Question

A manufacturer of television parts received an order for $52$ inch picture tubes, measure along the diagonal, as shown in Figure. The tubes are to be rectangular in shape and $4$ inches wide than they are high. Find the dimensions of the tube.

• A. $34.17,30.17$
• B. $34.71,38.71$
• C. $28.17,24.17$
• D. $32.71,28.71$

Answer 1

Answer: (B)

$34.71,38.71$

We need to find the height and width of the rectangular picture tube. We note that two adjacent sides of the picture tube and a diagonal form a right triangle.
We can let h represent the height of the picture tube. Then h + 4 will represent the width. Since two adjacent sides and a diagonal of the tube form a right triangle, we can use the Pythagorean theorem to form the equation.
$a^2+b^2=c^2$ [The Pythagorean theorem]
$h^2+(h+4{)}^2={52}^2$
$h^2+h^2+8h+16=2704$
$2h^2+8h-2688=0$
$h^2+4h-1344=0$
To solve $h^2+4h-1344=0$,
we cannot use the square root method, and the factoring method looks difficult because of the large number (1344) involved. We will use the quadratic formula.
$h=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ [The quadratic formula]
$=\frac{-4\pm \sqrt{(4{)}^2-4(1\left)\right(-1344)}}{2\left(1\right)}=\frac{-4\pm \sqrt{16+5736}}{2}$
$=\frac{-4\pm \sqrt{5492}}{2}\approx \frac{-4\pm 73.430239}{2}$
$h\approx \frac{-4\pm 73.430239}{2}$ (Negative value not possible)
$\approx \frac{69.430239}{2}\approx 34.7151195$

And $h+4\approx 38.71$