Question

In rectangle $ABCD$, diagonal $BD=26$ cm and cotangent of angle $ABD=15$. Find the area and the perimeter of the rectangle $ABCD$.

• A. Area$=716$ sq. cm and perimeter $=11\sqrt{2}$ cm and $DC=12$ cm.
• B. Area$=318$ sq. cm and perimeter $=10\sqrt{2}$ cm and $DC=12$ cm.
• C. Area$=113$ sq. cm and perimeter $=13\sqrt{2}$ cm and $DC=12$ cm.
• D. Area$=312$ sq. cm and perimeter $=12\sqrt{2}$ cm and $DC=12$ cm.

Answer 1

The correct option is D Area$=312$ sq. cm and perimeter $=12\sqrt{2}$ cm and $DC=12$ cm.

Cot $\angle ABD=1.5$

In right $\angle ABD$,

Cot $\angle ABD=1.5=\frac{3}{2}$

$\frac{AB}{AD}=\frac{3}{2}$

$\Rightarrow$ Let $AB=3K$ and $AD=2K$

${AB}^2+{AD}^2={BD}^2$ (Pythagoras theorem)

$\Rightarrow {\left(3k\right)}^2+{\left(2k\right)}^2={\left(26\right)}^2\Rightarrow 13k^2=26\times 26$

$\Rightarrow k^2=52\Rightarrow k=\sqrt{52}=2\sqrt{13}cm$

$\therefore AB=3\times 2\sqrt{13}=6\sqrt{13}cm,AD=2\times 2\sqrt{13}=4\sqrt{13}cm$

Area of rectangle $ABCD=AB\times AD=6\sqrt{13}\times 4\sqrt{13}=312{cm}^2$

Perimeter of rectangle $ABCD=2(AB+AD)=2(6\sqrt{13}+4\sqrt{13})=20\sqrt{13}cm$