Math  /  Geometry

QuestionImagine you need to purchase a laptop bag for your 14 -inch laptop. The only problem is you don't have your laptop with you, and it sure would be frustrating to buy a bag only to realize that your laptop doesn't quite fit.
You recall laptop computers are measured according to the diagonals of their screens, and you remember your 14 -inch laptop has a screen that is 8 inches tall. How wide is the screen?
Exact Answer (written as a simpified radical): \square in.
Approximate (decimal) Answer: \square in. Give your approximate answer accurate to 2 decimal places.

Studdy Solution

STEP 1

1. The laptop screen is a rectangle.
2. The diagonal of the laptop screen is 14 inches.
3. The height of the laptop screen is 8 inches.
4. We need to find the width of the screen.

STEP 2

1. Use the Pythagorean Theorem to set up an equation.
2. Solve for the width of the screen.
3. Simplify the expression to find the exact answer.
4. Calculate the approximate decimal value.

STEP 3

The Pythagorean Theorem relates the sides of a right triangle: a2+b2=c2 a^2 + b^2 = c^2 , where c c is the hypotenuse. In this case, the height and width of the screen are the two legs, and the diagonal is the hypotenuse.
Let w w be the width of the screen. Then, according to the Pythagorean Theorem:
82+w2=142 8^2 + w^2 = 14^2

STEP 4

Substitute the known values into the equation and solve for w2 w^2 :
64+w2=196 64 + w^2 = 196
Subtract 64 from both sides:
w2=19664 w^2 = 196 - 64

STEP 5

Simplify the expression:
w2=132 w^2 = 132

STEP 6

Take the square root of both sides to solve for w w :
w=132 w = \sqrt{132}
Simplify the radical expression by finding the largest perfect square factor:
w=433 w = \sqrt{4 \cdot 33} w=233 w = 2\sqrt{33}

STEP 7

Calculate the approximate decimal value of w w :
w2×33 w \approx 2 \times \sqrt{33}
Using a calculator, find 335.744 \sqrt{33} \approx 5.744 .
w2×5.744 w \approx 2 \times 5.744 w11.488 w \approx 11.488
Round to two decimal places:
w11.49 w \approx 11.49
The width of the screen is:
Exact Answer: 233 \boxed{2\sqrt{33}} in.
Approximate Answer: 11.49 \boxed{11.49} in.

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