Assertion (A): ABCD is a rectangle such that \[\angle CAB=60{}^\circ \]and \[\text{AC}=\text{a}\] units. The area of rectangle ABCD is \[\frac{\sqrt{3}}{2}{{a}^{2}}\]. |
Reason (R): The value of \[\sin 60{}^\circ \] is \[\frac{\sqrt{3}}{2}\] and \[\cos 60{}^\circ \]is \[\frac{1}{2}\]. |
A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A)
B) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A)
C) Assertion (A) is true but Reason (R) is false
D) Assertion (A) is false but Reason (R) is true
Correct Answer: D
Solution :
[d] In \[\Delta ABC,AC=a\ \operatorname{units}\ \angle A={{60}^{\operatorname{o}}}\] |
\[\sin {{60}^{\operatorname{o}}}=\frac{BC}{AC}=\frac{BC}{a}\] |
\[\Rightarrow \ \ \ \frac{\sqrt{3}}{2}=\frac{BC}{a}\ \ \ \Rightarrow \ \ BC=\frac{\sqrt{3}a}{2}\] |
Also \[\cos {{60}^{\operatorname{o}}}\frac{AB}{AC}\ \ \ \Rightarrow \ \ \frac{1}{2}=\frac{AB}{a}\] |
\[\Rightarrow \ \ \ AB=a/2\] |
\[\therefore \ \operatorname{Area}\ of\ rectangle\ ABCD=AB\times BC\] |
\[=\frac{a}{2}\times \ \ \frac{\sqrt{3}a}{2}=\frac{\sqrt{3}{{a}^{2}}}{4}\] |
\[\therefore \] Assertion False; Reason; True. |
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