Question : $a, b, c$ are the lengths of three sides of a triangle ABC. If $a, b, c$ are related by the relation $a^{2}+b^{2}+c^{2}=ab+bc+ca$, then the value of $\sin^{2}A+\sin^{2}B+\sin^{2}C$ is:
Option 1: $\frac{3}{2}$
Option 2: $\frac{3\sqrt3}{2}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{9}{4}$
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Correct Answer: $\frac{9}{4}$
Solution : Given $a^{2}+b^{2}+c^{2}=ab+bc+ca$ $⇒2a^{2}+2b^{2}+2c^{2}=2ab+2bc+2ca$ $⇒(a^{2}+b^{2}-2ab)+(b^2+c^{2}-2bc)+(a^2+c^2-2ac)=0$ $⇒(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$ $⇒a=b=c$ Which means the triangle is equilateral. So, all angles are of 60º each Therefore, $\sin^{2}A+\sin^{2}B+\sin^{2}C=\sin^{2}60º+\sin^{2}60º+\sin^{2}60º = \frac{3}{4}+ \frac{3}{4}+ \frac{3}{4}= \frac{9}{4}$ Hence, the correct answer is $\frac{9}{4}$.
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Question : D, E, and F are the midpoints of the sides BC, CA, and AB, respectively of a $\triangle ABC$. Then the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is:
Question : In $\triangle ABC$, right angled at B, if $\tan A=\frac{1}{2}$, then the value of $\frac{\sin A(\cos C+\cos A)}{\cos C(\sin C-\sin A)}$ is:
Question : $\triangle ABC$ is a right triangle. If $\angle B=90^{\circ}$ and $\tan A=\frac{1}{\sqrt{2}}$, then the value of $\sin A \cos C + \cos A \sin C$ is:
Question : If $\frac{2+a}{a}+\frac{2+b}{b}+\frac{2+c}{c}=4$, then the value of $\frac{ab+bc+ca}{abc}$ is:
Question : Three sides of a triangle are $\sqrt{a^2+b^2}, \sqrt{(2 a)^2+b^2}$, and $\sqrt{a^2+(2 b)^2}$ units. What is the area (in unit squares) of the triangle?
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