Question

Let $ABCD$ be a convex quadrilateral such that all of its sides and diagonals have integer lengths. Given $\angle ABC=\angle ADC={90}^{\circ },$ $AB=BD$ and $CD=41.$ The length of $BC$ is

• A. $579$
• B. $580$
• C. $627$
• D. $630$

Answer 1

The correct option is B $580$


$(AC{)}^2-(AD{)}^2={41}^2$
As $AC$ and $AD$ are integers,
$AC=841,AD=840$
In $△ABD,$
$\frac{p}{\sin \theta }=\frac{AD}{\sin \left(2\theta \right)}$
$\Rightarrow p=\frac{840}{2\cos \theta }=\frac{420}{\cos \theta }\dots \left(1\right)$

In $△BCD,$
$\frac{41}{\sin (2\theta -\frac{\pi }{2})}=\frac{p}{\sin \theta }$
$\Rightarrow p=\frac{41\sin \theta }{-\cos 2\theta }\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$
$\frac{420}{\cos \theta }=-\frac{41\sin \theta }{\cos 2\theta }$
$\Rightarrow \frac{-420}{41}=\frac{\tan 2\theta }{2}\Rightarrow \cos 2\theta =-\frac{41}{841}$
$p^2=\frac{(420{)}^2}{{\cos }^2\theta }=\frac{(420{)}^2\times 2}{1+\cos 2\theta }=\frac{2\times (420{)}^2}{1-\frac{41}{841}}=\frac{2\times (420{)}^2}{800}\times 841\Rightarrow p^2=441\times 841$

$(BC{)}^2=(AC{)}^2-p^2=(841{)}^2-441\times 841=841(841-441)=841\times 400\Rightarrow BC=20\times 29=580$