<h2>
Answer:</h2>
The ratio of the area of region R to the area of region S is:
![\dfrac{24}{25}](https://tex.z-dn.net/?f=%5Cdfrac%7B24%7D%7B25%7D)
<h2>
Step-by-step explanation:</h2>
The sides of R are in the ratio : 2:3
Let the length of R be: 2x
and the width of R be: 3x
i.e. The perimeter of R is given by:
![Perimeter\ of\ R=2(2x+3x)](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D2%282x%2B3x%29)
( Since, the perimeter of a rectangle with length L and breadth or width B is given by:
)
Hence, we get:
![Perimeter\ of\ R=2(5x)](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D2%285x%29)
i.e.
![Perimeter\ of\ R=10x](https://tex.z-dn.net/?f=Perimeter%5C%20of%5C%20R%3D10x)
Also, let " s " denote the side of the square region.
We know that the perimeter of a square with side " s " is given by:
![\text{Perimeter\ of\ square}=4s](https://tex.z-dn.net/?f=%5Ctext%7BPerimeter%5C%20of%5C%20square%7D%3D4s)
Now, it is given that:
The perimeters of square region S and rectangular region R are equal.
i.e.
![4s=10x\\\\i.e.\\\\s=\dfrac{10x}{4}\\\\s=\dfrac{5x}{2}](https://tex.z-dn.net/?f=4s%3D10x%5C%5C%5C%5Ci.e.%5C%5C%5C%5Cs%3D%5Cdfrac%7B10x%7D%7B4%7D%5C%5C%5C%5Cs%3D%5Cdfrac%7B5x%7D%7B2%7D)
Now, we know that the area of a square is given by:
![\text{Area\ of\ square}=s^2](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20square%7D%3Ds%5E2)
and
![\text{Area\ of\ Rectangle}=L\times B](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3DL%5Ctimes%20B)
Hence, we get:
![\text{Area\ of\ square}=(\dfrac{5x}{2})^2=\dfrac{25x^2}{4}](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20square%7D%3D%28%5Cdfrac%7B5x%7D%7B2%7D%29%5E2%3D%5Cdfrac%7B25x%5E2%7D%7B4%7D)
and
![\text{Area\ of\ Rectangle}=2x\times 3x](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3D2x%5Ctimes%203x)
i.e.
![\text{Area\ of\ Rectangle}=6x^2](https://tex.z-dn.net/?f=%5Ctext%7BArea%5C%20of%5C%20Rectangle%7D%3D6x%5E2)
Hence,
Ratio of the area of region R to the area of region S is:
![=\dfrac{6x^2}{\dfrac{25x^2}{4}}\\\\=\dfrac{6x^2\times 4}{25x^2}\\\\=\dfrac{24}{25}](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B6x%5E2%7D%7B%5Cdfrac%7B25x%5E2%7D%7B4%7D%7D%5C%5C%5C%5C%3D%5Cdfrac%7B6x%5E2%5Ctimes%204%7D%7B25x%5E2%7D%5C%5C%5C%5C%3D%5Cdfrac%7B24%7D%7B25%7D)
Given that ∠B ≅ ∠C.
to prove that the sides AB = AC
This can be done by the method of contradiction.
If possible let AB
=AC
Then either AB>AC or AB<AC
Case i: If AB>AC, then by triangle axiom, Angle C > angle B.
But since angle C = angle B, we get AB cannot be greater than AC
Case ii: If AB<AC, then by triangle axiom, Angle C < angle B.
But since angle C = angle B, we get AB cannot be less than AC
Conclusion:
Since AB cannot be greater than AC nor less than AC, we have only one possibility. that is AB =AC
Hence if angle B = angle C it follows that
AB = AC, and AB ≅ AC.
Answer:
K would equal to:
K= 50
Hope that helps! :)
Step-by-step explanation:
Answer:
The dilation on any point of the rectangle is
.
Step-by-step explanation:
From Linear Algebra, we define the dilation of a point by means of the following definition:
(1)
Where:
- Coordinates of the point G, dimensionless.
- Center of dilation, dimensionless.
- Scale factor, dimensionless.
- Coordinates of the point G', dimensionless.
If we know that
,
and
, then scale factor is:
![(5,-5) = (0,0) +k\cdot [(2,-2)-(0,0)]](https://tex.z-dn.net/?f=%285%2C-5%29%20%3D%20%280%2C0%29%20%2Bk%5Ccdot%20%5B%282%2C-2%29-%280%2C0%29%5D)
![(5,-5) = (2\cdot k, -2\cdot k)](https://tex.z-dn.net/?f=%285%2C-5%29%20%3D%20%282%5Ccdot%20k%2C%20-2%5Ccdot%20k%29)
![k = \frac{5}{2}](https://tex.z-dn.net/?f=k%20%3D%20%5Cfrac%7B5%7D%7B2%7D)
The dilation on any point of the rectangle is:
![P'(x,y) = (0,0) + \frac{5}{2}\cdot [P(x,y)-(0,0)]](https://tex.z-dn.net/?f=P%27%28x%2Cy%29%20%3D%20%280%2C0%29%20%2B%20%5Cfrac%7B5%7D%7B2%7D%5Ccdot%20%5BP%28x%2Cy%29-%280%2C0%29%5D)
(2)
The dilation on any point of the rectangle is
.