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Sunil Kumar

· started a discussion

· 1 Months ago

s (Question: 107629.0 )

how can you say that k=a/2

Question:
If \(\cfrac{x}{a}=\cfrac{y}{b}=\cfrac{z}{c}\) then xy + yz + zx is equal to:
Options:
A) \(\cfrac{(a + b + c)^2}{x^2 + y^2+ z^2}\)
B) \(\cfrac{x^2 (a + b + c)^2 - a^2 (x^2+y^2+z^2)}{2a^2}\)
C) \(\cfrac{ax+by+cz}{(a + b + c)^2}\)
D) None of the above
Solution:
Ans: (b) x = ak, y = bk, z = ck

x+y+z = k (a+b+c)

x2+y2+z2 + 2 (xy+yz+zx) = k2 (a+b+c)2

\(\left ( xy+yz+zx \right )=\cfrac{k^2}{2}(a+b+c)^2-\cfrac{1}{2}(x^2+y^2+z^2)\)

\(=\cfrac{x^2}{2a^2}(a+b+c)^2-\cfrac{1}{2}(x^2+y^2+z^2)\)

\(=\cfrac{x^2(a+b+c)^2-a^2(x^2+y^2+z^2)}{2a^2}\)

Talented Boy

· commented

· 1 Months ago

k=x/a is taken there bro because x=ak is considered......see again.........
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