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Preeti Dabral 2 years, 8 months ago
Let the radius of base of hemisphere and cone, each be r cm. Let the height of the cone be h cm.
Volume of the cone = {tex}{1\over3}\pi r^2{/tex} hcm3
Volume of the hemisphere = {tex}{2\over3}\pi{/tex}r3 cm3
According to the question, {tex}{1\over3}\pi r^2h={2\over3}\pi r^3{/tex}
{tex}\Rightarrow{/tex} h = 2r
{tex}\Rightarrow{/tex} Height of the cone = 2r cm.
Height of the hemisphere = r cm
{tex}\therefore{/tex} Ratio of their heights = 2r : r = 2 : 1
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Preeti Dabral 2 years, 8 months ago
Given: ABCD is a rhombus
In {tex}\triangle{/tex}ABC and {tex}\triangle{/tex}ADC,
AB = CD [Sides of a rhombus]
BC = DA [Sides of a rhombus]
AC = AC [Common]
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC {tex}\cong{/tex} {tex}\triangle{/tex}ADC [By SSS Congruency]
{tex}\therefore{/tex} {tex}\angle {/tex}CAB = {tex}\angle {/tex}CAD And {tex}\angle {/tex}ACB = {tex}\angle {/tex}ACD
Hence AC bisects {tex}\angle {/tex}A as well as {tex}\angle {/tex}C
Similarly, by joining B to D, we can prove that {tex}\triangle{/tex}ABD {tex}\cong{/tex} {tex}\triangle{/tex}CBD
Hence BD bisects {tex}\angle {/tex}B as well as {tex}\angle {/tex}D
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Rashmi Mirchandani 2 years, 9 months ago
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Nasima Praveen 2 years, 8 months ago
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