All categories

1 year ago

What is the area in square meters of 6 triangles. how many triangles are needed to compose a region that is 1 1\2 square meters?

Mathematics

1 answer:

Delicious77 [7]1 year ago

8 0

To create an area that is 1.5 m^2 in size, you will need to make use of a total of 27 triangles.

<h3>How many triangles are needed to compose a region that is 1.5 square meters?</h3>

A square that has a size of one square meter is divided into nine smaller squares that are similar to each other. Each of the little squares is divided into two triangles that are similar to one another.

There are nine smaller squares contained inside one square meter, since 1 square meter may be broken down into nine identical smaller squares. Each of the little squares is divided into two triangles that are similar to one another. 9 smaller squares may be broken down into the following:

9*2=18 (shows identical triangles)

Hence, 1 square meter is decomposed into 18 identical triangles.

We need to find the number of triangles needed to compose a region that is square meters

where

m^2 = 1.5 m^2

Where

1 m^2 = 18 identical triangles.

1.5 m^2 = 1.5 * 18

1.5 m^2 = 27 identical triangles.

In conclusion, To create an area that is 1.5 m^2 in size, you will need to make use of a total of 27 triangles.

You might be interested in

Can someone make sure that I’m right

WARRIOR [948]

Answer:

Step-by-step explanation: You are correct!

5 0

3 years ago

Read 2 more answers

What is 20 times 20plz tell me

vladimir1956 [14]

Answer:

400

Step-by-step explanation:

8 0

3 years ago

Fabian is purchasing a home for $89,500 and is financing 75% of it. The documentary stamp tax on the deed in his state is $0.70

ss7ja [257]

Answer: C) $462.96
Hope this will helpful.
Thank you.

6 0

3 years ago

A hallow metal pipe has an outer diameter of 3 inches and a thickness of 1/4 inch. The length of the pipe is 12 inches. Compute

saveliy_v [14]

Answer:

Step-by-step explanation:

I don"t really know the answer but i might have and explanation to help. Hollow everyday circular objects appear differently than drawn two-dimensional circles. Objects like pipes and hoses have two different diameters. The outside diameter measures the distance of a straight line from one point on the outside of the object, through its center, and to an opposite point on the outside. The internal diameter measures the inside of the object. Calculating the internal diameter depends on the outside diameter and the thickness of the outer circle.

3 0

3 years ago

use the general slicing method to find the volume of The solid whose base is the triangle with vertices (0 comma 0 ), (15 comma

lyudmila [28]

Answer:

volume V of the solid

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

Step-by-step explanation:

The situation is depicted in the picture attached

(see picture)

First, we divide the segment [0, 5] on the X-axis into n equal parts of length 5/n each

[0, 5/n], [5/n, 2(5/n)], [2(5/n), 3(5/n)],..., [(n-1)(5/n), 5]

Now, we slice our solid into n slices.

Each slice is a quarter of cylinder 5/n thick and has a radius of

-k(5/n) + 5 for each k = 1,2,..., n (see picture)

So the volume of each slice is

$\displaystyle\frac{\pi(-k(5/n) + 5 )^2*(5/n)}{4}$

for k=1,2,..., n

We then add up the volumes of all these slices

$\displaystyle\frac{\pi(-(5/n) + 5 )^2*(5/n)}{4}+\displaystyle\frac{\pi(-2(5/n) + 5 )^2*(5/n)}{4}+...+\displaystyle\frac{\pi(-n(5/n) + 5 )^2*(5/n)}{4}$

Notice that the last term of the sum vanishes. After making up the expression a little, we get

$\displaystyle\frac{5\pi}{4n}\left[(-(5/n)+5)^2+(-2(5/n)+5)^2+...+(-(n-1)(5/n)+5)^2\right]=\\\\\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2$

But

$\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}(-k(5/n)+5)^2=\displaystyle\frac{5\pi}{4n}\displaystyle\sum_{k=1}^{n-1}((5/n)^2k^2-(50/n)k+25)=\\\\\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)$

we also know that

$\displaystyle\sum_{k=1}^{n-1}k^2=\displaystyle\frac{n(n-1)(2n-1)}{6}$

and

$\displaystyle\sum_{k=1}^{n-1}k=\displaystyle\frac{n(n-1)}{2}$

so we have, after replacing and simplifying, the sum of the slices equals

$\displaystyle\frac{5\pi}{4n}\left((5/n)^2\displaystyle\sum_{k=1}^{n-1}k^2-(50/n)\displaystyle\sum_{k=1}^{n-1}k+25(n-1)\right)=\\\\=\displaystyle\frac{5\pi}{4n}\left(\displaystyle\frac{25}{n^2}.\displaystyle\frac{n(n-1)(2n-1)}{6}-\displaystyle\frac{50}{n}.\displaystyle\frac{n(n-1)}{2}+25(n-1)\right)=\\\\=\displaystyle\frac{125\pi}{24}.\displaystyle\frac{n(n-1)(2n-1)}{n^3}$

Now we take the limit when n tends to infinite (the slices get thinner and thinner)

$\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}\displaystyle\frac{n(n-1)(2n-1)}{n^3}=\displaystyle\frac{125\pi}{24}\displaystyle\lim_{n \rightarrow \infty}(2-3/n+1/n^2)=\\\\=\displaystyle\frac{125\pi}{24}.2=\displaystyle\frac{125\pi}{12}$

and the volume V of our solid is

$\boxed{V=\displaystyle\frac{125\pi}{12}}$

3 0

3 years ago