Question

A company is replacing cables with fiber optic lines in rectangular casing BCDE. If line segment DE = 3 cm and line segment BE = 3 cm, what is the smallest diameter of pipe that will fit the fiber optic line? Round your answer to the nearest hundredth.
3.54 cm

3.91 cm

4.24 cm

4.95 cm

Answer 1

Answer:

The smaller diameter is $4.24\ cm$

Step-by-step explanation:

we know that

The diameter of the circle is equal to BD

Applying the Pythagoras Theorem

$BD^{2} =BE^{2} +DE^{2}$

substitute the given values

$BD^{2} =3^{2} +3^{2}$

$BD^{2} =18$

$BD=4.24\ cm$

Answer 2

Answer:

The correct option is C.

Step-by-step explanation:

Given information: BCDE is a rectangular casing, DE = 3 cm and BE = 3 cm.

We need to find the smallest diameter of pipe that will fit the fiber optic line. It means we have to find the measure of DB.

The measure of all interior angles of a rectangle or square is 90°.

$\angle DEB=90^{\circ}$

It means the DEB is right angled triangle.

According to the Pythagoras theorem:

$hypotenuse^2=leg_1^2+leg_2^2$

In triangle DEB,

$(DB)^2=(DE)^2+(BE)^2$

$(DB)^2=(3)^2+(3)^2$

$(DB)^2=9+9$

$(DB)^2=18$

Taking square root both sides.

$DB=\sqrt{18}$

$DB=4.24264068712$

$DB\approx 4.24$

Therefore the correct option is C.