Question

Javier has four cylindrical models. The heights, radii, and diagonals of the vertical cross-sections of the models are shown in the table.
Model 1
radius: 14 cm
height: 48 cm
diagonal: 50 cm
Model 2
radius: 6 cm
height: 35 cm
diagonal: 37 cm
Model 3
radius: 20 cm
height: 40 cm
diagonal: 60 cm
Model 4
radius: 24 cm
height: 9 cm
diagonal: 30 cm



In which model does the lateral surface meet the base at a right angle?
a. Model 1
b. Model 2
c. Model 3
d. Model 4

Answer 1

Given the heights, radii, and diagonals of the vertical cross-sections of the models, the model in which the lateral surface meet the base at a right angle is the model in which the height, the diameter and the diagonal of the vertical cross-section forms a right triangle.

i.e. the sum of the squares of the height (h) and the diameter (d) gives the square of the diagonal vertical cross-section (l).

For model 1:

radius: 14 cm, thus diameter = 2(14) = 28 cm
height: 48 cm
diagonal: 50 cm

$d^2+h^2=28^2+48^2 \\ \\ =784+2,304=3,090\neq50^2=l^2$

Thus, the lateral surface of model 1 does not meets the base at right angle.

For model 2:

radius: 6 cm, thus diameter = 2(6) = 12 cm
height: 35 cm
diagonal: 37 cm

[tex]d^2+h^2=12^2+35^2 \\ \\ =144+1,225=1,369=37^2=l^2[/tex]

Thus, the lateral surface of model 2 meets the base at right angle.

For model 3:

radius: 20 cm, thus, diameter = 2(20) = 40 cm
height: 40 cm
diagonal: 60 cm

$d^2+h^2=40^2+40^2 \\ \\ =1,600+1,600=3,200\neq60^2=l^2$

Thus, the lateral surface of model 3 does not meets the base at right angle.

For model 4:

radius: 24 cm, thus, diameter = 2(24) = 48 cm
height: 9 cm
diagonal: 30 cm

$d^2+h^2=48^2+9^2 \\ \\ =2,304+81=2,385\neq30^2=l^2$

Thus, the lateral surface of model 3 does not meets the base at right angle.

Therefore, the model in which the lateral surface meets the base at a right angle is model 2 (option b)

Answer 2

The Answer is in fact B or "Model 2"