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Yogita Ingle 6 years, 2 months ago
As a cylinder can be seen as a collection of multiple congruent disks stacked one above the other. In order to calculate the space occupied by a cylinder, we calculate the space occupied by each disk and then add them up. Thus, the volume of the cylinder can be given by the product of the area of base and height.
The volume of a cylinder = πr2h cubic units.
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Sia ? 4 years, 6 months ago
Complete Step-by-step answer:
</div>
<div itemprop="description">The difference of outer curved surface area and inner curved surface area is 88<nobr aria-hidden="true">cm2</nobr><script type="math/tex" id="MathJax-Element-5">cm^2</script> given in the question.So,
Outer curved surface area – Inner curved surface area <nobr aria-hidden="true">=88cm2</nobr><script type="math/tex" id="MathJax-Element-6">=88c{{m}^{2}}</script>.
Since, the curved surface area of a cylinder is <nobr aria-hidden="true">2πrh</nobr><script type="math/tex" id="MathJax-Element-7">2\pi rh</script>, where r is the radius and h is the height of the cylinder.
In our problem, the height of the cylinder is 14 cm.
<nobr aria-hidden="true">2πRh − 2πrh = 882πh (R−r) = 88R − r = 882 × 227 × 14R − r = 88 × 72 × 22 × 14R−r = 1...(1)</nobr>
Also, the difference of volume of outer part and inner part of cylinder is given.
So, Volume of outer part – Volume of inner part<nobr aria-hidden="true">=176</nobr><script type="math/tex" id="MathJax-Element-9">=176</script>.
Since, the volume of a cylinder is <nobr aria-hidden="true">πr2h</nobr><script type="math/tex" id="MathJax-Element-10">\pi {{r}^{2}}h</script>, where r is the radius and h is the height of the cylinder.
<nobr aria-hidden="true">πR2h − πr2h = 176πh(R2 − r2) = 176</nobr>
We expand the <nobr aria-hidden="true">(R2−r2)</nobr><script type="math/tex" id="MathJax-Element-12">({{R}^{2}}-{{r}^{2}})</script> using the identity <nobr aria-hidden="true">(a2−b2)=(a−b)(a+b)</nobr>.
<nobr aria-hidden="true">πh(R−r)(R+r) = 176π × 14(R+r) = 176R+r = 176×722×14R+r = 4 ...(2)</nobr>
Adding equation (1) and equation (2), we get
<nobr aria-hidden="true">R+r+R−r=52R=5R=52D=5</nobr>
Putting in equation (2),
<nobr aria-hidden="true">52+r=4r=4−52r=8−52r=32d=3.</nobr>
Hence, the diameter of outer cylinder is 5cm and the diameter of inner cylinder is 3cm.</div> </div>
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As the share of his both sons,i.e., Elder and Younger One are same,viz., 5/11 and 5/11 and the rest is for widow,i.e., 1/11.
And as the widow gets Rs. 3600 of her share....
Therefore, the sharing of both the boys will be...... => 5 into 3600 = Rs. 18000
And as the sharing of both the son is same...... Thus, Both will get Rs. 18000. That's It. :)
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Sia ? 6 years, 2 months ago
Consider two right triangles ABC and PQR in which {tex} \angle B{/tex} and {tex}\angle Q{/tex} are the right angles.
We have,
In {tex}\triangle ABC{/tex}
{tex}\sin B=\frac{AC}{AB}{/tex}
and, In {tex}\triangle PQR{/tex}
{tex}\sin Q=\frac{PR}{PQ}{/tex}
{tex} \because \quad \sin B = \sin Q{/tex}
{tex} \Rightarrow \quad \frac { A C } { A B } = \frac { P R } { P Q }{/tex}
{tex} \Rightarrow \quad \frac { A C } { P R } = \frac { A B } { P Q } = k{/tex}(say) ...... (i)
{tex} \Rightarrow {/tex} AC = kPR and AB = kPQ .....(ii)
Using Pythagoras theorem in triangles ABC and PQR, we obtain
AB2 = AC2 + BC2 and PQ2 = PR2 + QR2
{tex} \Rightarrow \quad B C = \sqrt { A B ^ { 2 } - A C ^ { 2 } } \text { and } Q R = \sqrt { P Q ^ { 2 } - P R ^ { 2 } }{/tex}
{tex} \Rightarrow \quad \frac { B C } { Q R } = \frac { \sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \frac { \sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } }{/tex} [ using (ii) ]
{tex} \Rightarrow \quad \frac { B C } { Q R } = \frac { k \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k{/tex}...(iii)
From (i) and (iii), we get
{tex} \frac { A C } { P R } = \frac { A B } { P Q } = \frac { B C } { Q R }{/tex}
{tex} \Rightarrow \quad \Delta A C B - \Delta P R Q{/tex} [By S.A.S similarity]
{tex} \therefore \quad \angle B = \angle Q{/tex}
Hence proved.
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