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3 years ago

Hello :) Please help me solve this Explain answer

Mathematics

1 answer:

USPshnik [31]3 years ago

5 0

Answer:

$Scale\ Factor = \frac{1}{2}$

ABC dilated

Step-by-step explanation:

Given

Triangles ABC and DEF

Required

Determine the scale of dilation

First, we pick a side on ABC

$AB = 6$

Pick a corresponding side on DEF

$DE = 3$

The scale factor is then calculated as:

$Scale\ Factor = \frac{DE}{AB}$

$Scale\ Factor = \frac{1}{2}$

i.e. ABC was dilated by 1/2

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What is the area of a parallelogram if the coordinates of its vertices are (0, -2), (3, 2), (8,2), and (5,-2)?

Llana [10]

Answer:

The area of the parallelogram is 20 units²

Step-by-step explanation:

* Lets explain how to solve the problem

- The vertices of the parallelogram are (0 , -2) , (3 , 2) , (8 , 2) , (5 , -2)

- The side joining the points (0 , -2) and (5 , -2) is a horizontal side

because the points have same y-coordinates

- The side joining the points (3 , 2) and (8 , 2) is a horizontal side

because the points have same y-coordinates

∴ These two sides are parallel bases of the parallelogram

∵ The length of any horizontal side = x2 - x1

∴ The length of the side = 5 - 0 = 5 or 8 - 3 = 5

∴ The length of one base of the parallelogram is 5 units

- The height of this base is the vertical distance between these two

parallel bases

∵ The length of any vertical distance = y2 - y1

∵ y2 = 2 and y1 = -2 ⇒ the y-coordinates of the parallel bases

∴ The length of the vertical distance = 2 - (-2) = 2 + 2 = 4 units

∵ This vertical distance between the two parallel bases is the height

of these bases

∴ The height of the parallelogram is 4 units

- The area of the parallelogram = base × height

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∴ The area = 5 × 4 = 20 units²

* The area of the parallelogram is 20 units²

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3 years ago

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,

So width of pool B = (2x +6 ) – 4 = 2x + 2

Area of rectangle = length x width

So area of pool B =(2x+6)(2x+2) ------(2)

Since area of pool A is equal to area of pool B, so from equation (1) and (2)

(x+6) (x+3) =(2x+6) (2x+2)

On solving above equation for x

(x+6) (x+3) =2(x+3) (2x+2)

x+6 = 4x + 4

x-4x = 4 – 6

x = $\frac{2}{3}$

Dimension of pool A

Length = x+6 = ( $\frac{2}{3}$ ) +6 = 6.667m

Width = x +3 = ( $\frac{2}{3}$ ) +3 = 3.667m

Dimension of pool B

Length = 2x +6 = 2( $\frac{2}{3}$ ) + 6 = $\frac{22}{3}$ = 7.333m

Width = 2x + 2 = 2( $\frac{2}{3}$ ) + 2 = $\frac{10}{3}$ = 3.333m

Verifying the area:

Area of pool A = ( $\frac{20}{3}$ ) x ( $\frac{11}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Area of pool B = ( $\frac{22}{3}$ ) x ( $\frac{10}{3}$ ) = $\frac{220}{9}$ = 24.44 meter

Summarizing the results:

(a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

(b)Area of pool A is equal to Area of pool B equal to 24.44 meters.

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