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Lucifer ? Morningstar? 6 years, 3 months ago
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Unnati Agarwal 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
Let the two similar triangles ABC and DEF in which AP and DQ are the medians respectively.
we have to prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
If ΔABC~ΔDEF
{tex}⇒ \frac{ar(BCA)}{ar(EFD)}=\frac{AB^2}{DE^2} {/tex} → (1)
As ΔABC~ΔDEF
{tex}\frac{AB}{DE}=\frac{BC}{EF}=\frac{2BP}{2EQ}{/tex}
Hence,{tex} \frac{AB}{DE}=\frac{BP}{EQ}{/tex}
In ΔABP and ΔDEQ
{tex}\frac{AB}{DE}=\frac{BP}{EQ}{/tex}
∠B=∠E (∵ΔABC~ΔDEF)
By SAS rule, ΔABP~ΔDEQ
{tex}⇒ \frac{AB}{DE}=\frac{AP}{DQ}{/tex}
Squaring, we get
{tex}\frac{AB^2}{DE^2}=\frac{AP^2}{DQ^2} {/tex} → (2)
Comparing (1) and (2), we get
{tex}\frac{ar(BCA)}{ar(EFD)}=\frac{AP^2}{DQ^2} {/tex}
Hence, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Posted by Rajendra Rajendra P G 6 years, 3 months ago
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Aadya Singh ? 6 years, 3 months ago
Smallest composite number = 4.
So, HCF = 2.
Posted by Vikas Jha 6 years, 3 months ago
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Yogita Ingle 6 years, 3 months ago
Let the speed of the stream be x km/hr
Then,
Speed downstream = (15 + x) km/hr
Speed upstream = (15 - x) km/hr
∴ 30/(15+x) + 30/ (15−x) = 4 1/2
⇒ 900/(225−x2) =9/ 2
⇒ 9x2 = 225
⇒x2 = 25
⇒x = 5km/hr
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Sneha Tyagi 6 years, 3 months ago
Yogita Ingle 6 years, 3 months ago
let root 5 be rational
then it must in the form of p/q [q is not equal to 0][p and q are co-prime]
root 5=p/q
=> root 5 × q = p
squaring on both sides
=> 5 ×q ×q = p ×p ------> 1
p ×p is divisible by 5
p is divisible by 5
p = 5c [c is a positive integer] [squaring on both sides ]
p ×p = 25c ×c --------- > 2
sub p ×p in 1
5 ×q ×q = 25 ×c ×c
q ×q = 5 ×c ×c
=> q is divisble by 5
thus q and p have a common factor 5
there is a contradiction
as our assumsion p &q are co prime but it has a common factor
so √5 is an irrational
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Yogita Ingle 6 years, 3 months ago
Integral coordinates are coordinates that are whole numbers. Integral coordinates cannot be fractional or have decimals.
Posted by Ankesh .. ??Now I Am Fully Ready? 6 years, 3 months ago
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Ankesh .. ??Now I Am Fully Ready? 6 years, 3 months ago
Kashvi ? 6 years, 3 months ago
Kashvi ? 6 years, 3 months ago
Kashvi ? 6 years, 3 months ago
Ankesh .. ??Now I Am Fully Ready? 6 years, 3 months ago
Kashvi ? 6 years, 3 months ago
Ankesh .. ??Now I Am Fully Ready? 6 years, 3 months ago
Ankesh .. ??Now I Am Fully Ready? 6 years, 3 months ago
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Krishna Sharma 6 years, 3 months ago
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