Found 13033 Articles for Mathematics

What is a regular polygon?
State the name of a regular polygon of
(i) 3 sides
(ii) 4 sides
(iii) 6 sides.

Tutorialspoint
Updated on 10-Oct-2022 13:47:56
To do:  We have to define regular polygon and state the name of a regular polygon of(i) 3 sides(ii) 4 sides(iii) 6 sides.Solution:A polygon is a closed plane or two-dimensional figure whose sides are straight line segments.Regular polygon:A regular polygon is a polygon that has equal sides and equal angles.(i) A regular polygon of 3 sides is called an equilateral triangle.(ii) A regular polygon of 4 sides is called a square.(iii) A regular polygon of 6 sides is called regular hexagon.

How many diagonals does each of the following have?
(a) A convex quadrilateral
(b) A regular hexagon
(c) A triangle.

Tutorialspoint
Updated on 10-Oct-2022 13:47:54
To do:  We have to find the number of diagonals each of the below shapes have.(a) A convex quadrilateral(b) A regular hexagon(c) A triangle.Solution:(a) A convex quadrilateral is shown below.From the above figure, we can observe that,A convex quadrilateral has two diagonals.(b) A regular hexagon is shown below.From the above figure, we can observe that,A regular hexagon has nine diagonals.(c) A triangle is shown below.From the above figure, we can observe that,A triangle has no diagonals.

Given here are some figures.

Classify each of them on the basis of the following.
(a) Simple curve
(b) Simple closed curve
(c) Polygon
(d) Convex polygon
(e) Concave polygon."

Tutorialspoint
Updated on 10-Oct-2022 13:47:53
To do:  We have to classify each of the given figures on the basis of the following.(a) Simple curve(b) Simple closed curve(c) Polygon(d) Convex polygon(e) Concave polygon.Solution:We know that, Simple curves:Simple curves do not cross or intersect.Simple closed curve:Does not have end points and does not intersect.Polygon:A polygon is a closed plane or two-dimensional figure whose sides are straight line segments. Convex Polygon: A polygon is convex if it contains all the line segments connecting any pair of its points.Concave Polygon: A polygon that is not convex is said to be a concave polygon.Therefore, (a) Figures 1, 2, 5, 6 and 7 are simple ... Read More

If $(2x-1,3y+1)$ and $(x+3,y-4)$ are equal ordered pairs. Find the values of $x$ and $y$.

Tutorialspoint
Updated on 10-Oct-2022 13:47:41
Given:$(2x-1,3y+1)$ and $(x+3,y-4)$ are equal ordered pairs.To do:We have to find the values of $x$ and $y$.Solution:We know that,Two ordered pairs are equal if and only if the corresponding first components are equal and the second components are equal. Therefore,$2x-1 = x+3$ and $3y+1 = y-4$$2x-x=3+1$ and $3y-y=-4-1$$x=4$ and $2y=-5$$x=4$ and $y=\frac{-5}{2}$The values of $x$ and $y$ are $4$ and $\frac{-5}{2}$ respectively. 

An oil funnel made of tin sheet consists of a \\( 10 \\mathrm{~cm} \\) long cylindrical portion attached to a frustum of a cone. If the total height is \\( 22 \\mathrm{~cm} \\), diameter of the cylindrical portion is \\( 8 \\mathrm{~cm} \\) and the diameter of the top of the funnel is \\( 18 \\mathrm{~cm} \\), find the area of the tin sheet required to make the funnel (see Fig. 13.25).
"

Tutorialspoint
Updated on 10-Oct-2022 13:47:39
Given:An oil funnel made of tin sheet consists of a \( 10 \mathrm{~cm} \) long cylindrical portion attached to a frustum of a cone. The total height is \( 22 \mathrm{~cm} \), diameter of the cylindrical portion is \( 8 \mathrm{~cm} \) and the diameter of the top of the funnel is \( 18 \mathrm{~cm} \).To do:We have to find the area of the tin sheet required to make the funnel.Solution:Diameter of the upper circular end of frustum $= 18\ cm$This implies, Radius of the upper circular end of frustum $(r_1) = 9\ cm$The radius of the lower circular end ... Read More

Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Tutorialspoint
Updated on 10-Oct-2022 13:47:39
To do:We have to derive the formula for the volume of the frustum of a cone.Solution:Let $ABC$ be a cone.From the cone the frustum $DECB$ is cut by a plane parallel to its base.$r_1$ and $r_2$ be the radii of the frustum ends of the cone and $h$ be the height of the frustum.In $\triangle ABG$ and $\triangle ADF$$DF \| BG$Therefore, $\triangle ABG \sim \triangle ADF$This implies, $\frac{D F}{B G}=\frac{A F}{A G}=\frac{A D}{A B}$$\frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}}$$\frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{l}{l_{1}}$$1-\frac{h}{\mathrm{h}_{1}}=\frac{r_{2}}{r_{1}}$$\frac{h}{\mathrm{h}_{1}}=1-\frac{r_{2}}{r_{1}}$$\frac{h}{\mathrm{h}_{1}}=\frac{r_1-r_{2}}{r_{1}}$$\frac{h_1}{h}=\frac{r_1}{r_1-r_2}$$h_1=\frac{r_1h}{r_1-r_2}$Volume of frustum of the cone $=$ Volume of cone $ABC -$ Volume of cone $ADE$$=\frac{1}{3}\pi r_1^2h_1 -\frac{1}{3}\pi r_2^2(h_1 - h)$$= \frac{\pi}{3}[r_1^2h_1-r_2^2(h_1 - h)]$$=\frac{\pi}{3}[r_{1}^{2}(\frac{h r_{1}}{r_{1}-r_{2}})-r_{2}^{2}(\frac{h r_{1}}{r_{1}-r_{2}}-h)]$$=\frac{\pi}{3}[(\frac{h r_{1}^{3}}{r_{1}-r_{2}})-r_{2}^{2}(\frac{h ... Read More

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

Tutorialspoint
Updated on 10-Oct-2022 13:47:38
To do:We have to derive the formula for the curved surface area and total surface area of the frustum of the cone.Solution:Let $ABC$ be a cone. From the cone the frustum $DECB$ is cut by a plane parallel to its base.$r_1$ and $r_2$ be the radii of the frustum ends of the cone and $h$ be the height of the frustum.In $\triangle ABG$ and $\triangle ADF$$DF \| BG$Therefore, $\triangle ABG \sim \triangle ADF$This implies, $\frac{D F}{B G}=\frac{A F}{A G}=\frac{A D}{A B}$$\frac{r_{2}}{r_{1}}=\frac{h_{1}-h}{h_{1}}=\frac{l_{1}-l}{l_{1}}$$\frac{r_{2}}{r_{1}}=1-\frac{h}{h_{1}}=1-\frac{l}{l_{1}}$$1-\frac{l}{\mathrm{l}_{1}}=\frac{r_{2}}{r_{1}}$$\frac{l}{l_{1}}=1-\frac{r_{2}}{r_{1}}$$\frac{l}{l_{1}}=\frac{r_{1}-r_{2}}{r_{1}}$$\Rightarrow l_{1}=\frac{r_{1} l}{r_{1}-r_{2}}$..............(i)Curved surface area of frustum $D E C B=$ Curved surface area of cone $A B C$-Curved surface ... Read More

In one fortnight of a given month, there was a rainfall of \( 10 \mathrm{~cm} \) in a river valley. If the area of the valley is \( 7280 \mathrm{~km}^{2} \), show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each \( 1072 \mathrm{~km} \) long, \( 75 \mathrm{~m} \) wide and \( 3 \mathrm{~m} \) deep.

Tutorialspoint
Updated on 10-Oct-2022 13:47:38
Given:In one fortnight of a given month, there was a rainfall of \( 10 \mathrm{~cm} \) in a river valley. The area of the valley is \( 7280 \mathrm{~km}^{2} \).To do:We have to show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each \( 1072 \mathrm{~km} \) long, \( 75 \mathrm{~m} \) wide and \( 3 \mathrm{~m} \) deep.Solution:Height of the rain $=10\ cm$$=\frac{10}{100}\ m$$=\frac{10}{100\times1000}\ km$Total rainfall in the river valley $=$ area of the valley $\times$ height of the rain$=7280\times\frac{10}{100\times1000}$$=0.7280\ km^3$Volume of normal water in each river $=1072\times\frac{75}{1000}\times\frac{3}{1000}$$= 0.2412\ km^3$Volume ... Read More

A cistern, internally measuring \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \), has \( 129600 \mathrm{~cm}^{3} \) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being \( 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} \) ?

Tutorialspoint
Updated on 10-Oct-2022 13:47:38
Given:A cistern, internally measuring \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \), has \( 129600 \mathrm{~cm}^{3} \) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water.Dimensions of each is \( 22.5 \mathrm{~cm} \times 7.5 \mathrm{~cm} \times 6.5 \mathrm{~cm} \).To do:We have to find the number of bricks that can be put in without overflowing the water.Solution:The dimensions of the cistern are \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \)This implies, Volume of the cistern $= 1980000\ ... Read More

A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of \( \pi \) as found appropriate.)

Tutorialspoint
Updated on 10-Oct-2022 13:47:38
Given:A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse. To do:We have to find the volume and surface area of the double cone so formed.Solution:Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of the right triangle $ABC$.This implies, using Pythagoras theorem, $AC^2=AB^2+BC^2$$AC^2=3^2+4^2$$=9+16$$=25$$\Rightarrow AC=\sqrt{25}$$=5\ cm$When the triangle $ABC$ is revolved about the hypotenuse $AC$, the following double cone is formed.In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{BDC}$, $\angle \mathrm{ABC}=\angle \mathrm{CDB}=90^{\circ}$           ($\mathrm{BD} \perp \mathrm{AC})$)$\angle \mathrm{BCA}=\angle \mathrm{BCD}$           (Common)Therefore, by AA similarity, ... Read More
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