87062
The area of the circle passes through the point \((4,6)\) and whose centre is \((1,2)\) is
1 \(5 \pi\) sq units
2 \(10 \pi\) sp units
3 \(25 \pi\) sq units
4 \(35 \pi\) sq units
Explanation:
(C) : Given, \(\text { Point }(p)=(4,6)\) Center \((\mathrm{c})=(1,2)\) So, radius \((\mathrm{r})=\sqrt{(1-4)^{2}+(2-6)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}\) \(=\sqrt{9+16}=\sqrt{25}=5\) Hence, area \((\mathrm{A})=\pi \mathrm{r}^{2}=\pi \times(5)^{2}=25 \pi\) square units
Jamia Millia Islamia-2012
Application of the Integrals
87064
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
(C) : \(x=30 \cos \theta \quad\) and \(y=30 \sin \theta\) Area of trapezium \(A B C D\) is \(A=\frac{1}{2} \times(30+30+2 x)(y) \Rightarrow A=\frac{1}{2} \times(60+2 x) y\) \(A=(30+x) y\) \(A=(30+30 \cos \theta)(30 \sin \theta)\) \(\mathrm{A}=900(\sin \theta+\sin \theta \times \cos \theta)\) Differentiating w.r.t, \(\theta_{1}\) we get - \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=900\left(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta\right)\) For maxima or minima \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=0\) \(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta=0\) \(2 \cos ^{2} \theta+\cos \theta-1=0\) \((2 \cos \theta-1)(\cos \theta+1)=0\) \(\cos \theta=\frac{1}{2} \text { or } \cos \theta=-1\) \(\quad \theta=\frac{\pi}{3} \text { or } \pi\) Hence, smallest angle is \(\frac{\pi}{3}\).
87062
The area of the circle passes through the point \((4,6)\) and whose centre is \((1,2)\) is
1 \(5 \pi\) sq units
2 \(10 \pi\) sp units
3 \(25 \pi\) sq units
4 \(35 \pi\) sq units
Explanation:
(C) : Given, \(\text { Point }(p)=(4,6)\) Center \((\mathrm{c})=(1,2)\) So, radius \((\mathrm{r})=\sqrt{(1-4)^{2}+(2-6)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}\) \(=\sqrt{9+16}=\sqrt{25}=5\) Hence, area \((\mathrm{A})=\pi \mathrm{r}^{2}=\pi \times(5)^{2}=25 \pi\) square units
Jamia Millia Islamia-2012
Application of the Integrals
87064
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
(C) : \(x=30 \cos \theta \quad\) and \(y=30 \sin \theta\) Area of trapezium \(A B C D\) is \(A=\frac{1}{2} \times(30+30+2 x)(y) \Rightarrow A=\frac{1}{2} \times(60+2 x) y\) \(A=(30+x) y\) \(A=(30+30 \cos \theta)(30 \sin \theta)\) \(\mathrm{A}=900(\sin \theta+\sin \theta \times \cos \theta)\) Differentiating w.r.t, \(\theta_{1}\) we get - \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=900\left(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta\right)\) For maxima or minima \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=0\) \(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta=0\) \(2 \cos ^{2} \theta+\cos \theta-1=0\) \((2 \cos \theta-1)(\cos \theta+1)=0\) \(\cos \theta=\frac{1}{2} \text { or } \cos \theta=-1\) \(\quad \theta=\frac{\pi}{3} \text { or } \pi\) Hence, smallest angle is \(\frac{\pi}{3}\).
87062
The area of the circle passes through the point \((4,6)\) and whose centre is \((1,2)\) is
1 \(5 \pi\) sq units
2 \(10 \pi\) sp units
3 \(25 \pi\) sq units
4 \(35 \pi\) sq units
Explanation:
(C) : Given, \(\text { Point }(p)=(4,6)\) Center \((\mathrm{c})=(1,2)\) So, radius \((\mathrm{r})=\sqrt{(1-4)^{2}+(2-6)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}\) \(=\sqrt{9+16}=\sqrt{25}=5\) Hence, area \((\mathrm{A})=\pi \mathrm{r}^{2}=\pi \times(5)^{2}=25 \pi\) square units
Jamia Millia Islamia-2012
Application of the Integrals
87064
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
(C) : \(x=30 \cos \theta \quad\) and \(y=30 \sin \theta\) Area of trapezium \(A B C D\) is \(A=\frac{1}{2} \times(30+30+2 x)(y) \Rightarrow A=\frac{1}{2} \times(60+2 x) y\) \(A=(30+x) y\) \(A=(30+30 \cos \theta)(30 \sin \theta)\) \(\mathrm{A}=900(\sin \theta+\sin \theta \times \cos \theta)\) Differentiating w.r.t, \(\theta_{1}\) we get - \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=900\left(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta\right)\) For maxima or minima \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=0\) \(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta=0\) \(2 \cos ^{2} \theta+\cos \theta-1=0\) \((2 \cos \theta-1)(\cos \theta+1)=0\) \(\cos \theta=\frac{1}{2} \text { or } \cos \theta=-1\) \(\quad \theta=\frac{\pi}{3} \text { or } \pi\) Hence, smallest angle is \(\frac{\pi}{3}\).
87062
The area of the circle passes through the point \((4,6)\) and whose centre is \((1,2)\) is
1 \(5 \pi\) sq units
2 \(10 \pi\) sp units
3 \(25 \pi\) sq units
4 \(35 \pi\) sq units
Explanation:
(C) : Given, \(\text { Point }(p)=(4,6)\) Center \((\mathrm{c})=(1,2)\) So, radius \((\mathrm{r})=\sqrt{(1-4)^{2}+(2-6)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}\) \(=\sqrt{9+16}=\sqrt{25}=5\) Hence, area \((\mathrm{A})=\pi \mathrm{r}^{2}=\pi \times(5)^{2}=25 \pi\) square units
Jamia Millia Islamia-2012
Application of the Integrals
87064
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
(C) : \(x=30 \cos \theta \quad\) and \(y=30 \sin \theta\) Area of trapezium \(A B C D\) is \(A=\frac{1}{2} \times(30+30+2 x)(y) \Rightarrow A=\frac{1}{2} \times(60+2 x) y\) \(A=(30+x) y\) \(A=(30+30 \cos \theta)(30 \sin \theta)\) \(\mathrm{A}=900(\sin \theta+\sin \theta \times \cos \theta)\) Differentiating w.r.t, \(\theta_{1}\) we get - \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=900\left(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta\right)\) For maxima or minima \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=0\) \(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta=0\) \(2 \cos ^{2} \theta+\cos \theta-1=0\) \((2 \cos \theta-1)(\cos \theta+1)=0\) \(\cos \theta=\frac{1}{2} \text { or } \cos \theta=-1\) \(\quad \theta=\frac{\pi}{3} \text { or } \pi\) Hence, smallest angle is \(\frac{\pi}{3}\).
87062
The area of the circle passes through the point \((4,6)\) and whose centre is \((1,2)\) is
1 \(5 \pi\) sq units
2 \(10 \pi\) sp units
3 \(25 \pi\) sq units
4 \(35 \pi\) sq units
Explanation:
(C) : Given, \(\text { Point }(p)=(4,6)\) Center \((\mathrm{c})=(1,2)\) So, radius \((\mathrm{r})=\sqrt{(1-4)^{2}+(2-6)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}\) \(=\sqrt{9+16}=\sqrt{25}=5\) Hence, area \((\mathrm{A})=\pi \mathrm{r}^{2}=\pi \times(5)^{2}=25 \pi\) square units
Jamia Millia Islamia-2012
Application of the Integrals
87064
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the nonparallel sides are all equal to 30 . In order to maximize the area of the trapezium, the smallest angle should be
1 \(\frac{\pi}{6}\)
2 \(\frac{\pi}{4}\)
3 \(\frac{\pi}{3}\)
4 \(\frac{\pi}{2}\)
Explanation:
(C) : \(x=30 \cos \theta \quad\) and \(y=30 \sin \theta\) Area of trapezium \(A B C D\) is \(A=\frac{1}{2} \times(30+30+2 x)(y) \Rightarrow A=\frac{1}{2} \times(60+2 x) y\) \(A=(30+x) y\) \(A=(30+30 \cos \theta)(30 \sin \theta)\) \(\mathrm{A}=900(\sin \theta+\sin \theta \times \cos \theta)\) Differentiating w.r.t, \(\theta_{1}\) we get - \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=900\left(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta\right)\) For maxima or minima \(\frac{\mathrm{dA}}{\mathrm{d} \theta}=0\) \(\cos \theta+\cos ^{2} \theta-\sin ^{2} \theta=0\) \(2 \cos ^{2} \theta+\cos \theta-1=0\) \((2 \cos \theta-1)(\cos \theta+1)=0\) \(\cos \theta=\frac{1}{2} \text { or } \cos \theta=-1\) \(\quad \theta=\frac{\pi}{3} \text { or } \pi\) Hence, smallest angle is \(\frac{\pi}{3}\).
