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A Semicircle is inscribed in a right Triangle as shown in figure. Calculate the value of \(X?\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
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Screenshot 2022-07-21 210930.png [ 44.13 KiB | Viewed 1032 times ]
24 Jul 2022, 02:06
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A Semicircle is inscribed in a right Triangle as shown in figure. Calculate the value of \(X?\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
a) \(\frac{1}{4}\)
b) \(\frac{2}{7}\)
c) \(\frac{3}{7}\)
d) \(\frac{1}{2}\)
e) \(\frac{4}{7}\)
Join Center of the semicircle, D, of say radius r to side BC at E and AC at F.
As DE and DF are tangents, angles BED and BFD are 90, making BEDF a square with side of length r.
Now take \(\triangle CDE\) and \(\triangle CAB\)
They are similar triangles, so \(\frac{DE}{AB}=\frac{CE}{CB}\)
=> \(\frac{r}{4}=\frac{3-r}{3}…….3r=12-4r…….r=\frac{12}{7}\)
Once again take \(\triangle CDE\) to find x
\((3-r)^2+r^2=(r+x)^2………(3-\frac{12}{7})^2+\frac{12}{7}^2=(r+x)^2\)
\(\frac{9}{7}^2+\frac{12}{7}^2=(r+x)^2\)
\(\frac{225}{49}=(r+x)^2…….\frac{15}{7}^2=(r+x)^2\)
\(\frac{15}{7}=r+x=\frac{12}{7}+x\)
\(x=\frac{3}{7}\)
C
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CE7AF80A-385C-4CB5-82F7-8D424CC6C4A6.jpeg [ 19.19 KiB | Viewed 875 times ]
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