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Ask QuestionPosted by Roshan Jha 7 years, 9 months ago
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Cutiee Princess 7 years, 9 months ago
Posted by Abhishek Patel 7 years, 9 months ago
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Posted by Jyotirmay Paul 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Construction : Draw {tex}A L \perp B C{/tex}
or, AL is median of BC (Isosceles triangle)
or, BL = LC = 6 cm.
In right {tex}\triangle A L B,{/tex} by Pythagoras theorem, AL = 8 cm.
In {tex}\triangle B P Q \text { and } \triangle B L A{/tex}
{tex}\angle B = \angle C{/tex} (Isosceles triangle)
{tex}\angle B P Q = \angle B L A = 90 ^ { \circ }{/tex}
{tex}\triangle B P Q \sim \triangle B L A{/tex} (AA similarity)
or, {tex}\frac { B P } { P Q } = \frac { B L } { A L }{/tex}
or, {tex}\frac { 6 - x } { y } = \frac { 6 } { 8 } \text { or, } x = 6 - \frac { 3 y } { 4 }{/tex}
Hence proved.
Posted by Jitender Shaw 7 years, 9 months ago
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Lavisha Nawal 7 years, 9 months ago
Posted by Jitender Shaw 7 years, 9 months ago
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Jitender Shaw 7 years, 9 months ago
Cutiee Princess 7 years, 9 months ago
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Cutiee Princess 7 years, 9 months ago
Lavisha Nawal 7 years, 9 months ago
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Cutiee Princess 7 years, 9 months ago
Posted by Khushboo Pandit 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
For a silver coin Diameter = 1.75 cm
{tex}\therefore {/tex} Radius (r)={tex}\frac{{1.75}}{2}cm = \frac{7}{8}cm{/tex}
Thickness (h) = 2 mm ={tex}\frac{1}{5}cm{/tex}
{tex}\therefore {/tex} The volume of a silver coin
={tex}\pi {r^2}h = \pi {\left( {\frac{7}{8}} \right)^2}\left( {\frac{1}{5}} \right){/tex}
={tex}\frac{{49}}{{320}}\pi c{m^3}{/tex}
Let n coins be melted.
Then, volume of n coin={tex}n\frac{{49}}{{320}}\pi c{m^3}{/tex}
For cuboid
length(L) = 5.5 cm
breath(B) = 10 cm
height(H) = 3.5 cm
{tex}\therefore {/tex} The volume of the cuboid = lbh
=5.5 × 10 × 3.5=192.5
={tex}\frac{{1925}}{{10}} = \frac{{385}}{2}{/tex}cm2
According to the question,
{tex}n\frac{{49\pi }}{{320}} = \frac{{385}}{2} \Rightarrow n = \frac{{385}}{2}.\frac{{320}}{{49\pi }}{/tex}
{tex}\Rightarrow {/tex} {tex}n = \frac{{385}}{2}.\frac{{320}}{{49}}.\frac{7}{{22}} \Rightarrow n = 400{/tex}
Hence, 400 coins must be melted
Posted by Ashok Jat 7 years, 9 months ago
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Sia ? 6 years, 5 months ago
ABC is an isosceles triangle right angled at B,
Let AB=BC=x cm

By Pythagoras theorem In {tex}\triangle {/tex} ABC
AC2 = AB2 + BC2
AC2 = x2 + x2
AC2 = 2x2
AC={tex}\sqrt 2 {/tex} x
{tex}\triangle{/tex}ACD{tex}\sim{/tex}{tex}\triangle{/tex}ABE (Given)
Thus, using Area Theorem
{tex}\therefore{/tex}{tex}\frac { \operatorname { ar } \Delta \mathrm { ABE } } { \operatorname { ar } \Delta A \mathrm { CD } } = \frac { \mathrm { AB } ^ { 2 } } { \mathrm { AC } ^ { 2 } } = \frac { x ^ { 2 } } { ( \sqrt { 2 } x ) ^ { 2 } }{/tex}
={tex}\frac { x ^ { 2 } } { 2 x ^ { 2 } }{/tex}={tex}\frac{1}{2}{/tex}=1:2.
Hence, Areas of the two given triangles are in the ratio 1:2
Posted by ∂Αяѕнαиαα ყα∂Αѵ ??? 7 years, 9 months ago
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Online Only On Brnly Geniusvatsone 7 years, 9 months ago
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Jitender Shaw 7 years, 9 months ago
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