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Ask QuestionPosted by Mohit Raajz 6 years, 11 months ago
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Raunak Pandey ?? 6 years, 11 months ago
Posted by Madhu Kumari 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Given: A trapezium ABCD, In which AB {tex}\parallel{/tex} CD and its diagonals AC and BD intersect at O.
To Prove: {tex}\frac{{AO}}{{OC}} = \frac{{BO}}{{OD}}{/tex}
Construction: Through O draw OE||AB
proof: In {tex}\triangle {/tex}ADC, OE {tex}\parallel{/tex} DC,
Hence {tex}\frac{{AE}}{{ED}} = \frac{{AO}}{{OC}}{/tex} .....(i).....[By BPT]
Again in {tex}\triangle {/tex}ABD, we have OE {tex}\parallel{/tex} AB,
hence {tex}\frac{{DE}}{{EA}} = \frac{{DO}}{{OB}}{/tex} ........[By BPT]
{tex}\Rightarrow {/tex} {tex}\frac{{EA}}{{ED}} = \frac{{OB}}{{OD}}{/tex} .....(ii)......[Using invertendo]
{tex}\therefore {/tex} From (i) and (ii), we.{tex}\frac{{OB}}{{OD}} = \frac{{AO}}{{OC}}{/tex}H once proved

Posted by Arpita Joshi 6 years, 11 months ago
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Anushka Jugran ? 6 years, 11 months ago
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Roman Empire 6 years, 11 months ago
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Archana☺ Singh 6 years, 11 months ago
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Twinkle Star ( Atul )....... ?? 6 years, 11 months ago
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Vishal Gorana 6 years, 11 months ago
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Ram Kushwah 6 years, 11 months ago
Volume of sphere=volume of cyl
4/3π(4.2)3 =π(6)2h
On solving
h=2.74 cm
Posted by Mukesh Baro 6 years, 11 months ago
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Saniya Sarfaraz 6 years, 11 months ago
Mukesh Baro 6 years, 11 months ago
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Saniya Sarfaraz 6 years, 11 months ago
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Posted by Manish Bishnoi 6 years, 11 months ago
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Honey ☺☺☺ 6 years, 11 months ago
Gaurav Seth 6 years, 11 months ago
Arithmetic Progression. An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1.
Posted by Rishab Anand 6 years, 11 months ago
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Puja Sahoo? 6 years, 11 months ago
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Yaar 6 years, 11 months ago
Posted by Aachal Singh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let P be the position of the pole and A & B be the opposite fixed gates. Let, BP= x metres.
{tex}\therefore{/tex} AP = x + 7
In right triangle APB,
AP2 + BP2 = AB2
{tex}\Rightarrow{/tex}(x + 7)2 + x2 = 132
{tex}\Rightarrow{/tex}x2 + 49 + 14x + x2 = 169
{tex}\Rightarrow{/tex} 2x2 + 14x + 49 - 169 = 0
{tex}\Rightarrow{/tex}2x2 + 14x - 120 = 0
{tex}\Rightarrow{/tex}2(x2 + 7x - 60) = 0
{tex}\Rightarrow{/tex} x2 + 7x - 60 = 0
{tex}\Rightarrow{/tex}x2 + 12x - 5x - 60 = 0
{tex}\Rightarrow{/tex} x(x + 12) - 5(x + 12) = 0
{tex}\Rightarrow{/tex} (x + 12)(x - 5) = 0
{tex}\Rightarrow x=5 \ or \ -12{/tex}
As x can not be negative. So, x = 5.
Therefore, AP = 7+5 = 12
Hence, AP = 12 m and BP = 5 m

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Mohit Raajz 6 years ago
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