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Ask QuestionPosted by Shashank Sahu 7 years, 2 months ago
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Posted by H. Singh Singh 7 years, 2 months ago
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Posted by Shivansh Upadhyay 7 years, 2 months ago
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Posted by Bandi Harini 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
AD2 = BD {tex}\times{/tex} CD
or, {tex}\frac { A D } { C D } = \frac { B D } { A D }{/tex}
Therefore, {tex}\triangle A D C \sim \triangle B D A{/tex} (by SAS)
or, {tex}\angle{/tex}BAD = {tex}\angle{/tex}ACD;
{tex}\angle{/tex}DAC = {tex}\angle{/tex}DBA (Corresponding angles of similar triangles)
{tex}\angle{/tex}BAD + {tex}\angle{/tex}ACD + {tex}\angle{/tex}DAC + {tex}\angle{/tex}DBA = 180o [sum of angles of ∆]
or, 2{tex}\angle{/tex}BAD + 2{tex}\angle{/tex}DAC = 180o
or, {tex}\angle{/tex}BAD + {tex}\angle{/tex}DAC = 90o
Therefore, {tex}\angle{/tex}A = 90o
Posted by Dhiraj Singh Rajput 7 years, 2 months ago
- 1 answers
Deep Keshari 7 years, 2 months ago
Posted by Satyam Aryaa 7 years, 2 months ago
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Posted by Achal Jogi 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let us assume that point P divides the line segment AB in the ratio (k : 1)
Using section formula,
P(4, m) = {tex}\frac{6k+2}{k+1} , {/tex}{tex}\frac{-3k+3}{k+1}{/tex}
{tex}\therefore{/tex} {tex}\frac{6 \mathrm{k}+2}{\mathrm{k}+1}{/tex} = 4 {tex}\Rightarrow 6k+2=4k+4 \Rightarrow 2k=2{/tex}
{tex}\Rightarrow{/tex} k = 1
{tex}\therefore{/tex} The ratio is 1 : 1
Now, m = {tex}\frac{-3k+3}{k+1}{/tex} = {tex}\frac{-3+3}{2}{/tex} = 0
Posted by Bhumika Goswami 7 years, 2 months ago
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Raunak _ Pandey ?? 7 years, 2 months ago
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Madhu V.D? 7 years, 2 months ago
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Posted by Piyush Rai 7 years, 2 months ago
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Posted by Bhavesh Pushkar 7 years, 2 months ago
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Posted by John Cena 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let the original list price of the toy be Rs. x.
{tex}\therefore{/tex} Number of toys can be bought for Rs 360 = {tex}\frac{360}{x}{/tex} toys
Now, Reduced list price of the toy = Rs(x - 2)
{tex}\therefore{/tex} Number of toys can be bought with new reduced list price for Rs 360 = {tex}\frac{360}{x-2}{/tex} toys
According to the question:
{tex}\frac{\displaystyle360}{\displaystyle x-2}=2+\frac{\displaystyle360}{\displaystyle x}{/tex} (2 extra toys can be bought if price reduces by 2 rupees)
{tex}\therefore \frac{360}{x - 2 } - \frac{360}{x }{/tex} = 2
{tex}\Rightarrow \frac{ 360x - 360(x - 2 )}{x (x - 2)}{/tex} = 2
{tex}\Rightarrow \frac{360x - 360x + 720}{x^2 - 2x }{/tex} = 2
{tex}\Rightarrow{/tex} 720 = 2(x2 - 2x)
{tex}\Rightarrow x^2 - 2x = \frac{720 }{2}{/tex}
{tex}\Rightarrow{/tex} x2 - 2x = 360
{tex}\Rightarrow{/tex} x2 - 2x - 360 = 0
{tex}\Rightarrow{/tex} x2 - 20x + 18x - 360 = 0
{tex}\Rightarrow{/tex} x(x - 20) + 18(x - 20) = 0
{tex}\Rightarrow{/tex} (x - 20)(x + 18) = 0
{tex}\Rightarrow{/tex} x - 20 = 0 [{tex}{/tex} Since, Cost cannot be negative. {tex}\therefore{/tex},x + 18 {tex}\neq{/tex} 0]
{tex}\Rightarrow{/tex} x = 20
Hence, the original list price of the toy is x = Rs 20.
Posted by Nitish Kumar 7 years, 2 months ago
- 2 answers
Aryan Yadav 7 years, 2 months ago
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Dhiraj Singh Rajput 7 years, 2 months ago
1Thank You