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3 years ago

A cuboid measures 60cm by 40cm by 20cm.

Mathematics

1 answer:

riadik2000 [5.3K]3 years ago

8 0

Answer:

a) The area of the largest face = 2400 cm², The area of the largest face = 0.24 m²

b) The volume of the cuboid = 48000 cm³, The volume of the cuboid = 0.048 m³

Step-by-step explanation:

The volume of any cuboid = L × W × H, where

L is its length

W is its width

H is its height

∵ A cuboid measures 60 cm by 40 cm by 20 cm.

→ Assume that L = 60 cm, W = 40 cm, H = 20 cm

∵ The largest face is the face who has the largest dimensions

∵ L and W are the largest dimensions

∴ The area of the largest face = L × W

∴ The area of the largest face = 60 × 40

∴ The area of the largest face = 2400 cm²

∵ 1 cm = 1/100 m

∴ 1 cm² = (1/100)² m² = 1/10000 m²

∴ The area of the largest face = 2400 × 1/10000

∴ The area of the largest face = 0.24 m²

∵ The volume of the cuboid = L × W × H

∴ The volume of the cuboid = 60 × 40 × 20

∴ The volume of the cuboid = 48000 cm³

∵ 1 cm = 1/100 m

∴ 1 cm3 = (1/100)³ m³ = 1/1000000 m³

∴ The volume of the cuboid = 48000 × 1/1000000

∴ The volume of the cuboid = 0.048 m³

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2 years ago

the height h(t) of a trianle is increasing at 2.5 cm/min, while it's area A(t) is also increasing at 4.7 cm2/min. at what rate i

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Answer:

The base of the triangle decreases at a rate of 2.262 centimeters per minute.

Step-by-step explanation:

From Geometry we understand that area of triangle is determined by the following expression:

$A = \frac{1}{2}\cdot b\cdot h$ (Eq. 1)

Where:

$A$ - Area of the triangle, measured in square centimeters.

$b$ - Base of the triangle, measured in centimeters.

$h$ - Height of the triangle, measured in centimeters.

By Differential Calculus we deduce an expression for the rate of change of the area in time:

$\frac{dA}{dt} = \frac{1}{2}\cdot \frac{db}{dt}\cdot h + \frac{1}{2}\cdot b \cdot \frac{dh}{dt}$ (Eq. 2)

Where:

$\frac{dA}{dt}$ - Rate of change of area in time, measured in square centimeters per minute.

$\frac{db}{dt}$ - Rate of change of base in time, measured in centimeters per minute.

$\frac{dh}{dt}$ - Rate of change of height in time, measured in centimeters per minute.

Now we clear the rate of change of base in time within (Eq, 2):

$\frac{1}{2}\cdot\frac{db}{dt}\cdot h = \frac{dA}{dt}-\frac{1}{2}\cdot b\cdot \frac{dh}{dt}$

$\frac{db}{dt} = \frac{2}{h}\cdot \frac{dA}{dt} -\frac{b}{h}\cdot \frac{dh}{dt}$ (Eq. 3)

The base of the triangle can be found clearing respective variable within (Eq. 1):

$b = \frac{2\cdot A}{h}$

If we know that $A = 130\,cm^{2}$ , $h = 15\,cm$ , $\frac{dh}{dt} = 2.5\,\frac{cm}{min}$ and $\frac{dA}{dt} = 4.7\,\frac{cm^{2}}{min}$ , the rate of change of the base of the triangle in time is:

$b = \frac{2\cdot (130\,cm^{2})}{15\,cm}$

$b = 17.333\,cm$

$\frac{db}{dt} = \left(\frac{2}{15\,cm}\right)\cdot \left(4.7\,\frac{cm^{2}}{min} \right) -\left(\frac{17.333\,cm}{15\,cm} \right)\cdot \left(2.5\,\frac{cm}{min} \right)$