Question

A huge suspended LED screen is the centerpiece of The Place, a popular mall in Beijing, China. Find the length and width of a similar rectangular screen if the length is 5 meters more than 6 times its width, and the viewable area is 7,525 square meters.

Answer 1

The length and width of the similar rectangular screen are 110 meters and 35 meters respectively.

SOLUTION:

Given, A huge suspended LED screen is the centerpiece of The Place, a popular mall in Beijing, China.  

We have to find the length and width of a similar rectangular screen  

We are also given that the length is 5 meters more than 3 times its width,  

Let width of rectangle be "b" meters, the its length will be 5 + 3b meters

And, the viewable area is 3,850 square meters.

$\begin{array}{l}{\text { So, area }=3850} \\\\ {\text { Length } \times \text { width }=3850} \\\\ {(5+3 b) \times b=3850} \\\\ {5 b+3 b^{2}=3850} \\\\ {3 b^{2}+5 b-3850=0}\end{array}$

Now,let us use quadratic formula:

$\begin{array}{l}{\mathrm{x}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}} \\\\ {\text { Then, } \mathrm{b}=\frac{-5 \pm \sqrt{5^{2}-4 \times 3 \times(-3850)}}{2 \times 3}}\end{array}$

$\begin{array}{l}{\mathrm{b}=\frac{-5 \pm \sqrt{25+46200}}{6}} \\\\ {\mathrm{b}=\frac{-5 \pm \sqrt{46225}}{6}} \\\\ {\mathrm{b}=\frac{-5 \pm 215}{6}} \\\\ {\mathrm{b}=\frac{-5+215}{6} \text { or } \frac{-5-215}{6}} \\\\ {\mathrm{b}=\frac{210}{6} \mathrm{or} \frac{-220}{6}}\end{array}$

Neglect negative values as width can’t be negative

$b=\frac{210}{6}=35$

so, width = 35 meters, then length = 5 + 3(35) = 5 + 105 = 110 meters

hence, the length and width of the similar rectangular screen are 110 meters and 35 meters respectively.