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Learn UNIT VI: Mensuration with Free Lessons & Tips

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To find the length of the wire formed by the frustum of the cone, we'll first calculate the dimensions of the frustum. read more

To find the length of the wire formed by the frustum of the cone, we'll first calculate the dimensions of the frustum.$Given:&space;-&space;Height&space;of&space;the&space;original&space;cone,&space;$$&space;h&space;=&space;20&space;$$&space;cm&space;-&space;Vertical&space;angle&space;of&space;the&space;cone,&space;$$&space;\theta&space;=&space;60^\circ&space;$$&space;-&space;Diameter&space;of&space;the&space;wire,&space;$$&space;d&space;=&space;\frac{1}{16}&space;$$&space;cm&space;We'll&space;use&space;trigonometry&space;to&space;find&space;the&space;dimensions&space;of&space;the&space;frustum.&space;1.&space;Radius&space;of&space;the&space;base&space;of&space;the&space;original&space;cone,&space;$$&space;R&space;$$:&space;$&space;R&space;=&space;\frac{h}{\tan(\theta)}&space;$&space;$&space;R&space;=&space;\frac{20}{\tan(60^\circ)}&space;$&space;$&space;R&space;=&space;\frac{20}{\sqrt{3}}&space;$&space;$&space;R&space;=&space;\frac{20\sqrt{3}}{3}&space;$&space;2.&space;Radius&space;of&space;the&space;top&space;of&space;the&space;frustum,&space;$$&space;r&space;$$:&space;$&space;r&space;=&space;\frac{h/2}{\tan(\theta)}&space;$&space;$&space;r&space;=&space;\frac{10}{\tan(60^\circ)}&space;$&space;$&space;r&space;=&space;\frac{10}{\sqrt{3}}&space;$&space;$&space;r&space;=&space;\frac{10\sqrt{3}}{3}&space;$$

$Now,&space;let's&space;find&space;the&space;slant&space;height&space;of&space;the&space;frustum,&space;$$&space;l&space;$$:&space;$&space;l&space;=&space;\sqrt{h^2&space;+&space;(R&space;-&space;r)^2}&space;$&space;$&space;l&space;=&space;\sqrt{20^2&space;+&space;\left(\frac{20\sqrt{3}}{3}&space;-&space;\frac{10\sqrt{3}}{3}\right)^2}&space;$&space;$&space;l&space;=&space;\sqrt{400&space;+&space;\left(\frac{10\sqrt{3}}{3}\right)^2}&space;$&space;$&space;l&space;=&space;\sqrt{400&space;+&space;\frac{100&space;\times&space;3}{9}}&space;$&space;$&space;l&space;=&space;\sqrt{400&space;+&space;\frac{100}{3}}&space;$&space;$&space;l&space;=&space;\sqrt{\frac{1200&space;+&space;100}{3}}&space;$&space;$&space;l&space;=&space;\sqrt{\frac{1300}{3}}&space;$&space;$&space;l&space;=&space;\frac{\sqrt{1300}}{\sqrt{3}}&space;$&space;$&space;l&space;=&space;\frac{10\sqrt{13}}{\sqrt{3}}&space;$&space;$&space;l&space;=&space;\frac{10\sqrt{39}}{3}&space;$$

$The&space;length&space;of&space;the&space;wire&space;is&space;the&space;circumference&space;of&space;the&space;top&space;circle&space;of&space;the&space;frustum,&space;which&space;is&space;given&space;by&space;$$&space;2\pi&space;r&space;$$:&space;$&space;\text{Length&space;of&space;wire}&space;=&space;2\pi&space;\times&space;\frac{10\sqrt{3}}{3}&space;$&space;$&space;\text{Length&space;of&space;wire}&space;=&space;\frac{20\pi\sqrt{3}}{3}&space;$&space;$&space;\text{Length&space;of&space;wire}&space;\approx&space;\frac{20&space;\times&space;3.14&space;\times&space;1.73}{3}&space;$&space;$&space;\text{Length&space;of&space;wire}&space;\approx&space;36.34&space;\,&space;\text{cm}&space;$&space;So,&space;the&space;length&space;of&space;the&space;wire&space;formed&space;by&space;the&space;frustum&space;of&space;the&space;cone&space;is&space;approximately&space;$$&space;36.34&space;$$&space;cm.$

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To find the volume of water left in the cylinder after placing the cone inside, we need to consider the volumes of both the cone and the cylinder. read more

To find the volume of water left in the cylinder after placing the cone inside, we need to consider the volumes of both the cone and the cylinder.
$Given:&space;-&space;Height&space;of&space;the&space;cone,&space;$$&space;h_{\text{cone}}&space;=&space;120&space;$$&space;cm&space;-&space;Radius&space;of&space;the&space;cone,&space;$$&space;r_{\text{cone}}&space;=&space;60&space;$$&space;cm&space;-&space;Height&space;of&space;the&space;cylinder,&space;$$&space;h_{\text{cylinder}}&space;=&space;180&space;$$&space;cm&space;-&space;Radius&space;of&space;the&space;cylinder,&space;$$&space;r_{\text{cylinder}}&space;=&space;r_{\text{cone}}&space;=&space;60&space;$$&space;cm&space;First,&space;let's&space;calculate&space;the&space;volume&space;of&space;the&space;cone:&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;r_{\text{cone}}^2&space;h_{\text{cone}}&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;(60)^2&space;\times&space;120&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{22}{7}&space;\times&space;60&space;\times&space;60&space;\times&space;120&space;$&space;$&space;V_{\text{cone}}&space;=&space;1440000&space;\text{&space;cm}^3&space;$$

$Now,&space;let's&space;calculate&space;the&space;volume&space;of&space;the&space;cylinder:&space;$&space;V_{\text{cylinder}}&space;=&space;\pi&space;r_{\text{cylinder}}^2&space;h_{\text{cylinder}}&space;$&space;$&space;V_{\text{cylinder}}&space;=&space;\pi&space;(60)^2&space;\times&space;180&space;$&space;$&space;V_{\text{cylinder}}&space;=&space;\frac{22}{7}&space;\times&space;60&space;\times&space;60&space;\times&space;180&space;$&space;$&space;V_{\text{cylinder}}&space;=&space;7920000&space;\text{&space;cm}^3&space;$&space;Now,&space;the&space;volume&space;of&space;water&space;left&space;in&space;the&space;cylinder&space;after&space;placing&space;the&space;cone&space;inside&space;is&space;the&space;difference&space;between&space;the&space;volume&space;of&space;the&space;cylinder&space;and&space;the&space;volume&space;of&space;the&space;cone:&space;$&space;V_{\text{left}}&space;=&space;V_{\text{cylinder}}&space;-&space;V_{\text{cone}}&space;$&space;$&space;V_{\text{left}}&space;=&space;7920000&space;-&space;1440000&space;$&space;$&space;V_{\text{left}}&space;=&space;6480000&space;\text{&space;cm}^3&space;$&space;Therefore,&space;the&space;volume&space;of&space;water&space;left&space;in&space;the&space;cylinder&space;is&space;$$&space;6480000&space;$$&space;cubic&space;centimeters.$

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To find the volume of the wood in the entire pen stand, we need to calculate the volume of the cuboid and the volumes of the four conical depressions and the cubical depression, and then subtract these volumes from the volume of the cuboid. read more

To find the volume of the wood in the entire pen stand, we need to calculate the volume of the cuboid and the volumes of the four conical depressions and the cubical depression, and then subtract these volumes from the volume of the cuboid.

$Given:&space;-&space;Dimensions&space;of&space;the&space;cuboid:&space;$$&space;10&space;\times&space;5&space;\times&space;4&space;$$&space;cm&space;-&space;Radius&space;of&space;each&space;conical&space;depression:&space;$$&space;r&space;=&space;0.5&space;$$&space;cm&space;-&space;Depth&space;of&space;each&space;conical&space;depression:&space;$$&space;h&space;=&space;2.1&space;$$&space;cm&space;-&space;Edge&space;of&space;the&space;cubical&space;depression:&space;$$&space;a&space;=&space;3&space;$$&space;cm&space;1.&space;Volume&space;of&space;the&space;cuboid:&space;$&space;V_{\text{cuboid}}&space;=&space;\text{length}&space;\times&space;\text{width}&space;\times&space;\text{height}&space;$&space;$&space;V_{\text{cuboid}}&space;=&space;10&space;\times&space;5&space;\times&space;4&space;$&space;$&space;V_{\text{cuboid}}&space;=&space;200&space;\,&space;\text{cm}^3&space;$$

$2.&space;Volume&space;of&space;one&space;conical&space;depression:&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;r^2&space;h&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;\times&space;0.5^2&space;\times&space;2.1&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{22}{7}&space;\times&space;0.25&space;\times&space;2.1&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{22&space;\times&space;0.25&space;\times&space;2.1}{7}&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{11&space;\times&space;0.25&space;\times&space;2.1}{7}&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{11&space;\times&space;0.525}{7}&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{5.775}{7}&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;0.825&space;$&space;$&space;V_{\text{cone}}&space;\approx&space;0.275&space;\,&space;\text{cm}^3&space;$&space;3.&space;Volume&space;of&space;the&space;cubical&space;depression:&space;$&space;V_{\text{cube}}&space;=&space;a^3&space;$&space;$&space;V_{\text{cube}}&space;=&space;3^3&space;$&space;$&space;V_{\text{cube}}&space;=&space;27&space;\,&space;\text{cm}^3&space;$$

$Now,&space;let's&space;calculate&space;the&space;total&space;volume&space;of&space;the&space;wood&space;in&space;the&space;pen&space;stand:&space;$&space;V_{\text{total}}&space;=&space;V_{\text{cuboid}}&space;-&space;4&space;\times&space;V_{\text{cone}}&space;-&space;V_{\text{cube}}&space;$&space;$&space;V_{\text{total}}&space;=&space;200&space;-&space;4&space;\times&space;0.275&space;-&space;27&space;$&space;$&space;V_{\text{total}}&space;=&space;200&space;-&space;1.1&space;-&space;27&space;$&space;$&space;V_{\text{total}}&space;=&space;171.9&space;\,&space;\text{cm}^3&space;$&space;Therefore,&space;the&space;volume&space;of&space;the&space;wood&space;in&space;the&space;entire&space;pen&space;stand&space;is&space;$$&space;171.9&space;\,&space;\text{cm}^3&space;$$.$

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To find the volume of the remaining solid after hollowing out a conical cavity from a cube, we'll calculate the volumes of the cube and the conical cavity, and then subtract the volume of the conical cavity from the volume of the cube. read more

To find the volume of the remaining solid after hollowing out a conical cavity from a cube, we'll calculate the volumes of the cube and the conical cavity, and then subtract the volume of the conical cavity from the volume of the cube.

$Given:&space;-&space;Side&space;length&space;of&space;the&space;cube,&space;$$&space;s&space;=&space;7&space;$$&space;cm&space;-&space;Height&space;of&space;the&space;conical&space;cavity,&space;$$&space;h&space;=&space;7&space;$$&space;cm&space;-&space;Radius&space;of&space;the&space;conical&space;cavity,&space;$$&space;r&space;=&space;3&space;$$&space;cm&space;1.&space;Volume&space;of&space;the&space;cube:&space;$&space;V_{\text{cube}}&space;=&space;s^3&space;$&space;$&space;V_{\text{cube}}&space;=&space;7^3&space;$&space;$&space;V_{\text{cube}}&space;=&space;343&space;\,&space;\text{cm}^3&space;$$

$2.&space;Volume&space;of&space;the&space;conical&space;cavity:&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;r^2&space;h&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;\times&space;3^2&space;\times&space;7&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\pi&space;\times&space;9&space;\times&space;7&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;\frac{22}{7}&space;\times&space;9&space;\times&space;7&space;$&space;$&space;V_{\text{cone}}&space;=&space;\frac{1}{3}&space;\times&space;22&space;\times&space;3&space;$&space;$&space;V_{\text{cone}}&space;=&space;22&space;\,&space;\text{cm}^3&space;$&space;Now,&space;let's&space;calculate&space;the&space;volume&space;of&space;the&space;remaining&space;solid:&space;$&space;V_{\text{remaining}}&space;=&space;V_{\text{cube}}&space;-&space;V_{\text{cone}}&space;$&space;$&space;V_{\text{remaining}}&space;=&space;343&space;-&space;22&space;$&space;$&space;V_{\text{remaining}}&space;=&space;321&space;\,&space;\text{cm}^3&space;$&space;Therefore,&space;the&space;volume&space;of&space;the&space;remaining&space;solid&space;after&space;hollowing&space;out&space;the&space;conical&space;cavity&space;from&space;the&space;cube&space;is&space;$$&space;321&space;\,&space;\text{cm}^3&space;$$.$

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To find the radius of the circle formed by bending a wire into an arc subtending an angle of 60âˆ˜, we can use the formula for the length of an arc of a circle: read more

To find the radius of the circle formed by bending a wire into an arc subtending an angle of 60âˆ˜, we can use the formula for the length of an arc of a circle:
$$&space;\text{Length&space;of&space;arc}&space;=&space;\frac{\theta}{360^\circ}&space;\times&space;2\pi&space;r&space;$&space;Given:&space;Length&space;of&space;wire&space;($$&space;l&space;$$)&space;=&space;22&space;cm&space;Angle&space;subtended&space;($$&space;\theta&space;$$)&space;=&space;$$&space;60^\circ&space;$$&space;We&space;can&space;substitute&space;these&space;values&space;into&space;the&space;formula:&space;$&space;22&space;=&space;\frac{60}{360}&space;\times&space;2\left(\frac{22}{7}\right)&space;r&space;$$

$Let's&space;solve&space;for&space;$$&space;r&space;$$:&space;$&space;22&space;=&space;\frac{1}{6}&space;\times&space;\frac{44}{7}&space;r&space;$&space;$&space;22&space;=&space;\frac{22}{7}&space;r&space;$&space;Now,&space;let's&space;solve&space;for&space;$$&space;r&space;$$:&space;$&space;r&space;=&space;\frac{22&space;\times&space;7}{22}&space;$&space;$&space;r&space;=&space;7&space;\,&space;\text{cm}&space;$&space;So,&space;the&space;radius&space;of&space;the&space;circle&space;is&space;$$&space;7&space;\,&space;\text{cm}&space;$$.$

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To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze. read more

To solve this problem, we first find the area the calf can graze with a rope of length 6 m and then find the area it can graze with a rope of length 11.5 m. The difference between these areas will give us the increase in the area of the grassy lawn the calf can graze.
$1.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m:**&space;When&space;the&space;calf&space;is&space;tied&space;with&space;a&space;rope&space;of&space;length&space;6&space;m,&space;it&space;can&space;graze&space;within&space;a&space;circle&space;whose&space;radius&space;is&space;6&space;m.&space;This&space;forms&space;a&space;circular&space;grazing&space;area.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;is&space;given&space;by&space;$$&space;\pi&space;r^2&space;$$,&space;where&space;$$&space;r&space;$$&space;is&space;the&space;radius&space;of&space;the&space;circle&space;(6&space;m).&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;6&space;m&space;is&space;$$&space;\pi&space;\times&space;6^2&space;=&space;36\pi&space;$$&space;square&space;meters.&space;2.&space;**Area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m:**&space;When&space;the&space;length&space;of&space;the&space;rope&space;is&space;increased&space;to&space;11.5&space;m,&space;the&space;radius&space;of&space;the&space;circular&space;grazing&space;area&space;also&space;increases&space;to&space;11.5&space;m.&space;The&space;area&space;of&space;the&space;circular&space;grazing&space;area&space;with&space;a&space;radius&space;of&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.&space;Therefore,&space;the&space;area&space;the&space;calf&space;can&space;graze&space;with&space;a&space;rope&space;of&space;length&space;11.5&space;m&space;is&space;$$&space;\pi&space;\times&space;11.5^2&space;$$&space;square&space;meters.$

$Now,&space;let's&space;find&space;the&space;difference&space;between&space;the&space;areas&space;the&space;calf&space;can&space;graze&space;with&space;these&space;two&space;rope&space;lengths:&space;$&space;\text{Increase&space;in&space;area}&space;=&space;(\pi&space;\times&space;11.5^2)&space;-&space;(\pi&space;\times&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(11.5^2&space;-&space;6^2)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;(132.25&space;-&space;36)&space;$&space;$&space;\text{Increase&space;in&space;area}&space;=&space;\pi&space;\times&space;96.25&space;$&space;Using&space;$$&space;\pi&space;\approx&space;\frac{22}{7}&space;$$:&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;\frac{22}{7}&space;\times&space;96.25&space;$&space;$&space;\text{Increase&space;in&space;area}&space;\approx&space;303.75&space;\,&space;\text{square&space;meters}&space;$&space;So,&space;the&space;increase&space;in&space;the&space;area&space;of&space;the&space;grassy&space;lawn&space;in&space;which&space;the&space;calf&space;can&space;graze&space;is&space;approximately&space;$$&space;303.75&space;\,&space;\text{square&space;meters}&space;$$.$

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The circumference of a circle is given by the formula: read more

The circumference of a circle is given by the formula:${Circumference}&space;=&space;2&space;\pi&space;r&space;\]&space;Given&space;that&space;the&space;sum&space;of&space;the&space;circumferences&space;of&space;two&space;circles&space;with&space;radii&space;15&space;cm&space;and&space;18&space;cm&space;is&space;equal&space;to&space;the&space;circumference&space;of&space;another&space;circle,&space;we&space;can&space;write&space;the&space;equation:&space;$&space;2\pi&space;\times&space;15&space;+&space;2\pi&space;\times&space;18&space;=&space;2\pi&space;\times&space;r&space;$&space;Let's&space;solve&space;for&space;$$&space;r&space;$$:&space;$&space;2\pi&space;\times&space;(15&space;+&space;18)&space;=&space;2\pi&space;\times&space;r&space;$&space;$&space;2\pi&space;\times&space;33&space;=&space;2\pi&space;\times&space;r&space;$&space;$&space;33&space;=&space;r&space;$&space;So,&space;the&space;radius&space;of&space;the&space;circle&space;whose&space;circumference&space;is&space;equal&space;to&space;the&space;sum&space;of&space;the&space;circumferences&space;of&space;two&space;circles&space;with&space;radii&space;15&space;cm&space;and&space;18&space;cm&space;is&space;$$&space;33&space;$$&space;cm.$

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To find the area of the shaded region, we need to subtract the area of the square OABC from the area of the quadrant OABQ. read more

To find the area of the shaded region, we need to subtract the area of the square OABC from the area of the quadrant OABQ.
$Given:&space;Radius&space;of&space;the&space;circle,&space;OA&space;=&space;14&space;cm&space;1.&space;**Area&space;of&space;the&space;square&space;OABC:**&space;Since&space;OABC&space;is&space;a&space;square&space;inscribed&space;in&space;the&space;circle,&space;its&space;diagonal&space;is&space;the&space;diameter&space;of&space;the&space;circle,&space;which&space;is&space;$$2&space;\times&space;14&space;=&space;28$$&space;cm.&space;The&space;area&space;of&space;a&space;square&space;can&space;be&space;found&space;using&space;the&space;formula:&space;Area&space;=&space;side^2.&space;So,&space;the&space;area&space;of&space;OABC&space;is&space;$$28^2&space;=&space;784$$&space;square&space;cm.&space;2.&space;**Area&space;of&space;the&space;quadrant&space;OABQ:**&space;The&space;area&space;of&space;a&space;quadrant&space;of&space;a&space;circle&space;with&space;radius&space;$$r$$&space;is&space;given&space;by&space;$$&space;\frac{1}{4}&space;\times&space;\pi&space;r^2&space;$$,&space;where&space;$$r$$&space;is&space;the&space;radius&space;of&space;the&space;circle.$

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To find the area of the shaded region, we need to subtract the area of the semi-circle from the area of the quadrant. read more

To find the area of the shaded region, we need to subtract the area of the semi-circle from the area of the quadrant.
$Given:&space;Radius&space;of&space;the&space;circle,&space;$$r&space;=&space;14$$&space;cm&space;1.&space;**Area&space;of&space;the&space;quadrant&space;ABDC:**&space;The&space;area&space;of&space;a&space;quadrant&space;of&space;a&space;circle&space;with&space;radius&space;$$r$$&space;is&space;$$&space;\frac{1}{4}&space;\times&space;\pi&space;\times&space;r^2&space;$$.&space;So,&space;the&space;area&space;of&space;the&space;quadrant&space;ABDC&space;is&space;$$&space;\frac{1}{4}&space;\times&space;\pi&space;\times&space;(14)^2&space;$$&space;square&space;cm.&space;2.&space;**Area&space;of&space;the&space;semi-circle&space;with&space;diameter&space;BC:**&space;The&space;radius&space;of&space;the&space;semi-circle&space;is&space;half&space;the&space;diameter,&space;which&space;is&space;$$&space;\frac{1}{2}&space;\times&space;14&space;=&space;7&space;$$&space;cm.&space;The&space;area&space;of&space;a&space;semi-circle&space;with&space;radius&space;$$r$$&space;is&space;$$&space;\frac{1}{2}&space;\times&space;\pi&space;\times&space;r^2&space;$$.&space;So,&space;the&space;area&space;of&space;the&space;semi-circle&space;with&space;diameter&space;BC&space;is&space;$$&space;\frac{1}{2}&space;\times&space;\pi&space;\times&space;(7)^2&space;$$&space;square&space;cm.$

$Now,&space;let's&space;calculate&space;the&space;area&space;of&space;the&space;shaded&space;region&space;by&space;subtracting&space;the&space;area&space;of&space;the&space;semi-circle&space;from&space;the&space;area&space;of&space;the&space;quadrant:&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;=&space;\text{Area&space;of&space;quadrant&space;ABDC}&space;-&space;\text{Area&space;of&space;semi-circle}&space;$&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;=&space;\left(\frac{1}{4}&space;\times&space;\pi&space;\times&space;14^2\right)&space;-&space;\left(\frac{1}{2}&space;\times&space;\pi&space;\times&space;7^2\right)&space;$&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;=&space;\left(\frac{1}{4}&space;\times&space;196&space;\pi\right)&space;-&space;\left(\frac{1}{2}&space;\times&space;49&space;\pi\right)&space;$&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;=&space;49&space;\pi&space;-&space;24.5&space;\pi&space;$&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;=&space;24.5&space;\pi&space;$&space;Using&space;$$&space;\pi&space;\approx&space;\frac{22}{7}&space;$$:&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;\approx&space;24.5&space;\times&space;\frac{22}{7}&space;$&space;$&space;\text{Area&space;of&space;shaded&space;region}&space;\approx&space;77&space;\,&space;\text{cm}^2&space;$&space;So,&space;the&space;area&space;of&space;the&space;shaded&space;region&space;is&space;approximately&space;$$&space;77&space;\,&space;\text{cm}^2&space;$$.$

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To solve this problem, we first need to find the areas of the two circles with diameters 10 cm and 24 cm. Then, we'll add these areas together to find the total area. Finally, we'll find the diameter of the larger circle whose area is equal to the sum of the areas of the two smaller circles. read more

To solve this problem, we first need to find the areas of the two circles with diameters 10 cm and 24 cm. Then, we'll add these areas together to find the total area. Finally, we'll find the diameter of the larger circle whose area is equal to the sum of the areas of the two smaller circles.

$Given:&space;Diameters&space;of&space;the&space;circles:&space;$$d_1&space;=&space;10$$&space;cm&space;and&space;$$d_2&space;=&space;24$$&space;cm&space;1.&space;**Area&space;of&space;the&space;circle&space;with&space;diameter&space;10&space;cm:**&space;Radius&space;of&space;the&space;first&space;circle&space;$$r_1&space;=&space;\frac{d_1}{2}&space;=&space;\frac{10}{2}&space;=&space;5$$&space;cm&space;Area&space;of&space;the&space;first&space;circle&space;$$A_1&space;=&space;\pi&space;\times&space;r_1^2$$&space;2.&space;**Area&space;of&space;the&space;circle&space;with&space;diameter&space;24&space;cm:**&space;Radius&space;of&space;the&space;second&space;circle&space;$$r_2&space;=&space;\frac{d_2}{2}&space;=&space;\frac{24}{2}&space;=&space;12$$&space;cm&space;Area&space;of&space;the&space;second&space;circle&space;$$A_2&space;=&space;\pi&space;\times&space;r_2^2$$$

$Now,&space;let's&space;find&space;the&space;total&space;area&space;$$A_{\text{total}}$$&space;of&space;the&space;two&space;smaller&space;circles:&space;$&space;A_{\text{total}}&space;=&space;A_1&space;+&space;A_2&space;$&space;$&space;A_{\text{total}}&space;=&space;\pi&space;\times&space;5^2&space;+&space;\pi&space;\times&space;12^2&space;$&space;Now,&space;let's&space;solve&space;for&space;$$A_{\text{total}}$$:&space;$&space;A_{\text{total}}&space;=&space;\pi&space;\times&space;25&space;+&space;\pi&space;\times&space;144&space;$&space;$&space;A_{\text{total}}&space;=&space;\pi&space;\times&space;(25&space;+&space;144)&space;$&space;$&space;A_{\text{total}}&space;=&space;\pi&space;\times&space;169&space;$$

$Now,&space;we&space;know&space;that&space;the&space;area&space;of&space;the&space;larger&space;circle&space;is&space;equal&space;to&space;$$A_{\text{total}}$$,&space;so&space;we&space;can&space;use&space;the&space;formula&space;for&space;the&space;area&space;of&space;a&space;circle&space;to&space;find&space;its&space;diameter:&space;$&space;A_{\text{larger}}&space;=&space;\pi&space;\times&space;\left(\frac{d_{\text{larger}}}{2}\right)^2&space;$&space;Given&space;$$A_{\text{larger}}&space;=&space;A_{\text{total}}&space;=&space;\pi&space;\times&space;169$$,&space;let's&space;solve&space;for&space;$$d_{\text{larger}}$$:&space;$&space;\pi&space;\times&space;\left(\frac{d_{\text{larger}}}{2}\right)^2&space;=&space;\pi&space;\times&space;169&space;$&space;$&space;\left(\frac{d_{\text{larger}}}{2}\right)^2&space;=&space;169&space;$&space;$&space;\frac{d_{\text{larger}}^2}{4}&space;=&space;169&space;$&space;$&space;d_{\text{larger}}^2&space;=&space;676&space;$&space;$&space;d_{\text{larger}}&space;=&space;\sqrt{676}&space;$&space;$&space;d_{\text{larger}}&space;=&space;26&space;\,&space;\text{cm}&space;$&space;So,&space;the&space;diameter&space;of&space;the&space;larger&space;circle&space;is&space;$$26$$&space;cm.$

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