Question : Let ABC and PQR be two congruent triangles such that $\angle A = \angle P = 90^{\circ}$. If BC = 17 cm, PR = 8 cm, find AB (in cm).
Option 1: 12
Option 2: 15
Option 3: 14
Option 4: 9
Correct Answer: 15
Solution : $\angle A = \angle P = 90^{\circ}$ BC = 17 cm PR = 8 cm Since $\triangle ABC$ is congruent to $\triangle PQR$, AC = PR = 8 cm By Pythagoras theorem, $BC^2 = AB^2 + AC^2$ ⇒ $17^2 = AB^2 + 8^2$ ⇒ $289 = AB^2 + 64$ ⇒ $AB^2 = 289 - 64 = 225$ ⇒ $AB = 15$ Hence, the correct answer is 15.
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Question : In a $\triangle \mathrm{PQR}$ and $\triangle\mathrm{ABC}$, $\angle$P = $\angle$A and AC = PR. Which of the following conditions is true for $\triangle$PQR and $\triangle$ABC to be congruent?
Option 1: AB = PQ by SSS
Option 2: AB = PQ by SAS
Option 3: BC = QR by ASS
Option 4: $\angle$Q = $\angle$B by AAA
Question : In ΔABC with sides 8 cm, 9 cm and 12 cm, the angle bisector of the largest angle divides the opposite sides into two segments. What is the length of the shorter segment?
Option 1: $5 \frac{11}{17} {~cm}$
Option 2: $4\frac{11}{17} {~cm}$
Option 3: $6\frac{13}{17} {~cm}$
Option 4: $3\frac{9}{17} {~cm}$
Question : In $\triangle$ ABC, $\angle$ C = 90$^{\circ}$. M and N are the midpoints of sides AB and AC, respectively. CM and BN intersect each other at D and $\angle$ BDC = 90$^{\circ}$. If BC = 8 cm, then the length of BN is:
Option 1: $6 \sqrt{3} {~cm}$
Option 2: $6 \sqrt{6} {~cm}$
Option 3: $4 \sqrt{6} {~cm}$
Option 4: $8 \sqrt{3} {~cm}$
Question : In $\triangle$PQR, $\angle$ PQR = $90^{\circ}$, PQ = 5 cm and QR = 12 cm. What is the radius (in cm) of the circumcircle of $\triangle$PQR?
Option 1: 6.5
Option 2: 7.5
Option 3: 13
Option 4: 15
Question : $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$ are two triangles such that $\triangle \mathrm{ABC} \cong \triangle \mathrm{FDE}$. If AB = 5 cm, $\angle$B = 40° and $\angle$A = 80°, then which of the following options is true?
Option 1: DF = 5 cm, $\angle$E = 60°
Option 2: DE = 5 cm, $\angle$F = 60°
Option 3: DE = 5 cm, $\angle$D = 60°
Option 4: DE = 5 cm, $\angle$E = 60°
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