Question : In $\triangle A B C$, M is the midpoint of the side AB. N is a point in the interior of $\triangle A B C$ such that CN is the bisector of $\angle C$ and $C N \perp N B$. What is the length (in cm) of MN, if BC = 10 cm and AC = 15 cm?
Option 1: 5
Option 2: 4
Option 3: 2.5
Option 4: 2
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Correct Answer: 2.5
Solution :
In $\triangle ABC, AM = MB$ In $\triangle NPC$ and $\triangle NBC,$ ⇒ $\angle BCN = \angle NCP$ ⇒ $\angle CNB = \angle CNP = 90^{\circ}$ ⇒ NC = NC (common side) $\therefore \triangle NPC$ is congruent to $\triangle NBC$ NB = NP BC = PC = 10 cm AP = AC – PC = 15 – 10 = 5 cm In $\triangle ABP$, M and N are midpoints of AB and BP By the midpoint theorem, $MN = \frac{AP}{2} = \frac{5}{2} = 2.5$ cm Hence, the correct answer is 2.5.
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Question : In triangle ABC, AD is the angle bisector of angle A. If AB = 8.4 cm, AC = 5.6 cm and DC = 2.8 cm, then the length of side BC will be:
Option 1: 4.2 cm
Option 2: 5.6 cm
Option 3: 7 cm
Option 4: 2.8 cm
Question : Suppose $\triangle A B C$ be a right-angled triangle where $\angle A=90^{\circ}$ and $A D \perp B C$. If area $(\triangle A B C)$ $=80 \mathrm{~cm}^2$ and $BC=16$ cm, then the length of $AD$ is:
Option 1: 10 cm
Option 2: 24 cm
Option 3: 18 cm
Option 4: 12 cm
Question : O is a point in the interior of $\triangle \mathrm{ABC}$ such that OA = 12 cm, OC = 9 cm, $\angle A O B=\angle B O C=\angle C O A$ and $\angle A B C=60^{\circ}$. What is the length (in cm) of OB?
Option 1: $6 \sqrt{3}$
Option 2: $4 \sqrt{6}$
Option 3: $4 \sqrt{3}$
Option 4: $6 \sqrt{2}$
Question : Suppose $\triangle ABC$ be a right-angled triangle where $\angle A=90°$ and $AD\perp BC$. If the area of $\triangle ABC =40$ cm$^{2}$ and $\triangle ACD =10$ cm$^{2}$ and $\overline{AC}=9$ cm, then the length of $BC$ is:
Option 1: 12 cm
Option 2: 18 cm
Option 3: 4 cm
Option 4: 6 cm
Question : In triangle ABC, $\angle$ B = 90°, and $\angle$C = 45°. If AC = $2 \sqrt{2}$ cm then the length of BC is:
Option 1: 3 cm
Option 2: 2 cm
Option 3: 1 cm
Option 4: 4 cm
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