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In the above figure, what are the values
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12 Dec 2024, 00:38
12 Dec 2024, 00:38
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In the above figure, what are the values of $\(x\)$ and $\(y\)$ ?
(A) $\(x=\frac{a k}{b} ; y=b-k\)$
(B) $\(x=a ; y=b\)$
(C) $\(x=\frac{a}{b} ; y=b-k\)$
(D) $\(x=\frac{a}{b k} ; y=b\)$
(E) $\(x=\frac{a k}{b} ; y=b+k\)$
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Re: In the above figure, what are the values
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22 Dec 2024, 04:06
22 Dec 2024, 04:06
Carcass how do we get the value of x?
In the above figure, what are the values
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22 Dec 2024, 08:52
22 Dec 2024, 08:52
1
Expert Reply
In the figure above, we have $\(\mathrm{OA}=\mathrm{a}, \mathrm{OB}=\mathrm{b}, \mathrm{BA}=\sqrt{\mathrm{a}^2+\mathrm{b}^2}, \mathrm{BC}=\mathrm{k}, \mathrm{CD}=\mathrm{x}\)$ \& $\(\mathrm{BD}=\sqrt{\mathrm{x}^2+\mathrm{y}^2}\)$ (Note: - $\(\mathrm{BA} \& \mathrm{BD}\)$ are found using Pythagoras theorem in triangle ABC \& triangle $\(B C D\)$ respectively)
Clearly $\(\triangle \mathrm{BOA} \sim \triangle \mathrm{BCD}\)$ (as $\(C D\)$ is parallel with $\(O A\)$ ), so using the similar triangle property (i.e. in similar triangles, the ratio of the corresponding sides is same), we get the $\(\frac{O B}{B C}=\frac{O A}{C D}=\frac{B A}{B D}\)$
i.e. $\(\frac{b}{k}=\frac{a}{x}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}}\)$.
Now, $\(\frac{b}{k}=\frac{a}{x} \Rightarrow x=\frac{a k}{b}\)$ \& from the figure $\(C O=y=B O-B C=b-k\)$
Hence we get $\(x=\frac{a k}{b} \& y=b-k\)$, so the answer is $\((A)\)$.
Clearly $\(\triangle \mathrm{BOA} \sim \triangle \mathrm{BCD}\)$ (as $\(C D\)$ is parallel with $\(O A\)$ ), so using the similar triangle property (i.e. in similar triangles, the ratio of the corresponding sides is same), we get the $\(\frac{O B}{B C}=\frac{O A}{C D}=\frac{B A}{B D}\)$
i.e. $\(\frac{b}{k}=\frac{a}{x}=\frac{\sqrt{a^2+b^2}}{\sqrt{x^2+y^2}}\)$.
Now, $\(\frac{b}{k}=\frac{a}{x} \Rightarrow x=\frac{a k}{b}\)$ \& from the figure $\(C O=y=B O-B C=b-k\)$
Hence we get $\(x=\frac{a k}{b} \& y=b-k\)$, so the answer is $\((A)\)$.
