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.

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.

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

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. 

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

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

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

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

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

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