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

  • Thread starter Thread starter Tadayen
  • Start date Start date
AI Thread Summary
The discussion revolves around finding the dimensions of a 7-inch diagonal tablet screen, where the height is 2.7 inches longer than the width. The Pythagorean theorem is applied to relate the width and height, leading to the equation w^2 + (w + 2.7)^2 = 7^2. A quadratic equation is derived from this relationship, which is solved using the quadratic formula to find the width (w) and height (h) in terms of square roots. The final expressions for the width and height are given as w = (-27 + √9071)/20 and h = (27 + √9071)/20. The discussion effectively demonstrates the mathematical process to determine the dimensions of the tablet screen.
Tadayen
Messages
2
Reaction score
0
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.
 
Mathematics news on Phys.org
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?
 
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.
 
Tadayen said:
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 $x$?
 
Tadayen said:
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:

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

Attachments

  • tablet.png
    tablet.png
    1.1 KB · Views: 70
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Back
Top