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Sia ? 6 years, 6 months ago
Let the present age of father be x years and sum of present age of two son's be y years.
According to question,
after five years
x + 5 = 2(y + 5 + 5)
x + 5 = 2y + 20
x - 2y = 15 ....(i)
and x = 3y ....(ii)
{tex}\therefore{/tex} 3y - 2y = 15
or y = 15
{tex}\therefore{/tex} age of father x = 3y
= 3 × 15
= 45 years
Posted by Abhyuday Pratapsingh 7 years, 9 months ago
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Posted by Sahil Indre 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
According to the question,we are given that,
Side of cube = 21 cm
Diameter of hemisphere = 21 cm
Then ,radius of hemisphere = {tex}\frac { 21 } { 2 }{/tex} cm
We have to find the volume and total surface area of the remaining block.
{tex}{/tex}Now, Volume of remaining block = Volume of cube - Volume of hemisphere
= {tex}( s i d e ) ^ { 3 } - \frac { 2 } { 3 } \pi r ^ { 3 }{/tex}
{tex}= 21 \times 21 \times 21 - \frac { 2 } { 3 } \times \frac { 22 } { 7 } \times \frac { 21 } { 2 } \times \frac { 21 } { 2 } \times \frac { 21 } { 2 }{/tex}
= 9261 - 2425.5
= 6835.5 cm3
Therefore,Total surface area of remaining block = Total surface area of cube - Area of top of depression + Curved surface area of hemispherical depression
{tex}= 6 ( side ) ^ { 2 } - \pi r ^ { 2 } + 2 \pi r ^ { 2 }{/tex}
{tex}= 6 ( side ) ^ { 2 } + \pi r ^ { 2 }{/tex}
{tex}= 6 ( 21 ) ^ { 2 } + \frac { 22 } { 7 } \times \frac { 21 } { 2 } \times \frac { 21 } { 2 }{/tex}
= 2646 + 346.5
= 2992.5 cm2
Posted by Abc Def 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let APB be the tangent and take O as centre of the circle.
Let us suppose that MP{tex}\bot{/tex}AB does not pass through the centre.
Then,
{tex}\angle OPA = 90^\circ{/tex} [{tex}\because{/tex} Tangent is perpendicular to the radius of circle]
But {tex}\angle MPA = 90^\circ{/tex} [Given]
{tex}\therefore \angle OPA = \angle MPA{/tex}
This is only possible when point O and point M coincide with each other.
Hence, the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
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