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JEE Advanced 2007Paper 2
Exam
JEE Advanced
Year
2007
Paper
Paper 2
Subject
Mathematics
Chapter
Circle

JEE Advanced 2007 · Paper 2 · Mathematics · Circle

Let A B C D be a quadrilateral with area 18 , with side A B...

Let ABCDA B C D be a quadrilateral with area 18 , with side ABA B parallel to the side CDC D and AB=2CDA B=2 C D. Let ADA D be perpendicular to ABA B and CDC D. If a circle is drawn inside the quadrilateral ABCDA B C D touching all the sides, then its radius isLet A B C D be a quadrilateral with area 18 , with side A B parallel to the side C D and A B=2 C D . Let A D beLet A B C D be a quadrilateral with area 18 , with side A B parallel to the side C D and A B=2 C D . Let A D beLet A B C D be a quadrilateral with area 18 , with side A B parallel to the side C D and A B=2 C D . Let A D beLet A B C D be a quadrilateral with area 18 , with side A B parallel to the side C D and A B=2 C D . Let A D be

Options

  1. A3
  2. B2
  3. C32\frac{3}{2}
  4. D1

Correct answer

B. 2

Step-by-step solution

18=12(\frac{1}{2}(3 α)(\alpha)(2 r) \Rightarrow α\alphar=6 <imgsrc="/api/proxy−image?url=https<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252Ft_pMgp4UiyFCXUuzb7KvmMfhX6yIJXigJy_YvZXXXZU.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2Ft_pMgp4UiyFCXUuzb7KvmMfhX6yIJXigJy_YvZXXXZU.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2Ft_pMgp4UiyFCXUuzb7KvmMfhX6yIJXigJy_YvZXXXZU.original.fullsize.png?alt=media"> \\ Line, y=−2rα(x−2α)y=-\frac{2 r}{\alpha}(x-2 \alpha) is tangent to circle (x−r)2+(y−r)2=r22α=3r and  α=6r=2\begin{aligned} (x-r)^2+(y-r)^2 & =r² \\ 2 \alpha& =3 r \text{\text{ and } } \alpha=6 \\ r & =2 \end{aligned}12(x+2x)×2r=18xr=6tan⁡θ=x−rrtan⁡(90∘−θ)=2x−rrx−rr=r2x−r×(2x−3r)=0\begin{aligned}\frac{1}{2}(x+2 x) \times2 r & =18 \\ x r & =6 \\ \tan \theta& =\frac{x-r}{r}\\ \tan \left(90^{{}^{\circ}}-\theta\right)& =\frac{2 x-r}{r}\\ \frac{x-r}{r}& =\frac{r}{2 x-r} \times(2 x-3 r)=0 \end{aligned}Let A B C D be a quadrilateral with area 18 , with side A B parallel to the side C D and A B=2 C D . Let A D be$\\ x=\frac{3 r}{2}From Eqs. (i) and (ii), we get r=2
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