Find dimensions of a 7" tablet screen where length is 2.7" longer than width

[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:

$$h=w+2.7\tag{1}$$

And by the Pythagorean theorem, we may write:

$$w^2+h^2=7^2\tag{2}$$

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

$$w^2+(w+2.7)^2=7^2$$

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

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

Adding within the parentheses, we have:

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

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

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

Squaring the binomial on the left, we get:

$$(10w)^2+(10w)^2+2(10w)(27)+27^2=(70)^2$$

Simplify further:

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

Factor difference of squares:

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

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

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

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

$$2u^2+54u-4171=0$$

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

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

Hence:

$$10w=\frac{-27+\sqrt{9071}}{2}\implies w=\frac{-27+\sqrt{9071}}{20}$$

And so:

$$h=\frac{-27+\sqrt{9071}}{20}+\frac{27}{10}=\frac{-27+\sqrt{9071}+54}{20}=\frac{27+\sqrt{9071}}{20}$$