Question : If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, $\angle$BGC = $120^{\circ}$, BC = 10 cm, then the area of the triangle ABC is:
Option 1: $50\sqrt{3}$ $cm^2$
Option 2: $60$ $cm^2$
Option 3: $25$ $cm^2$
Option 4: $25\sqrt{3}$ $cm^2$
Correct Answer: $25\sqrt{3}$ $cm^2$
Solution : In $\triangle$ BGC, BG = CG, $\angle$BGC = $120^{\circ}$, BC = 10 cm $\triangle$ BGC is an isosceles triangle. ⇒ $\angle$GBC = $\angle$GCB Sum of angles of a triangle = $180^{\circ}$ $\angle$GBC + $\angle$GCB + $\angle$BGC = $180^{\circ}$ $2\angle$GBC = $180^{\circ}-120^{\circ}$ = $60^{\circ}$ $\angle$GBC = $\frac{60^{\circ}}{2}$ = $30^{\circ}$ GD $\perp$ BC $\angle$BGD = $\angle$CGD = $60^{\circ}$ $\cos 30^{\circ} = \frac{BD}{BG}$ ⇒ BG = $5 × \frac{2}{\sqrt3}$ = $\frac{10}{\sqrt3}$ ⇒ GD = BG × $\sin 30^{\circ}$ = $\frac{10}{\sqrt3}$ × $\frac{1}{2}$ = $\frac{5}{\sqrt3}$ Since G is the centroid of $\triangle$ABC, AG : GD = 2 :1 AD : GD = 3 : 1 AD = 3 ×$\frac{5}{\sqrt3}$ = $\frac{15}{\sqrt3}$ Since AD is the altitude of $\triangle$ABC, Area of $\triangle$ABC = $\frac{1}{2} × BC × AD = \frac{1}{2}×10×\frac{15}{\sqrt3} = \frac{75}{\sqrt3} = 25\sqrt3 \;cm^2$ Hence, the correct answer is $25\sqrt3$ $cm^2$.
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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 a triangle ABC, if $\angle B=90^{\circ}, \angle C=45^{\circ}$ and AC = 4 cm, then the value of BC is:
Option 1: $\sqrt{2} \mathrm{~cm}$
Option 2: $4 \mathrm{~cm}$
Option 3: $2 \sqrt{2} \mathrm{~cm}$
Option 4: $4 \sqrt{2} \mathrm{~cm}$
Question : If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB = 19 cm and AC = 22 cm then the length of BC is:
Option 1: 19.5 cm
Option 2: 26 cm
Option 3: 20.5 cm
Option 4: 13 cm
Question : Internal bisectors of $\angle$ B and $\angle$ C of $\triangle$ ABC meet at O. If $\angle$ BAC = $80^{\circ}$, then the value of $\angle$ BOC is:
Option 1: $120^{\circ}$
Option 2: $140^{\circ}$
Option 3: $110^{\circ}$
Option 4: $130^{\circ}$
Question : In $\triangle$ ABC, $\angle$ BCA = $90^{\circ}$, AC = 24 cm and BC = 10 cm. What is the radius (in cm) of the circumcircle of $\triangle$ ABC?
Option 1: 12.5
Option 2: 13
Option 3: 25
Option 4: 26
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