All categories

3 years ago

If the area of a square with side x is equal to the area of a triangle with base x, then find the altitude of the triangle.

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

3 0

Answer:

$h= 2x$

Step-by-step explanation:

Given

Square

$Side = x$

Triangle

$Base = x$

$Altitude = h$

Required

Find h, if both shapes have the same area

The area of the square is:

$Area = x * x$

$Area = x^2$

The area of the triangle is:

$Area = \frac{1}{2} * Base*Height$

$Area = \frac{1}{2} * x*h$

Equate both areas

$x^2 = \frac{1}{2} * x*h$

Divide both sides by x

$x = \frac{1}{2} *h$

Multiply both sides by 2

$2*x = \frac{1}{2} *h*2$

$2x = h$

$h= 2x$

You might be interested in

I do not know how to do this but I need a answer cuz u don’t know it

Law Incorporation [45]

You answer would be 16.2.

Hope this helps,
♥Nikki♥

3 0

3 years ago

Please help me with problem 22a I dont understand how to find the rate of change

Anna [14]

The rate of change is x2. So, since it's forty, it became 80 because 40x2=80!
Good luck for a good grade!

5 0

3 years ago

What is the image of (−3,−9) after a dilation by a scale factor of 1/3 centered at the origin.

denis-greek [22]

Answer:

(-1,-3)

Step-by-step explanation:

Multiply both x and y values by 1/3 or divide both by 3.

4 0

3 years ago

Given triangle TVU list the sides of the shortest to longest

Margaret [11]

Answer:

VU, VT, TU

Step-by-step explanation:

List the side lengths of triangle TUV from shortest to longest.

A. VU, VT, TU

B. VT, TU, VU

C. VT, VU, TU

D. VU, TU, VT"

8 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago