac{DF}{QP}$. Step 4: Option (B) says $\frac{DE}{PQ}$, which is incorrect (should be $QR$).