MHT CET · Maths · Properties of Triangles
The area of the triangle \(\mathrm{ABC}\) is \(10 \sqrt{3} \mathrm{~cm}^2\), angle \(\mathrm{B}\) is \(60^{\circ}\) and its perimeter is \(20 \mathrm{~cm}\), then \(\ell(\mathrm{AC})=\)
- A 10 cm
- B 8 cm
- C 5 cm
- D 7 cm
Answer & Solution
Correct Answer
(D) 7 cm
Step-by-step Solution
Detailed explanation
\(\text {Area }=\frac{\operatorname{acsin} B}{2}=\Rightarrow 10 \sqrt{3}=\frac{\operatorname{acsin} 60^{\circ}}{2} \\ 10 \sqrt{3}=\frac{a c \sqrt{3}}{4} \Rightarrow a c=40\)
Now
\(\cos B=\frac{a^2+c^2-b^2}{2 a c} \)
\( \cos 60^{\circ} \times 2 a c=a^2+c^2-b^2 \)
\( \therefore \frac{1}{2} \times 2 \times 40=\left(a^2+c^2\right)-2 a c-b^2 \)
\( \therefore 40=(20-b)^2-(2 \times 40)-b^2 \quad \ldots[a+b\) \(+c=20, \text { given }] \)
\( =400+b^2-40 b-80-b^2 \)
\( \therefore 40 b=280 \Rightarrow b=7\)
Now
\(\cos B=\frac{a^2+c^2-b^2}{2 a c} \)
\( \cos 60^{\circ} \times 2 a c=a^2+c^2-b^2 \)
\( \therefore \frac{1}{2} \times 2 \times 40=\left(a^2+c^2\right)-2 a c-b^2 \)
\( \therefore 40=(20-b)^2-(2 \times 40)-b^2 \quad \ldots[a+b\) \(+c=20, \text { given }] \)
\( =400+b^2-40 b-80-b^2 \)
\( \therefore 40 b=280 \Rightarrow b=7\)
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