on set 
.Solution: It is given that $x_{4}$ on set $S=\{0,1,2,3\}$ Here, $1 \times_{4} 1=$ remainder obtained by dividing $1 \times 1$ by 4 $=1$. $0 \times_{4} 1=$ remainder obtained by dividing $0 \times 1$...
![\[f(x)=\left\{\begin{array}{c} a x+5, \text { if } x \leq 2 \\ x-1, \text { if } x>2 \end{array} \text { is continuous at } x=2\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a09e82237af54cf19308b9d94b66dc50_l3.png "Rendered by QuickLaTeX.com")
so that the function 
is continuous at 
so that the function ![\[f(x)=\left\{\begin{array}{c} k x^{2}, \text { if } x \leq 2 \\ 3, \text { if } x>2 \end{array} \text { is continuous at } x=2\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-43406b4f8bc5a062aa3db52e684ee13b_l3.png "Rendered by QuickLaTeX.com")
is the function ![\[f(x)=\left\{\begin{array}{c} \frac{\sin 5 x}{3 x}, \text { if } x \neq 0 \\ k, \text { if } x=0 \end{array} \text { continuous at } x=0 ?\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f964ca249dd8e4fc2c980d1759cab1f3_l3.png "Rendered by QuickLaTeX.com")
so that the function ![\[f(x)=\left\{\begin{array}{c} \frac{x^{2}-3 x+2}{x-1}, x \neq 1 \\ k, x=1 \end{array} \text { continuous at } x=1\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f794bb359aab15e94ba6e8f267e25451_l3.png "Rendered by QuickLaTeX.com")
is the function ![\[f(x)=\left\{\begin{array}{c} \frac{x^{2}-1}{x-1}, x \neq 1 \\ k, x=1 \end{array} \text { continuous at } x=1 ?\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f21de786db7b02f065930decb0a2c7ae_l3.png "Rendered by QuickLaTeX.com")
![\[f(x)=\left\{\begin{array}{l} 2 x-1, x<0 \\ 2 x+1, x \geq 0 \end{array} \text { atx }=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b2c7b98d82b533cbe29d1e14bef0f65a_l3.png "Rendered by QuickLaTeX.com")
![\[f(x)=\left\{\begin{array}{c} x, 0 \leq x<\frac{1}{2} \\ 12, x=\frac{1}{2} \\ 1-x, \frac{1}{2}<x \leq 1 \end{array} \text { at the point } x=\frac{1}{2}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d7298f90996c8169bd410fb07fc7ae5f_l3.png "Rendered by QuickLaTeX.com")
![\[f(x)=\left\{\begin{array}{l} x, x>0 \\ 1, x=0 \\ -x, x<0 \end{array} \text { at the point } x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-540d79cc489467c89855e0b0dd0fc0ad_l3.png "Rendered by QuickLaTeX.com")
![\[f(x)=\left\{\begin{array}{c} 3 x-2, x \leq 0 \\ x+1, x>0 \end{array} \text { at } x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cdb6798d28f709abfc52b59a3fb65cd0_l3.png "Rendered by QuickLaTeX.com")
Hence f (x) is discontinuous at x = 0
defined by ![\[f(x)=\left\{\begin{array}{l} a \sin \frac{\pi}{2}(x+1), x \leq 0 \\ \frac{\tan x-\sin x}{x^{3}}, x>0 \end{array} \text { is continuous at } x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1dc7af2b6c496ba1c87e5440b53c4221_l3.png "Rendered by QuickLaTeX.com")
![\[\text { Show that } f(x)=\left\{\begin{array}{l} \frac{\sin 3 x}{\tan 2 x}, \text { if } x<0 \\ \frac{3}{2}, \text { if } x=0 \quad \text { is continuous at } x=0 \\ \frac{\log (1+3 x)}{e^{2 x}-1, \text { if } x>0} \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-87c4f235c04644629d21a9747fd1a6a8_l3.png "Rendered by QuickLaTeX.com")
![\[\text { Show that } f(x)=\left\{\begin{array}{c} 1+x^{2}, \text { if } 0 \leq x \leq 1 \\ 2-x, \text { if } x>1 . \end{array} \text { is discontinuous at } x=1\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6b32b24d736bab5b3fe1a44ea9d321f3_l3.png "Rendered by QuickLaTeX.com")
(viii) 
(vii) (viii)
![\[\text { (v) } f(x)=\left\{\begin{array}{ll} \frac{1-x^{n}}{1-x}, & x \neq 1 \\ n-1, & x=1 \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c0d055098c3d6ba1b6a3278422c737fa_l3.png "Rendered by QuickLaTeX.com")
atx 
(v) (vi)
![\[\text { (iii) } f(x)=\left\{\begin{array}{ll} (x-a) \sin \left(\frac{1}{x-a}\right), & x \neq a \\ 0, & x=a \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0acc6a26ee1703dab6936fc6b93d2816_l3.png "Rendered by QuickLaTeX.com")
![\[\text { (iv) } f(x)=\left\{\begin{array}{ll} \frac{e^{x}-1}{\log (1+2 x)}, \text { if } & x \neq a \\ 7, i f & x=0 \end{array} a t x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a06047faf6069d4f50e3af6ae1c5967c_l3.png "Rendered by QuickLaTeX.com")
(iii) (iv)
atx 
![\[\text { (ii) } f(x)=\left\{\begin{array}{ll} x^{2} \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0 \end{array} a t x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e49d0006007cb95bf2122b635601253e_l3.png "Rendered by QuickLaTeX.com")
(i) (ii)
![\[\text { Show that } f(x)=\left\{\begin{array}{c} \frac{|x-a|}{x-a}, \text { when } x \neq a \\ 1, \text { when } x=a . \end{array} \text { is discontinuous at } x=a\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4cf2ac13e8858ea7f8a6ae1ce50a383a_l3.png "Rendered by QuickLaTeX.com")
![\[\text { Show that } f(x)=\left\{\begin{array}{l} \frac{x-|x|}{2}, \text { when } x \neq 0 \\ 2, w h e n x=0 . \end{array} \text { is discontinuous at } x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b7a05e51c130c0deee261410e504c849_l3.png "Rendered by QuickLaTeX.com")
![\[\text { Let } f(x)=\left\{\begin{array}{c} \frac{1-\cos x}{x^{2}}, \text { when } x \neq 0 \\ 1, \text { when } x=0 . \end{array} \text { Show that } f(x) \text { is discontinuous at } x=0\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7aaf3f37d69a81ea01a54bfbbeef5227_l3.png "Rendered by QuickLaTeX.com")
Find whether 
is continuous at 
.
Find whether is continuous at .
Find whether 
is continuous at 
. is defined as ![\[f(x)=\left\{\begin{array}{c} \frac{x^{2}-9}{x-3}, \text { if } x \neq 3 \\ 6, \quad i f x=3 \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5f6c6df1ef1f45dde8fb6ece72556ff3_l3.png "Rendered by QuickLaTeX.com")
is continuous at 
. is defined as ![\[f(x)=\left\{\begin{array}{c} \frac{x^{2}-x-6}{x-3}, \quad i f x \neq 3 \\ 5, \text { if } x=3 \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e988142eefe0a70b52c97368c1e32a63_l3.png "Rendered by QuickLaTeX.com")
is continuous at 
. ![\[f(x)=\left\{\begin{array}{l} \frac{x}{|x|}, x \neq 0 \\ 1, x=0 \end{array}\right.\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c9880da205fe277409651c4d06a25f28_l3.png "Rendered by QuickLaTeX.com")
Consider LHL at x = 0
(viii) 
(vii) since, $\tan ^{-1}(\tan x)=x$ if $x \in[-\pi / 2, \pi / 2]$ But $x=4$ which does not belongs to above range => $\tan (\pi-\theta)=-\tan (\theta)$ => $\tan (\theta-\pi)=\tan (\theta)$...
(vi) 
(v) since,$\tan ^{-1}(\tan x)=x$ if $x \in[-\pi / 2, \pi / 2]$ substituting this condition in given question $\operatorname{Tan}^{-1}(\tan 1)=1$ (vi) since, $\tan ^{-1}(\tan x)=x$ if $x \in[-\pi /...
(iv) 
(iii) since, $\tan 7 \pi / 6=1 / \sqrt{3}$ substituting this value in $\tan ^{-1}(\tan 7 \pi / 6)$ $\operatorname{Tan}^{-1}(1 / \sqrt{3})$ let $\tan ^{-1}(1 / \sqrt{3})=\mathrm{y}$...
(ii) 
(i) since, $\tan ^{-1}(\tan \mathrm{x})=\mathrm{x}$ if $\mathrm{x} \in[-\pi / 2, \pi / 2]$ applying this condition in the given question $\operatorname{Tan}^{-1}(\tan \pi / 3)=\pi / 3$ (ii)since,...
(viii) 
(vii) since $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi] \approx[0,3.14]$ And here $x=5$ which does not lie in the above range. => $\cos (2 \pi-x)=\cos (x)$ => $\cos (2 \pi-5)=\cos (5)$ so $2...
(vi) 
(v) since $\cos ^{-1}(\cos \theta)=\theta$ if $0 \leq \theta \leq \pi$ applying this in given question $\cos ^{-1}(\cos 3)=3,3 \in[0, \pi]$ (vi) $\cos ^{-1}(\cos x)=x$ if $x \in[0, \pi]...
(iv) 
(iii) since, $\cos (4 \pi / 3)=-1 / 2$ substituting these values in $\cos ^{-1}\{\cos (4 \pi / 3)\}$ $\cos ^{-1}(-1 / 2)$ let $y=\cos ^{-1}(-1 / 2)$ => $\cos y=-1 / 2$ $ \begin{array}{l}...
(ii) 
(i) Since,t $\cos (-\pi / 4)=\cos (\pi / 4)[$ since $\cos (-\theta)=\cos \theta$ => $\cos (\pi / 4)=1 / \sqrt{2}$ substituting these values in $\cos ^{-1}\{\cos (-\pi / 4)\}$...
(ix) Since, $\sin ^{-1}(\sin x)=x$ with $x \in[-\pi / 2, \pi / 2]$ which is approximately equal to $[-1.57,1.57]$ But $x=12$, which does not lie on the above range, => $\sin (2 n \pi-x)=\sin...
(viii) 
(vii) since, $\sin ^{-1}(\sin x)=x$ with $x \in[-\pi / 2, \pi / 2]$ ~ But $x=3$, which does not lie on the above range, => $\sin (\pi-x)=\sin (x)$ Therefore, $\sin (\pi-3)=\sin (3)$ also $\pi-3...
![\[\left( \mathbf{v} \right)\text{ }\mathbf{si}{{\mathbf{n}}^{-\mathbf{1}}}(\mathbf{sin}\text{ }\mathbf{17\pi }/\mathbf{8})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1f606fd20e75984e3a8dbd38e96f3a99_l3.png "Rendered by QuickLaTeX.com")
(v) $\sin (2 \pi+\pi / 8)$ => $\sin (\pi / 8)$ substituting these values in $\sin ^{-1}(\sin 17 \pi / 8)$ =>$\sin ^{-1}(\sin \pi / 8)$ Since, $\sin ^{-1}(\sin x)=x$ with $x \in[-\pi / 2, \pi /...
(iv) 
(iii) Since, $\sin 5 \pi / 6$ is $1 / 2$ substituting this value in $\sin ^{-1}(\sin 5 \pi / 6)$ $\sin ^{-1}(1 / 2)$ let $y=\sin ^{-1}(1 / 2)$ $\sin (\pi / 6)=1 / 2$ The range of principal value of...
(ii) 
(i) Since, f $\sin \pi / 6$ is $1 / 2$ substituting this value in $\sin ^{-1}(\sin \pi / 6)$ => $\sin ^{-1}(1 / 2)$ let $y=\sin ^{-1}(1 / 2)$ $\operatorname{Sin}(\pi / 6)=1 / 2$ The range of...
(iv) 
(iii) \[\begin{array}{*{35}{l}} Let\text{ }y\text{ }=\text{ }co{{t}^{-1}}\left( -1/\surd 3 \right) \\ Cot\text{ }y\text{ }=\text{ }\left( -1/\surd 3 \right) \\ \text{ }-Cot\text{ }\left( \pi /3...
times the length of its shadow on the ground the angle of elevation of the sun at that time is (a) 
(b) 
(c) 
(d) 
AO be the pole; BO be its shadow and $\theta$ be the angle of elevation of the sun. Let $B O=x$ => $A 0=x \sqrt{3}$ In $\triangle A O B$, $\tan \theta=\frac{\mathrm{AO}}{\mathrm{B} 0}$...
with the ground. The distance from the foot of the tree to the point where the top touches the ground is 
meters. Find the height of the tree.Let us consider AC be the height of the tree which is (x + h) m It is given that; the broken portion of the tree is making an angle of 30o with the ground. In the fig. In ΔBCD, we get tan$30{}^\circ...
meters longer when the sun’s attitude is than when it was . Prove that the height of tower is 
meters.The figure is made by the given information in the question Assume the height of tower (AB)=h m Consider the distance BC=y m Now, in ΔABC tan =AB/BC $1=$h/y y=h Then, in ΔABD tan$30{}^\circ $=AB/BD...
m. From a point on the plane, the angle of elevation of the bottom of flag staff is and that of the top of the flag staff is 
.Find the height of the tower.\ As per the given information in the question, The length of the flag staff $=7$m Angles of elevation of the top and bottom of the flag staff from point D is$45{}^\circ $ and $30{}^\circ...
m on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are 
and respectively. Find the width of the river.From the given information we can say that, Assume width of river =PQ=(x+y)m Height of tree will be (AB) $=20$m Thus, in ΔABP tan$60{}^\circ $=AB/ BP $\sqrt{3}=20/x$ $x=20\sqrt{3}m$ In ΔABQ, tan...
and respectively. Find the height of the poles and the distances of the point from the poles.According to the question it is given that, Distance between the poles $=80$m=BD Assume the point of observation of the angles be O. The angles of elevation to the top of the points is $60{}^\circ...
and respectively. If the bridge is at the height of 
m from the banks, find the width of the river.As per the question it is given that, The bridge is at a height of $30$m from the banks. Assume, A and B represent the points on the bank on opposite sides of the river. And, AB is the width of the...
and the angle of the top of the tower from the foot of the building is . If the tower is 
m high, find the height of the building.Let us consider AB is the building and CD is the tower. As per the question we can see that, The angle of elevation of the top of the building from the foot of the tower is $30{}^\circ $. Now, the...
m tall lighthouse, the angles of depression of two ships are and . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.As per the given information in the question we can say that, Height of the lighthouse $=75m$ = ‘h’ m = AB Angle of depression of ship 1, $\alpha =30{}^\circ $ Angle of depression of bottom of the...
m high building, the angle of elevation of the top of a cable is and the angel of depression of its foot is . Determine the height of the tower.According to the question given we can say that, Height of the building $=7$m=AB Height of the cable tower =CD Angle of elevation of the top of the cable tower from the top of the building...
. From a point m away this point on the same bank, the angle of elevation of the top of the tower is . Find the height of the tower and the width of the river.Assume AB be the T.V tower of height ‘h’ m on the bank of river and ‘D’ be the point on the opposite side of the river. An angle of elevation at the top of the tower is $30{}^\circ $ Let us consider...
First we will take L.H.S Using the trigonometric identity ${{\sin }^{2}}+{{\cos }^{2}}\theta =1$, we get $=\frac{{{\cos }^{2}}\theta }{\sin \theta }-\cos ec\theta +\sin \theta $ $=\left(...
(ii) 
(i) Solving L.H.S and using the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get $=\cot \theta -\tan \theta $ $=\frac{\cos \theta }{\sin \theta }-\frac{\sin \theta }{\cos...
Using trigonometric properties, ${{\cot }^{2}}A={{\cos }^{2}}A/{{\sin }^{2}}A$and ${{\tan }^{2}}A={{\sin }^{2}}A/{{\cos }^{2}}A$ Putting the above in L.H.S, we have $={{\sin }^{2}}A{{\cot...
Firstly we will solve for L.H.S $\left( 1+{{\tan }^{2}}\theta \right)\left( 1-\sin \theta \right)\left( 1+\sin \theta \right)$ Then, we all know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$and...
First solve L.H.S $=(cosecA–sinA)(secA–cosA)(tanA+cotA)$ Substituting $\cos ecA=\frac{1}{\sin A},\sec A=\frac{1}{\cos A},\tan A=\frac{\sin A}{\cos A},\cot A=\frac{\cos A}{\sin A}$ Putting the values...
Firstly we will solve L.H.S $\sec A\left( 1-\sin A \right)\left( \sec A+\tan A \right)=1$ Putting $secA=1/cosA$ and $tanA=sinA/cosA$ in the above we get, $=1/cosA(1–sinA)(1/cosA+sinA/cosA)$...
Firstly, we will solve L.H.S $\left( \sec \theta +\cos \theta \right)\left( \sec \theta -\cos \theta \right)$ After multiplying we have, $={{\sec }^{2}}\theta -{{\sin }^{2}}\theta $ $=\left(...
First solve L.H.S $\left( \cos ec\theta +\sin \theta \right)\left( \cos ec\theta -\sin \theta \right)$ After multiplying we have, $=\cos e{{c}^{2}}\theta -{{\sin }^{2}}\theta $ $=\left( 1+{{\cot...
Solving L.H.S $={{\tan }^{2}}\theta -{{\sin }^{2}}\theta $ As, ${{\tan }^{2}}\theta =\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }$ $=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }-{{\sin...
Solving L.H.S $=\frac{\left( 1+{{\cot }^{2}}\theta \right)\tan \theta }{{{\sec }^{2}}\theta }$ We know that, $1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta $ $=\frac{\cos e{{c}^{2}}\theta \times...
Solving L.H.S, $=\frac{1-\sin \theta }{1+\sin \theta }$ Multiply by its conjugate, we will get $=\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }$ Thus, $1-{{\sin...
Solving L.H.S $=\frac{\sin \theta }{1-\cos \theta }$ Multiply by its conjugate, we will get $\frac{\sin \theta }{1-\cos \theta }\times \frac{1+\cos \theta }{1+\cos \theta }$ $=\frac{\sin \theta...
We use trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Then, multiplying both the numerator and the denominator by $\left( 1+\cos \theta \right)$, we get $=\frac{1-{{\cos...
Using the trigonometric identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Solving the L.H.S $=\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}$ $=\sqrt{\frac{\left( 1-\cos \theta \right)\left(...
Using trigonometric identity ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Solving L.H.S $=\sin {{A}^{2}}+\frac{1}{1+{{\tan }^{2}}A}$ $=\sin...
Using trigonometric identity, $\cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta =1$and ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Solving L.H.S, $={{\cos }^{2}}A+\frac{1}{1+{{\cot }^{2}}A}$ $={{\cos...
Use trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Thus, multiply both the numerator and the denominator by $\left( 1-\sin \theta \right)$, we have $\frac{\cos \theta }{1+\sin...
We use trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ So, multiply both the numerator and the denominator by $\left( 1+\sin \theta \right)$, we have $\frac{\cos \theta...
We will solve L.H.S $\tan \theta +1/\tan \theta $ $=\left( {{\tan }^{2}}\theta +1 \right)/\tan \theta $ $={{\sec }^{2}}\theta /\tan \theta \left[ \because {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1...
We will use the trigonometric identities $\left( {{\sec }^{2}}\theta -{{\tan }^{2}}\theta \right)=1$and $\left( \cos e{{c}^{2}}\theta -{{\cot }^{2}}\theta \right)=1$ L.H.S $=\left( {{\sec...
We will use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ First solve L.H.S, $\cos ec\theta \sqrt{\left( 1-{{\cos...
We use the trigonometric identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ Solve L.H.S ${{\tan }^{2}}\theta {{\cos }^{2}}\theta $ $={{\left( \tan \theta \times \cos \theta \right)}^{2}}$...
By using the trigonometric identity $\cos e{{c}^{2}}A-{{\cot }^{2}}A=1\Rightarrow \cos e{{c}^{2}}A={{\cot }^{2}}A+1$ Take L.H.S $\left( 1+{{\cot }^{2}}A \right){{\sin }^{2}}A$ $=\cos...
First solve the L.H.S, $\left( 1-{{\cos }^{2}}A \right)\cos e{{c}^{2}}A$ $=\left( {{\sin }^{2}}A \right)\cos e{{c}^{2}}A\left[ \because {{\sin }^{2}}A+{{\cos }^{2}}A=1\Rightarrow 1-{{\sin...
m tall stands on the top of a pedestal. From a point on the ground, angle of elevation of the top of the statue is and from the same point the angle of elevation of the top of the pedestal is . Find the height of the pedestal.Consider the AB as the statue, BC be the pedestal and D be the point on ground from where elevation angles are measured. As per the question it is given that, Angle of elevation of the top of statue...
m high building are 
and 
respectively. Find the height of the transmission tower.As per the question it is given that, Height of the building $=20$ m = AB Assume height of tower above building = h = BC Height of tower + building $=(h+20)$m [from ground] = OA Angle of elevation...
is the centroid of a 
and two of its vertices are 
and 
, find the third vertex of the triangle.Two vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(1,-6)$ and $\mathrm{B}(-5,2)$. Let the third vertex be $\mathrm{C}(\mathrm{a}, \mathrm{b})$. => the coordinates of its centroid are...
Solution: Suppose the cost of 1 pen and 1 pencil are Rs.$x$ and Rs.$y$ respectively. Then according to the question $\begin{array}{ll} 5 x+8 y=120 \ldots \ldots \text { (i) } \\ 8 x+5 y=153 \ldots...
longer when Sun’s altitude is 
than when it was 
. Find the height of the tower.When the sun’s altitude is the angle of elevation of the top of the tower from the tip of the shadow. Consider AB be h m and BC be x m. From the question, DC is $40m$ longer than BC. Thus,...
tall boy is standing at some distance from a 
tall building. The angle of elevation from his eyes to the top of the building increases from to as he walks towards the building. Find the distance he walked towards the building.As per the question it is given that, The height of the tall boy (AS) $=1.5m$ The length of the building (PQ) $=30m$ Assume the initial position of the boy be S. And, then he walks towards the...
tall girl stands at a distance of 
from a lamp post and casts a shadow of 
on the ground. Find the height of the lamp post by using (i) trigonometric ratio (ii) properties of similar triangles.Consider AC be the lamp post of height ‘h’ DE is the tall girl and her shadow is BE. Thus, we have $ED=1.6m$, $BE=4.8m$ and $EC=3.2$ (i) By using trigonometric ratio In $\vartriangle BDE$ $\tan...
tall building is . A flag is hoisted at the top of the building and the angle of elevation of the top of the flag staff from P is 
. Find the length of the flag staff and the distance of the building from the point P.Consider the height of flag-staff(AB) $=hm$ Then, the distance $PQ=xm$ It is given in the question that, Angle of elevation of top of the building $={{30}^{\circ }}$ Angle of elevation of top of...
with the ground. The distance between the foot of the tree to the point where the top touches the ground is 
. Find the height of the tree.Assume initial height of the tree be AC. Then, due to storm the tree is broken at B. Consider bent portion of the tree be $AB=xm$ and the remaining portion $BC=hm$ Thus, the height of the tree...
away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are and respectively. Find the height of the tower and the flag pole mounted on it.Consider the BC be the tower and AB be the flag pole on the tower Distance of the point of observation from foot of the tower $DC=9m$ Angle of elevation of top of flag pole is ${{60}^{\circ }}$...
high the angle of elevation of the top of tower is found to be . From the bottom of the same building, the angle of elevation of the top of the tower is found to be . Find the height of the tower and the distance between the tower and the building.As per the question it is given that, The height of the building $=15m$ The angle of elevation from the top of the building to top of the tower $={{30}^{\circ }}$ The angle of elevation from the...
. On moving a distance of meters towards the foot of the tower to a point B the angle of elevation increases to . Find the height of the tower and the distance of the tower from the point A.According to the question it is given that, Angle of elevation of top of the tower from point A, $\alpha ={{30}^{\circ }}$ Angle of elevation of top of tower from point B, $\beta ={{60}^{\circ }}$...
. When the observer moves towards the tower a distance of 
, he finds the angle of elevation of the top to be 
. Find the height of the tower and the distance of the first position from the tower. [Take 
and 
]Assume the height of the tower $=hm$ Then the distance $BC=xm$ Now, from the fig. In $\vartriangle ABC$ $\tan {{63}^{\circ }}=AB/BC$ $1.9626=h/x$ $x=h/1.9626$ $x=0.5095h....(i)$ Next, in...
. On advancing 
meters towards the foot of the tower, the angle of elevation of the tower becomes . Show that the height of the tower is 
meters.As per the question it is given that, The angle of elevation of top tower from first point D, $\alpha ={{30}^{\circ }}$ On advancing through D to C by $150m$, then $CD=150m$ Angle of elevation of...
and . If the height of the tower is 150 m, find the distance between the objects.It is given in the question that, The height of the tower (AB) $=150m$ Angles of depressions of the two objects are ${{45}^{\circ }}$ and ${{60}^{\circ }}$. In $\vartriangle ABD$ $\tan {{45}^{\circ...
and at two observing points 100 m apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of point where he falls on the ground from the just observation point.Assume parachute at highest point A and assume C and D be points which are $100m$ apart on ground where from then $CD=100m$ Angle of elevation from point $D={{45}^{\circ }}=\alpha $ Angle of...
, is found to be longer than when it was . Find the height of the tower.Assume height of the tower(AB) = h m Consider the length of the shorter shadow be $xm$ Now, the longer shadow is $(10+x)m$ Thus, from the fig. $\vartriangle ABC$ $\tan {{60}^{\circ }}=AB/BC$...
.$p(x)=x^{3}+4 x^{2}-3 x-18$ $p(2)=2^{3}+4 \times 2^{2}-3 \times 2-18=0$ $\therefore 2$ is a zero of $p(x)$.
. He walked 
towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as . Find the height of the towerAccording to the question, it is given that The angle of elevation of the tower before he started walking $={{30}^{\circ }}$ Distance walked by the person towards the tower $=50m$ The angle of...
meters. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are respectively and . Find the height of the tower.As per the question it is given that, Height of the flag staff $=5m=AB$ Angle of elevation of the top of flag staff $={{60}^{\circ }}$ Angle of elevation of the bottom of the flagstaff...
high, is broken by the wind in such a way that its top just touches the ground and makes an angle of with the ground. At what height from the ground did the tree break?According to the question, The initial height of tree $H=15m=AB+AC$ Let us consider that it is broken at point A. Now, the angle made by broken part with the ground $\left( \theta ...
meters away from the tower, an observer notices that the angles of elevation of the top and bottom of the flag-staff are respectively and . Find the height of the flag staff and that of the tower.As per the question it is given that, A vertical tower is surmounted by flag staff. Distance between observer and the tower $=70m=DC$ Angle of elevation of bottom of the flag staff $={{45}^{\circ...
meters long reaches the top of a vertical wall. If the ladder makes an angle of with the wall, find the height of the wall.As per the question it is given that, The length of the ladder $=15m=AO$ Angle made by the ladder with the wall $={{60}^{\circ }}$ Assume the height of the wall be h meters. And the horizontal...
is equal to the product of its zeroes, then 
? (a) 
(b) 
(c) 
(d) 
The correct option is option (d) $\frac{-2}{3}$ $\alpha$ and $\beta$ be the zeroes of $\mathrm{kx}^{2}+2 \mathrm{x}+3 \mathrm{k}$. Then $\alpha+\beta=\frac{-2}{k}$ and $\alpha \beta=3$ $\Rightarrow...
meters from the ground level, attached to a string inclined at to the horizontal. Find the length of the string to the nearest meter.As per the question, Height of kite flying from the ground level $=75m=AB$ Angle of inclination of the string with the ground $\left( \theta \right)={{60}^{\circ }}$ Let us consider the length of...
high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of with the horizontal through the foot of the pole, find the length of the wire.As per the question it is given, Height of the electric pole $=10m=AB$ The angle made by steel wire with ground (horizontal) $\theta ={{45}^{\circ }}$ Assume length of wire $=L=AC$ Thus, from the...
away from the wall and the ladder is making an angle of with the level of the ground. Determine the height of the wall.As per the question, Distance between the wall and the foot of the ladder $=2m=BC$ Angle made by ladder with ground $\left( \theta \right)={{60}^{\circ }}$ Height of the wall (H) $=AB$ Then, the...
and the foot of the ladder is 
away from the wall. Find the length of the ladder.It is given in the question that, Distance between the wall and foot of the ladder $=9.5m$ Angle of elevation $\theta ={{60}^{\circ }}$ Length of the ladder $=L=AC$ Then, from fig. ABC $\vartriangle...
away from the foot of the tower, the angle of elevation of the top the tower is . What is the height of the tower?As per the question, Distance between the foot of the tower and point of observation $=20 m=BC$ Angle of elevation of the top of the tower $={{60}^{\circ }}=\theta $ Now, Height of tower (H) $=AB$...
$ The\text{ }given\text{ }AP\text{ }is: $ $ 5,4\frac{1}{2},4,3\frac{1}{2},3.......... $ $ First\,term=5 $ $ Common\,difference,\,d=4\frac{1}{2}-5=-\frac{1}{2} $ $ \therefore a=5\,and\,d=-\frac{1}{2}...
has been inscribed in a circle. Find the area of the shaded region. (Take 
).As per the question, Side of the equilateral triangle $=6cm$ The area of the equilateral triangle $=\sqrt{3}/4{{\left( side \right)}^{2}}$ $=\sqrt{3}/4{{\left( 6 \right)}^{2}}$ $=\sqrt{3}/4\left( 36...
, touching its sides (fig.,). Find the radius of the inscribed circle and the area of the shaded part.According to the question, An equilateral triangle of side $=12cm$ Area of the equilateral triangle $=\sqrt{3}/4{{\left( side \right)}^{2}}$ $=\sqrt{3}/4{{\left( 12...
, 
and O is the center of the circleAs per the question, it is given that, $AC=24cm$ and $BC=10cm$ Since, AB is the diameter of the circle Angle ACB $={{90}^{\circ }}$ So, using Pythagoras theorem...
. In sector OSFT, square OEFG is inscribed. Find the area of the shaded region.It’s seen that, OEFG is a square of side $20cm$. So its diagonal $=\sqrt{2}$ side $=20\sqrt{2}cm$ And, the radius of the quadrant = diagonal of the square Radius of the quadrant $=20\sqrt{2}cm$...
. If OAPC is a quadrant of a circle with center O, then find the area of the shaded region.According to the question, OABC is a square of side $7cm$ So, $OA=AB=BC=OC=7cm$ Area of square OABC $=sid{{e}^{2}}={{7}^{2}}=49c{{m}^{2}}$ And given, OAPC is a quadrant of a circle with center O....
, find the area of the shaded region.As per the question, Side of the square $=21cm=OA$ Area of the square $=O{{A}^{2}}={{21}^{2}}=441c{{m}^{2}}$ Diagonal of the square $OB=\sqrt{2}OA=21\sqrt{2}cm$ In the given figure we can see that...
. If 
, find the area of the (i) quadrant OACB (ii) shaded region.According to the question it is given that, Radius of small quadrant, $r=2cm$ Radius of big quadrant, $R=3.5cm$ (i) Area of quadrant OACB $=1/4\pi {{R}^{2}}$ $=1/4\left( 22/7 \right){{\left( 3.5...
)As per the question given, The side of the square $=14cm$ Thus, it’s area $={{14}^{2}}=196c{{m}^{2}}$ Let’s consider the radius of each semi-circle be r cm. Then, $r+2r+r=14–3–3$ $4r=8$ $r=2$ The...
, find: (i) the circumference of the central part. (ii) the perimeter of the part ABEF.According to the question, Side of the square $=22cm=AB$ Assume the radius of the center part be r cm. Now, area of the circle $=1/5\times $ area of the square $=\pi {{r}^{2}}=1/5\times {{22}^{2}}$...
and 
. Two sectors of 
have been cut off. Calculate: (i) the area of the shaded region. (ii) the length of the boundary of the shaded region.As per the question, Length of the rectangle $=AB=20cm$ Breadth of the rectangle $=BC=14cm$ (i) Area of the shaded region $=$ Area of rectangle $-2\times $ Area of the semi-circle $=280–154$...
and 
. Taking DC, BC and AD as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.As per the question, ABCD is a rectangle with $AB=14cm$ and $BC=7cm$ The area of shaded region $=$ Area of rectangle ABCD $+2\times $ area of semi-circle with AD and BC as diameters $–$ area of the...
, a quadrant of a circle of radius 
is cut. Another circle of radius 
is also cut from the center as shown in Fig. Find the area of the remaining (shaded) portion of the square. (Use 
)According to the question, Side of the square $=8cm$ Radius of circle $=4.2cm$ Radius of the quadrant $=1.4cm$ Therefore, Area of the shaded potion $=$ Area of square $–$ Area of circle $-2\times $...
, a right angle AED with 
and 
, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. 
A per the question, Length of the rectangle ABCD $=20cm$ $AE=9cm$ and $DE=12cm$ The radius of the semi-circle $=BC/2$ or $AD/2$ Now, using Pythagoras theorem in triangle AED $AD=\sqrt{\left(...
We can say that The radius of the semi-circle $=10/2=5cm$ The area of figure $=$ Area of square $+$ Area of semi-circle $–$ Area of triangle AED $=10\times 10+1/2\pi {{r}^{2}}-1/2\times 6\times 8$...
. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 
wide, find the area of the track. Also, find the length of the outer running trackAssume radius of the inner semi-circle $=r$ And that of the outer semi-circle $=R$ As per the question it is given that, Length of the straight portion $=90m$ Width of the track $=14m$ The inner...
by . It is surrounded by semi-circular flower beds all round. Find the cost of leveling the semi-circular flower beds at 
paisa per square meterAs per the question it is given that, Length of the park $=100m$ and the breadth of the park $=50m$ The radius of the semi-circular flower beds $=$ half of the corresponding side of the rectangular...
. There are four semi-circular grassy plots all around it. Find the cost of turfing the plot at 
per square meter.As per the question, Side of the square tank $=40m$ The diameter of the semi-circular grassy plot $=$ side of the square tank $=40m$ Radius of the grassy plot $=40/2=20m$ Now, The area of the four...
at the corner of a square grassy lawn of side . If the length of the rope is increased by 
, find the increase in area of the grassy lawn in which the calf can graze.According to the question we can say that The initial length of the rope $=6m$ Then the rope is said to be increased by $5.5m$ Thus, the increased length of the rope $=(6+5.5)=11.5m$ As we all know...
at the corner of a rectangle field of dimensions 
, find the area of the field in which the cow can graze.As per the question we can say that, The dotted portion indicated the area over which the cow can graze. It’s clearly seen that, the shaded area is the area of a quadrant of a circle of radius equal...
, so that they just cannot reach one another. What area will be left un-grazed?According to the question, Side of square plot $=50m$ Radius of a quadrant $=25m$ Thus, we can say that, Area of plot left un-grazed $=$ Area of the plot $-4\times $ (area of a quadrant)...
. Find the area of the shaded square.As per the question, We all know that, each quadrant is a sector of ${{90}^{\circ }}$ in a circle of $1cm$ radius. In other words its 1/4th of a circle. Thus, its area $=1/4\pi {{r}^{2}}$...
long and wide. From its four corners, quadrants of radii 3.5m have been cut. Find the area of the remaining part.According to the question, Length of the rectangle $=20m$ Breadth of the rectangle $=15m$ Radius of the quadrant $=3.5m$ Thus, Area of the remaining part $=$ Area of the rectangle $–$ $4\times $...
is inscribed.As per the question, it is given that Area of square inscribed the circle $=64c{{m}^{2}}$ $sid{{e}^{2}}=64$ Side $=8cm$ Thus, $AB=BC=CD=DA=8cm$ Using Pythagoras theorem in triangle ABC,...
and 
, find the area of the playground.According to the question it is given, Length of rectangle $=36m$ Breadth of rectangle $=24.5m$ Radius of the semi-circle $=breath/2=24.5/2=12.25 m$ Then, Area of the playground $=$ Area of the...
and 
, find the area of the plot.As per the question it is given ABCD is a rectangle So, $AB=CD=60m$ And, $BC=AD=28m$ For the radius of the semi-circle $=BC/2=28/2=14 m$ Then, Area of the plot $=$ Area of rectangle ABCD $+$ Area of...
mounted on a hemisphere of same radius. The total height of the toy is 
. Find the total surface area of the toy.According to the question, Radius of the conical portion of the toy $=3.5cm=r$ Total height of the toy $=15.5cm=H$ If H is the length of the conical portion Now, Length of the cone (h)...
and the total height of the vessel is 
. Find the inner surface area of the vessel.As per the question it is given that, Diameter of the hemisphere $=14cm$ Thus, the radius of the hemisphere $=7cm$ Total height of the vessel $=13cm=h+r$ Thus, Inner surface area of the vessel...
long. Its internal diameter is 
and the thickness of the iron sheet used in making roller is 
. Find the mass of the road roller, if 
of the iron has 
mass.As per the question it is given that, Height/length of the cylindrical road roller $=h=1m=100cm$ Internal Diameter of the cylindrical road roller $=54cm$ Thus, the internal radius of the cylindrical...
and height 
is fixed symmetrically inside a similar vessel of diameter 16cm and height of . The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic cms of the cork dust will be required?According to the question it is given that, Depth of the cylindrical vessel = Height of the cylindrical vessel $=h=42cm$ (common for both) Inner diameter of the cylindrical vessel $=14cm$ Thus, the...
and the radius of each of the hemispherical ends is , find the cost of polishing its surface at the rate of 
per 
.According to the question it is given that, Radius of the hemispherical end (r) $=7cm$ Height of the solid $=(h+2r)=104cm$ $\Rightarrow h+2r=104$ $\Rightarrow h=104-\left( 2\times 7 \right)$ Then,...
and the diameter of the hemisphere is 
. Calculate the volume and the internal surface area of the solid.As per the question it is given that, Diameter of the hemisphere $=3.5m$ Thus, the radius of the hemisphere (r) $=1.75m$ Height of the cylinder (h) $=14/3m$ We all know that, volume of the Cylinder...
long with hemispherical ends each of diameter. Find the volume of the boiler.According to the question, Diameter of the hemisphere $=2m$ So, the radius of the hemisphere (r) $=1m$ Height of the cylinder $\left( {{h}_{1}} \right)=2m$ Then, the volume of the Cylinder $=\pi...
and height 
, surmounted by a cone of equal base and height 
. Find the capacity of tent and the cost of the canvas at 
per square meter.As per the question, Diameter of the cylinder $=20m$ Thus, its radius of the cylinder (R) $=10m$ Height of the cylinder $\left( {{h}_{1}} \right)=2.5m$ Radius of the cone $=$ Radius of the cylinder...
and base radius 
. The height and base radius of the cone are also the same. Find the whole surface and volume of the remaining Cylinder.As per the question it is given that, Height of the circular Cylinder $\left( {{h}_{1}} \right)=12cm$ Base radius of the circular Cylinder (r) $=5cm$ Height of the conical hole $=$ Height of the...
and length 
fitted with the conical ends each of axis length . Determine the capacity of the tank.It is given that, Base diameter of the cylindrical base of the petrol tank $=21cm$ Thus, its radius (r) $=diameter/2=21/2=10.5cm$ Height of the Cylindrical portion of the tank $\left( {{h}_{1}}...
. The heights of the cylindrical and conical portions is and 
respectively. Find the volume of that tent.As per the question it is given, Radius of the cylindrical portion (R) $=20m$ Height of the cylindrical portion $\left( {{h}_{1}} \right)=4.2m$ Height of the conical portion $\left( {{h}_{2}}...
and its length 
. It is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in tub. If the radius of the hemisphere is and the height of the cone outside the hemisphere is , find the volume of water left in the tub.According to the question we have, The radius of the Cylindrical tub (r) $=5cm$ Height of the Cylindrical tub (H) $=9.8cm$ Height of the cone outside the hemisphere (h) $=5cm$ Radius of the...
and , respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 
.It is given in the question that, Height of the Cylindrical portion (H) $=13cm$ Radius of the Cylindrical portion (r) $=5cm$ Height of the whole solid $=30cm$ Now, The curved surface area of the...
and the height of the cylindrical and conical portions are 
and , respectively. Find the total surface area of the solid. (Use ).According to the question, Radius of the common base (r) $=3.5cm$ Height of the cylindrical part (h) $=10cm$ Height of the conical part (H) $=6cm$ Assume, ‘l’ be the slant height of the cone Now, we...
and , respectively. Determine the surface area of the toy.It is given in the question that, The height of the cone (h) $=4cm$ Diameter of the cone (d) $=6cm$ Then, its radius (r) $=3$ Assume, ‘l’ be the slant height of cone. Now, we all know that...
is in the form of a right circular cylinder of diameter and height 
surmounted by a right circular cone. Find the cost of the canvas at 
per 
According to the question, Height of the tent $=77dm$ Height of a surmounted cone $=44dm$ Height of the Cylindrical Portion $=$ Height of the tent $–$ Height of the surmounted Cone $=77–44$...
and height 
and the cone has the slant height . Calculate the total surface area and the volume of the rocket.According to the question it is given that, Radius of the cylindrical portion of the rocket (R) $=2.5m$ Height of the cylindrical portion of the rocket (H) $=21m$ Slant Height of the Conical surface...
. The height of the cylindrical portion is 
while the vertex of the cone is 
above the ground. Find the area of canvas required for the tent.As per the question, The diameter of the cylinder (also the same for cone) $=24m$. Thus, its radius (R) $=24/2=12m$ The height of the cylindrical part $\left( {{H}_{1}} \right)=11m$ Now, Height of...
, find the value of .$x=3$ is one zero of the polynomial $2 x^{2}+x+k$ Therefore, it will satisfy the above polynomial. Now, we have $2(3)^{2}+3+k=0$ $\Rightarrow 21+\mathrm{k}=0$ $\Rightarrow...
and 
is (x, y) and 
then find the value of k.As we know that (x,y) is the mid-point $x=(3+k)/2$ and $y=(4+7)/2=11/2$ Also it is given that the mid-point lies on the line $2x+2y+1=0$ $2[(3+k)/2]+2(11/2)+1=0$ $3+k+11+1=0$ Thus, $k=-15$
and 
internally in the ratio of 
.Let’s consider P(x, y) be the required point. By using section formula, we know that the coordinates are $x=\frac{m{{x}_{2}}+n{{x}_{1}}}{m+n}$ $y=\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n}$ Here,...
,
,
and 
are the vertices of a parallelogram.Let’s consider A$(3,-2)$, B$(4,0)$,C$(6,-3)$ and D$(5,-5)$ Let’s take P(x, y) be the point of intersection of diagonals AC and BD of ABCD. The mid-point of AC is provided that,...
,
,
and
meet.Let’s consider A$(-2,-1)$,B$(1,0)$,C$(4,3)$ and D$(1,2)$ are the given points. Let’s take P(x, y) be the point of intersection of the diagonals of the parallelogram formed by the given points. As We...
,-b) divides the line segment joining A
and B
in the ratio $3:1, find the values of a and b.Given that, P($9a–2$, -b) divides the line segment joining A$(3a+1,-3)$ and B$(8a,5)$ in the ratio$3:1$ Therefore, by using section formula The Coordinates of P are $9a-2=\frac{3(8a)+1(3a+1)}{3+1}$...
, B(k,
) and a
, find the value of k and the distance AB.As it is given (a, b) is the mid-point of the line segment A($10,-6$) and B(k,$4$) Therefore, (a, b) $=(10+k/2,-6+4/2)$ a $=(10+k)/2$ and b $=-1$ $2a=10+k$ $K=2a–10$ Given that, $a–2b=18$ By Using...
, y) divides the line segment joining the points A
and B
. Also find the value of y.Let’s consider the point P($2$, y) divide the line segment joining the points A$(-2,2)$ and B$(3,7)$ in the ratio k: 1 Now, the coordinates of P are given by $\left[ \frac{3k+(-2)\times...
, B
and C
are the vertices of a triangle ABC, find the length of median through A.Let’s consider AD be the median through A. As we know that, AD is the median and D is the mid-point of BC Therefore, the coordinates of D are $(1+5/2,-1+1/2)=(3,0)$ So, Length of median...
and divides by the y-axis? find out the coordinates of the point of division.(ii) At what ratio is the line segment joining 
and 
divided at the point 
?Let’s consider P$(-2,-3)$ and Q$(9,3)$ be the given points. Let’s Suppose we have the y-axis that divides PQ in the ratio k:$1$ at R($0$, y) So, the coordinates of R are as given below Now, on...
, B
, C
and D
are the vertices of a rhombus.Given that the points are A$(-3,2)$, B$(-5,5)$, C$(2,-3)$ and D$(4,4)$ So, Coordinates of the mid-point of AC are $(-3+2/2,2-3/2)=(-1/2,-1/2)$ And, The Coordinates of mid-point of BD are...
divides the line segments joining the point A
and B
.Given that, Points A$(1/2,3/2)$ and B$(2,-5)$ Let’s consider the point P$(3/4,5/12)$ divide the line segment AB in the ratio k:$1$ As, we know that P$(3/4,5/12)=(2k+1/2)/(k+1),(2k+3/2)/(k+1)$...
and 
is divided by (i) x-axis (ii) y-axis. Also, find that the coordinates of the point of division in each case.Let’s A $(-2,-3)$ and B$(5,6)$ be the given points. (i) Suppose that x-axis divides AB in the ratio k:$1$ at the point P Now, the coordinates of the point of division are $\left[...
,
,
, are the vertices of a parallelogram. Is it a rectangle?Let’s A$(4,5)$, B$(7,6)$,C$(6,3)$ and D$(3,2)$ be the given points. And, P be the point of intersection of AC and BD. Coordinates of the mid-point of AC are $(4+6/2,5+3/2)=(5,4)$ Coordinates of the...
,
, and 
are the angular points of a square.Let’s A$(4,3)$,B$(6,4)$,C$(5,6)$ and D$(3,5)$ be the given points. We know the distance formula is $D=\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}}$...
,
, and 
are the vertices of a rectangle.Let’s A$(-4,-1)$,B$(-2,-4)$,C$(4,0)$ and$(2,3)$ be the given points. Now we have, Coordinates of the mid-point of AC are $(-4+4/2,-1+0/2)$ =$(0,-1/2)$ Coordinates of the mid-point of BD are...
, B and C.Let’s AD, BF and CE be the medians of ΔABC The Coordinates of D are $(5+1/2,1–1/2)$ $=(3,0)$ Coordinates of E are $(-1+1/2,3–1/2)$ $=(0,1)$ Coordinates of F are $(5–1/2,1+3/2)$ $=(2,2)$ Now, Finding...
Solution:
Which shows more variability, heights or weights? Solution: Given: The mean and SD is given of 50 students. Let us find which shows more variability, height and weight. By using the formulas,...
Solution:
(i) Solution: (ii) Solution:
and 
. Verify the relation between the coefficients and the zeroes of the polynomial.Let $\alpha=\frac{2}{3}$ and $\beta=\frac{-1}{4}$. Sum of the zeroes $=(\alpha+\beta)=\frac{2}{3}+\left(\frac{-1}{4}\right)=\frac{8-3}{12}=\frac{5}{12}$ Product of the zeroes, $\alpha...
Calculate the mean and variance. Solution:
Let ‘E’ denotes the event that student passes in English examination. And ‘H’ be the event that student passes in Hindi exam. It is given that, P (E) = 0.75 P (passing both) = P (E ∩ H) = 0.5 P...
If a dice is thrown twice, it has a total of 36 possible outcomes. If S represents the sample space then, n (S) = 36 Let ‘A’ represent events the event such that 3 comes in the first throw. A =...
In a single throw of 2 die, we have total 36 outcomes possible. Say, n (S) = 36 Where, ‘S’ represents sample space Let ‘A’ denotes the event of getting a double. So, A = {(1,1), (2,2), (3,3), (4,4),...
Let A and B be two events. As, out of 2 events, only one can happen at a time which means no event have anything common. ∴ We can say that A and B are mutually exclusive events. So, by definition of...
A and B are two mutually exclusive events is given P (A) = 1/2 and P (B) = 1/3 We need to find P (A ‘or’ B). P (A or B) = P (A ∪ B) So by definition of mutually exclusive events we know, P (A ∪ B) =...
A and B are two events is given P (A) = 0.3, P (B) = 0.5 and P (A ∪ B) = 0.5 We need to find P (A ∩ B). By definition of P (A or B) under axiomatic approach we know, P (A ∪ B) = P (A) + P (B) – P (A...
and B 
is divided by x- axis. find the coordinates of the point of division.Let’s the point on the x-axis be (x, $0$). [y – coordinate is zero] And, let’s this point divides the line segment AB in the ratio of k :$1$. Now by using the section formula for the y-coordinate,...
and B 
. Also, find the value of x.Let’s P divide the line joining A and B and let it divide the segment in the ratio k:$1$ Now, by using the section formula for the y – coordinate we have $2=(-3k+5)/(k+1)$ $2(k+1)=-3k+5$...
. Find the perimeter of the triangle.Assume the circle inscribed in the equilateral triangle be with a center O and radius r. We know that, formula of area of a Circle $=\pi {{r}^{2}}$ Now, the given that area is $154c{{m}^{2}}$....
Assume side of square be x cm which is inscribed in a circle. It is given that, Radius of circle (r) $=1/2$ (diagonal of square) $=1/2\left( x\sqrt{2} \right)$ $r=x/\sqrt{2}$ We all know that, area...
. Find the area of the circumscribed and inscribed circles.For circumscribed circle: Radius $=$ diagonal of square $/2$ Diagonal of the square $=$ side $\times \sqrt{2}$ $=10\sqrt{2}cm$ Radius $=\left( 10\times 1.414 \right)/2=7.07cm$ Now, the radius of the...
. Find the cost of fencing this ground at the rate of 
per meter.According to the question it is given that, Area of the circular playground $=22176{{m}^{2}}$ And the cost of fencing per meter $=₹50$ Assume the radius of the ground as r. Now, its area $=\pi...
and respectively. Find the radius and area of the circle which has circumferences is equal to sum of the circumference of two circles.As, per the question it is given that Radius of circle $1={{r}_{1}}=19cm$ Radius of circle $2={{r}_{2}}=9cm$ We all know that formula for finding, the circumference of a circle (C) $=2\pi r$ Thus,...
and respectively. Find the radius of the circle having its area equal to the sum of the areas of two circles.According to the question, Radii of the two circles are $6cm$ and $8cm$ Area of circle with radius $8cm=\pi {{\left( 8 \right)}^{2}}=64\pi c{{m}^{2}}$ Area of circle with radius $6cm=\pi {{\left( 6...
and .As per the question it is given that, Radius of circle $1={{r}_{1}}=15cm$ Radius of circle $2={{r}_{2}}=18cm$ We know that, the circumference of a circle (C) $=2\pi r$ Thus, ${{C}_{1}}=2\pi r$ and...
and the difference of their circumference is 
. Find the diameters of the circles.Assume the radii of the two circles be ${{r}_{1}}$ and ${{r}_{2}}$ Then, the circumferences of the two circles be ${{C}_{1}}$ and ${{C}_{2}}$. We know that, circumference of circle (C) $=2\pi r$...
. Find the ratio of their areas.Let’s assume the radius of two circles ${{C}_{1}}$ and ${{C}_{2}}$ be ${{r}_{1}}$ and ${{r}_{2}}$ We all know that, Circumference of a circle (C) $=2\pi r$ And their circumference will be $2\pi...
. If the same wire is bent in the form of a circle, find the area of the circleAccording to the question, Area of the square $={{a}^{2}}=121c{{m}^{2}}$ As we all know that, formula for finding Area of the circle $=\pi {{r}^{2}}$ Area of a square $={{a}^{2}}$...
long string. Find the area where the horse can graze.As per the question Length of the string (l) $=28m$ Area the horse can graze is the area of the circle with a radius equal to the length of the string. Formula for finding Area of a Circle $=\pi...
. Find the circumference of the circle.Assume the radius of the circle be r cm Now, the diameter (d) $=2r$ [As radius is half the diameter] Formula for finding Circumference of a circle (C) $=2\pi r$ It is given Circumference of the...
.According to the question Circumference $=44cm$ Formula for finding Circumference of a circle $=2\pi r=44cm$ $2\times \left( 22/7 \right)\times r=44$ $r=7cm$ Then, Area of a Circle $=\pi {{r}^{2}}$...
.As per the question it is given that, Area of the circle $=301.84c{{m}^{2}}$ As we know that, Formula for area of a Circle $=\pi {{r}^{2}}=301.84c{{m}^{2}}$ $\left( 22/7 \right)\times...
.It is given in the question that, Radius (r) $=4.2cm$ We all know that, Formula for circumference of a circle $=2\pi r$ $=2\times \left( 22/7 \right)\times 4.2=26.4c{{m}^{2}}$ Area of a circle $=\pi...
with radii of its lower and upper ends as and respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container, at the rate of 
per litre.Let us assume R and r be the radii of the top and base of the bucket respectively, Let us assume h be its height of the bucket. Then, according to the question we have $R=20cm$, $r=10cm$, $h=30cm$...
is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as and respectively. Find the cost of milk at the rate of 
per liter which the container can hold.As per the given information, A milk container in a form of frustum of a cone with, Radius of the lower end $\left( {{r}_{1}} \right)=8cm$ And radius of the upper end $\left( {{r}_{2}} \right)=20cm$...
and 
, the height of frustum be and the slant height of the conical cap be 
, find the canvas required for the tent.According to the given data in the question, Height of frustum (h) $=8m$ (given) Bigger and smaller radii of the frustum cone are $13cm$ and $7cm$. Therefore, ${{r}_{1}}=13cm$ and ${{r}_{2}}=7cm$...
and 
and the height is . Find the total surface area and volume of frustum.The height of frustum cone $=12cm$ (given) Bigger and smaller radii of a frustum cone are $12cm$ and $3cm$ respectively. (given) Therefore , ${{r}_{1}}=12cm;{{r}_{2}}=3cm$ Let us assume that the...