Discover the Dimensions of a 7" Tablet Screen with a Missing Leg Solution

[Tadayen]

A tablet computer has a 7" diagonal screen. The length of the screen is 2.7 " longer than the width. Find the dimensions of the screen.

 

[MarkFL]

Hello and welcome to MHB! :D

Can you think of some way we can relate the width, length and the diagonal measure of the screen?

 

[Tadayen]

I know it's C^2=A^2+B^2 so I've tried 49=x^2+(x+2.7)^2 doesn't seem to give the right answer.

 

[MarkFL]

I know it's C^2=A^2+B^2 so I've tried 49=x^2+(x+2.7)^2 doesn't seem to give the right answer.

Yes, that's correct. What did you get for $x$?

 

[MarkFL]

A tablet computer has a 7" diagonal screen. The length of the screen is 2.7 " longer than the width. Find the dimensions of the screen.

I will use height rather than length, and this makes more sense to me. Let's draw a diagram first (where all measures are in inches):

View attachment 5036

We are given that the height $h$ is 2.7" more than the width $w$, so we may state:

\(\displaystyle h=w+2.7\tag{1}\)

And by the Pythagorean theorem, we may write:

\(\displaystyle w^2+h^2=7^2\tag{2}\)

Now, using (1) we may substitute for $h$ in (2) to get:

\(\displaystyle w^2+(w+2.7)^2=7^2\)

I don't like working with decimals, so let's instead write:

\(\displaystyle w^2+\left(w+\frac{27}{10}\right)^2=7^2\)

Adding within the parentheses, we have:

\(\displaystyle w^2+\left(\frac{10w+27}{10}\right)^2=7^2\)

Multiplying through by $10^2$, we obtain:

\(\displaystyle (10w)^2+(10w+27)^2=(7\cdot10)^2\)

Squaring the binomial on the left, we get:

\(\displaystyle (10w)^2+(10w)^2+2(10w)(27)+27^2=(70)^2\)

Simplify further:

\(\displaystyle 2(10w)^2+54(10w)+\left(27^2-70^2\right)=0\)

Factor difference of squares:

\(\displaystyle 2(10w)^2+54(10w)+(27+70)(27-70)=0\)

\(\displaystyle 2(10w)^2+54(10w)-97\cdot43=0\)

\(\displaystyle 2(10w)^2+54(10w)-4171=0\)

Let $u=10w$, and we have a quadratic in $u$ in standard form:

\(\displaystyle 2u^2+54u-4171=0\)

Applying the quadratic formula (and discarding the negative root), we obtain:

\(\displaystyle u=\frac{-54+\sqrt{54^2+4(2)(4171)}}{2(2)}=\frac{-2(27)+2\sqrt{27^2+(2)(4171)}}{2(2)}=\frac{-27+\sqrt{729+8342}}{2}=\frac{-27+\sqrt{9071}}{2}\)

Hence:

\(\displaystyle 10w=\frac{-27+\sqrt{9071}}{2}\implies w=\frac{-27+\sqrt{9071}}{20}\)

And so:

\(\displaystyle h=\frac{-27+\sqrt{9071}}{20}+\frac{27}{10}=\frac{-27+\sqrt{9071}+54}{20}=\frac{27+\sqrt{9071}}{20}\)