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2 years ago

8th grade math . Pythagorran practice

Mathematics

1 answer:

Serhud [2]2 years ago

4 0

The answers according to rows are:

ROW 1 : diameter of base = 10, base area = 25π, volume of cylinder = 175π volume of cone = $\frac{175}{3}$ π

ROW 2 : base radius = 3, base area = 9π, height = 40/3, volume of cylinder = 120π
ROW 3 : base radius = 6, base diameter = 12, height = 4, volume of cylinder = 144π

<h3>What is mensuration?</h3>

Mensuration is an aspect of geometry that deals with measurement of volumes and area.

Analysis:

Row 1

Diameter of base = 2 x radius = 2 x 5 = 10

Base area = π $r^{2}$ = π $(5)^{2}$ = 25 $\pi$

Volume of cylinder = base area x height = 25π x 7 = 175π

Volume of cone = 1/3 x volume of cylinder = 175π/3

Row 2

Base radius = base diameter/2 = 6/2 = 3

Base area = π $r^{2}$ = π $(3)^{2}$ = 9π

Volume of cone = 1/3 x base area x height

base area x height = 3 x volume of cone

base area x height = 3 x 40π

Height = 120π/9π = 40/3

Cylinder volume = 3 x cone volume = 3 x 40π = 120π

Row 3

Base area = π $r^{2}$

36π = π $r^{2}$

r2 = $\sqrt{36}$

r = 6

Base diameter = 2 x 6 = 12

Volume of cone = 1/3 x base area x height

48π = 1/3 x 36π x height

Height = 4

Cylinder volume = 3 x 48 = 144π

In conclusion,

ROW 1 : diameter of base = 10, base area = 25π, volume of cylinder = 175π

volume of cone = $\frac{175}{3}$ π

ROW 2 : base radius = 3, base area = 9π, height = 40/3, volume of cylinder = 120π

ROW 3 : base radius = 6, base diameter = 12, height = 4, volume of cylinder = 144π

Learn more about mensuration: brainly.com/question/18541562

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If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .