Can someone help me please :)

Help! This is due today and I'm stuck on this problem.

Varvara68 [4.7K]

Answer:

280.6 $m^2$

Step-by-step explanation:

find the length of each side:

you're only given h = 9 m

the formula for h is $h=\frac{\sqrt{3} }{2} *a$ , where a is a side length of the hexagon

plug h in:

$9=\frac{\sqrt{3}}{2}*a\\\\a=9*\frac{2}{\sqrt{3}}\\\\$

a ≈ 10.392 m

now find the area of a triangle:

1/2 (10.392)(9) = 46.764 $m^2$

multiply that by 6 and you get 280.584 ≈ 280.6 $m^2$

you can also use the regular hexagon area formula instead of finding the individual area and multiplying by 6: $\frac{3\sqrt{3} }{2} *a^2$

7 0

3 years ago

Read 2 more answers

A car depreciates in value by 17% per year.

otez555 [7]

Answer:

17%of 23000

17/100×23000

1 yrs depreciation is 3910

then

after 4 years depreciation is 4×3910

=15640

the worth of after 4 yrs is 23000-15640

=Rs 7360

7 0

3 years ago

Can anyone please help me with this ?

djverab [1.8K]

Answer:

The distance from point A to the top of the building is 5 $\sqrt{2}$ feet

The Height of the Skyscraper is 5 feet

Step-by-step explanation:

Given

Let Top point of building be point C

Also Let Base of the building be point B

Distance from point A to base B of the building AB= 5 feet

∴ It makes a Right angle triangle

Also ∠ACB = 45°

Also tan 45° = 1

Now tan ∠ACB = $\frac{AB}{BC}$

∴ AB= BC =5 feet

The Height of the Skyscraper is 5 feet

Now Triangle ABC is right angle triangle with right angled at B

So by Pythagoras theorem

AC= $\sqrt{AB^2+BC^2}=\sqrt{5^2+5^2} =\sqrt{50} =\sqrt{25\times2} =\sqrt{5^2\times2}=5\sqrt{2}$

The distance from point A to the top of the building is 5 $\sqrt{2}$ feet

5 0

3 years ago

Read 2 more answers

A bag of numbered lottery balls contains the numbers 1 through 40. What is the probability that a randomly selected ball will be

MA_775_DIABLO [31]

Answer:

3/40

Step-by-step explanation:

So we look for all the multiples of 13 not over 40

There are 3 which are; 13, 26, 39

So it will be 3/40

8 0

4 years ago

Given:

diamong [38]

Answer:

10-5 $\sqrt{2}$

Step-by-step explanation:

As per the attached figure, right angled $\triangle MDL$ has an inscribed circle whose center is $I$ .

We have joined the incenter $I$ to the vertices of the $\triangle MDL$ .

Sides MD and DL are equal because we are given that $\angle M = \angle L = 45 ^\circ.$

Formula for area of a $\triangle = \dfrac{1}{2} \times base \times height$

As per the figure attached, we are given that side a = 10.

Using pythagoras theorem, we can easily calculate that side ML = 10 $\sqrt{2}$

Points P,Q and R are at $90 ^\circ$ on the sides ML, MD and DL respectively so IQ, IR and IP are heights of $\triangle$ MIL, $\triangle$ MID and $\triangle$ DIL.

Also,

$\text {Area of } \triangle MDL = \text {Area of } \triangle MIL +\text {Area of } \triangle MID+ \text {Area of } \triangle DIL$

$\dfrac{1}{2} \times 10 \times 10 = \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10 + \dfrac{1}{2} \times r \times 10\sqrt2\\\Rightarrow r = \dfrac {10}{2+\sqrt2} \\\Rightarrow r = \dfrac{5\sqrt2}{\sqrt2+1}\\\text{Multiplying and divinding by }(\sqrt2 +1)\\\Rightarrow r = 10-5\sqrt2$

So, radius of circle = $10-5\sqrt2$

8 0

3 years ago