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Geometry basics Quiz 2

mathematics

 Aarti was making a rectangular wall-hanging. She pasted blue ribbons of width 1 cm on its borders and one of the diagonals. The difference between the length and the breadth of the wall-hanging is 1 cm. Had she pasted the ribbons only on the diagonals, she would have required $58\text{ }cm$ of ribbons. What is the approximate area left uncovered by the ribbons?

 First find the length and breadth of the rectangle by using the Pythagorean theorem with the given length of diagonal, then subtract the area covered by the ribbons.

  Let the length and breadth be $l$ and $\left( l-1 \right)$ respectively $ \\ $ Then, $\sqrt{{{l}^{2}}+{{\left( l-1 \right)}^{2}}}~=29\text{ }({{d}_{1}}+{{d}_{2}}=58\text{ }cm) \\  \Rightarrow 2{{l}^{2}}-2l+1\text{ }=\text{ }841 \\  \Rightarrow {{l}^{2}}-21l+20l-420=0 \\  \Rightarrow (l+20)(l-21)=0 \\  \text{ }\therefore l\text{ }=\text{ }21,-\text{ }20,\text{ }b\text{ }=\text{ }20 \\ $ Since, she pasted blue ribbons of width $1 cm$ on its borders and one of the diagonals $ \\ \text{Required area }=18\times 19-27\times 1=315\text{ }c{{m}^{2}} \\ $

 The radius of a closed right circular cylinder is twice that of a right circular cone while the height of the cone is thrice that of the cylinder. What will be the ratio of the total surface area of the cone to that of the cylinder when the radius and height of the cone are 9 cm and 12 cm respectively?

 First find the areas of the cone and cylinder by using the formula, $\text{ }\pi r\left( \sqrt{{{r}^{2}}+{{h}^{2}}}+r \right)$ and $\text{ 2}\pi r'\left( 2h+r' \right)$ respectively.

  $\text{For the cone: }r=9cm\text{ and }h=12cm \\  \text{So, for the cylinder: }{r}'=2r=18cm\text{ and }{h}'=\dfrac{h}{3}=4cm \\  \text{Hence, surface area of the cone = }\pi r\left( \sqrt{{{r}^{2}}+{{h}^{2}}}+r \right) \\  =9\pi \left( \sqrt{{{9}^{2}}+{{12}^{2}}}+9 \right)=216\pi c{{m}^{2}} \\  \text{Surface area of the cylinder = }2\pi {r}'\left( {h}'+{r}' \right) \\  =2\pi \times 18(4+18)=792\pi c{{m}^{2}} \\  \text{Ratio}=3:11 \\ $

 The area of a sector of a circle is $2.88\pi \text{ }c{{m}^{2}}$. The area of a sector of another circle with the angle being $5{}^\circ$ less than the first sector is $5.76\pi \text{ }c{{m}^{2}}$. If the radii of the circles are in the ratio $2:3$ respectively, then what will be the area of the equilateral triangle inscribed by the second circle?

 First find the radii of the two sectors by using the area of sector formula $\dfrac{\pi {{r}^{2}}d}{{{360}^{{}^\circ }}}$, where $''d''$ is the measure in degrees, then find the length of the triangle.

For the first circle, area of sector $=\dfrac{\pi 4{{r}^{2}}d}{360{}^\circ }=\dfrac{\pi {{r}^{2}}d}{90{}^\circ }\left[ {{r}_{1}}=2r \right] \\ \Rightarrow \dfrac{{{r}^{2}}d}{90{}^\circ }=2.88\Rightarrow {{r}^{2}}d=259.2 \\ $ For the second circle, area of the sector $=\dfrac{\pi 9{{r}^{2}}(d-5)}{360{}^\circ }=\dfrac{\pi {{r}^{2}}(d-5)}{40{}^\circ }\left[ {{r}_{2}}=3r \right] $ $ \\  \Rightarrow \dfrac{{{r}^{2}}(d-5)}{40{}^\circ }=5.76\Rightarrow {{r}^{2}}d-5{{r}^{2}}=230.4 \\  \Rightarrow 5{{r}^{2}}=259.2-230.4=28.8 \\  \Rightarrow r=2.4 \\ $ So, the radius of the second circle $= 7.2 cm$ $ \\ $ So, the length of each side of the equilateral triangle $ \\=2\sqrt{3}\times 7.2=14.4\sqrt{3}cm \\  \text{Hence, area = }\dfrac{\sqrt{3}}{4}\times {{\left( 14.4\sqrt{3} \right)}^{2}}=155.52\sqrt{3}c{{m}^{2}} \\ $

 The ratio of the radii of a cylinder, a hemisphere and a cone is $1:3:4$. The height of the cone is $150%$ more than that of the cylinder. The volumes of the cylinder and hemisphere are $76.8\pi \text{ }c{{m}^{3}}$ and $1152\pi \text{ }c{{m}^{3}}$ respectively. What is the ratio of the total surface area and the volume of the cone? 

 First find the radius of the hemisphere by using the volume formula, $\dfrac{2}{3}\pi {{\left( 3r \right)}^{3}}$ then find the radius of the cone, use the formula for total surface area and volume of the cone i.e. $\pi rl+\pi {{r}^{2}}~\text{ and }\dfrac{1}{3}\pi {{\left( r \right)}^{2}}h$.

 Let the radii of the cylinder, the hemisphere and the cone be $r, 3r$ and $4r$ respectively $ \\ $ The height of the cylinder and the cone be $h$ and $2.5h$ respectively. $ \\ $ Now, volume of cylinder $=\pi {{r}^{2}}h=76.8\pi$ $ \\  \Rightarrow {{r}^{2}}h=76.8 \\ $ And, the volume of hemisphere $= \dfrac{2}{3}\pi {{(3r)}^{3}}=1152\pi $ $ \\ \Rightarrow 27{{r}^{3}}=1728\Rightarrow r=4cm \\ $ Radius of the cone $= 16cm$ $ \\  \Rightarrow {{4}^{2}}h=76.8\Rightarrow h=4.8cm \\ $  Height of cone $=12cm$ $  \\ $ Total surface area of the cone $= 16\pi \left( \sqrt{{{16}^{2}}+{{12}^{2}}}+16 \right)=576\pi c{{m}^{2}}$ $ \\ $ Volume of the cone $= \dfrac{1}{3}\pi \times {{16}^{2}}\times 12=1024\pi c{{m}^{3}}$ $ \\  \therefore \text{Ratio = 9:16} \\ $

 A spherical iron ball of radius 72cm is to be painted. However, it is realized that the requirement is for balls of radius 12cm. So, the bigger ball is melted and made into smaller balls of radius 12cm each. The rate per sq. cm for painting is reduced to $\dfrac{2}{9}^{th}$ of the original rate. What will be the impact on the original painting cost?

 First find the number of balls that can be made then, find the surface area of the original sphere and the total surface area of the small spheres, and put the rate of painting, new area for the relation with the original painting cost.

 $  \text{Volume of sphere = }\left( \dfrac{4}{3} \right)\pi {{R}^{3}} \\  \text{Radius of big sphere = 72cm} \\  \text{Radius of smaller sphere = 12cm} \\  \text{Number of balls that can be made of radius 12cm } \\  \text{= }\dfrac{\left( \dfrac{4}{3} \right)\pi {{R}^{3}}}{\left( \dfrac{4}{3} \right)\pi {{\left( \dfrac{R}{6} \right)}^{3}}}=216~ \\  \text{Surface area of original sphere = }4\pi {{R}^{2}}~=4\pi {{\left( 72 \right)}^{2}} \\  \text{Total surface area of smaller spheres }=216\times 4\pi {{R}^{2}}~=216\pi \times 4{{\left( 12 \right)}^{2}} \\  \text{Therefore, }\dfrac{\left( \text{Total surface area of small spheres} \right)}{\left( \text{Surface area of big sphere} \right)}\text{ = 6 }\!\!~\!\!\text{ } \\  \text{Let, the original area }=A \\  \text{Original rate of painting per sq}\text{. cm }=Rs.p \\  \text{Original painting cost }=Rs.\text{ }Ap \\  \text{New area }=6A \\  \text{New rate = }Rs.\left( \dfrac{2}{9} \right)p \\  \text{New painting cost = }6A\times Rs.\left( \dfrac{2}{9} \right)p=\text{ }Rs.1.33Ap \\  \text{Increased by 33 }\!\!%\!\!\text{ }\text{.} \\ $