Find the area of the rectangle or square 8yd 8yd

Mathematics

1 answer:

miv72 [106K]3 years ago

3 0

The area of the square is 64yds :) 8 x 8 = 64! Please mark as brainliest!! :)

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The base of a solid right pyramid is a regular hexagon with a radius of 2x units and an apothem of units. A solid right pyramid

Xelga [282]

Answer:

$6 x^2*\sqrt{3}$

Step-by-step explanation:

The base of the pyramid we need to study is a hexagon.

Let's look at the attached image of an hexagon to understand how we are going to find the area of this figure.

Notice that an hexagon is the combination of 6 exactly equal equilateral triangles in our case of size "2x" (notice that the "radius" of the hexagon is given as "2x")

Therefore the area of the hexagon is going to be 6 times the area of one of those equilateral triangles.

We know the area of a triangle is the product of its base times its height, divided by 2: $\frac{base*height}{2} = \frac{2x*height}{2}$

We notice that the triangle's height is exactly what is called the "apothem" of the hexagon (depicted in green in our figure) which measures $x\sqrt{3}$ , so replacing this value in the formula above for the area of one of the triangles:

$\frac{2x*height}{2}= \frac{2x*x\sqrt{3} }{2}=x^2\sqrt{3}$

Then we multiply this area times 6 to get the total area of the hexagon (6 of these triangles):

Area of hexagon = $6x^2\sqrt{3}$

which is the last option given in the list.

5 0

3 years ago

Read 2 more answers

is this quadrilateral a parallelogram?If yes, state the definition or theorem that justifies your answer.

Xelga [282]

Answer:

Yes.

Step-by-step explanation:

Yes. Because opposite angles are congruent.

4 0

3 years ago

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The altitude of a triangle is increasing at a rate of 1 cm/ min while the area of the triangle is increasing at a rate of 2 cm2

serg [7]

Ya, calculus and related rates, such fun!

everything is changing with respect to t

altitude rate will be dh/dt and that is 1cm/min
dh/dt=1cm/min

area will be da/dt which is increasing at 2cm²/min
da/dt=2cm²/min

base=db/dt

alright

area=1/2bh
take dervitivie of both sides
da/dt=1/2((db/dt)(h)+(dh/dt)(b))
solve for db/dt
distribute
da/dt=1/2(db/dt)(h)+1/2(dh/dt)(b)
move
da/dt-1/2(dh/dt)(b)=1/2(db/dt)(h)
times 2 both sides
2da/dt-(dh/dt)(b)=(db/dt)(h)
divide by h
(2da/dt-(dh/dt)(b))/h=db/dt

ok
we know
height=10
area=100
so
a=1/2bh
100=1/2b10
100=5b
20=b

so
h=10
b=20
da/dt=2cm²/min
dh/dt=1cm/min

therefor
(2(2cm²/min)-(1cm/min)(20cm))/10cm=db/dt
(4cm²/min-20cm²/min)/10cm=db/dt
(-16cm²/min)/10cm=db/dt
-1.6cm/min=db/dt

the base is decreasing at 1.6cm/min

4 0

3 years ago

The height of the tunnel at the center is 35ft, and the vertical clearance must be 21 ft at a point of 8ft from the center. Find

Zinaida [17]

Answer:

y=-0.215x^2+35

Step by Step:

Let, $h=0$ , $k=35$ , $x=8$ , $y=21$