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yan [13]
3 years ago
5

Do you know how to do this question

Mathematics
1 answer:
Sever21 [200]3 years ago
8 0

Percentage of the area shaded = 55%

Solution:

Perimeter of the square = 40 cm

Each side of the square = \frac{40}{4} = 10 cm

Area of the square = \text{side}^2

                               = 10²

                               = 100

Total area of the square = 100 cm²

The unshaded portion is two triangles.

Base of the smaller triangle = 10 cm – 7 cm = 3 cm

Height of the smaller triangle = 10 cm

Area of the smaller triangle = \frac{1}{2}\times \text{base}\times \text{height}

                                              $=\frac{1}{2}\times 3\times 10

Area of the smaller triangle = 15 cm²

Base of the larger triangle = 10 cm – 4 cm = 6 cm

Height of the larger triangle = 10 cm

Area of the larger triangle = \frac{1}{2}\times \text{base}\times \text{height}

                                              $=\frac{1}{2}\times 6\times 10

Area of the larger triangle = 30 cm²

Area of the shaded portion = Area of the square – Area of the smaller

                                                   triangle – Area of the larger triangle

                                              = 100 cm² – 15 cm² – 30 cm²

Area of the shaded portion = 55 cm²

Percentage of the area shaded = \frac{\text{Area of the shaded portion}}{\text{Total area of the square}}\times100\%

                                                    $=\frac{55}{100}\times100\%

Percentage of the area shaded = 55%

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scZoUnD [109]

In order to check if the lines are perpendicular, we need to check if their slopes have the following relation (to find the slope we can use the slope-intercept form y = mx + b):

m_1=-\frac{1}{m_2}

A.

In this option, y = -5 is an horizontal line and x = 2 is a vertical line, therefore they are perpendicular.

B.

First let's find the slope of each line:

\begin{gathered} x+\frac{y}{5}=2 \\ 5x+y=10 \\ y=-5x+10\to m=-5 \\  \\ -\frac{x}{5}+y=3 \\ y=\frac{x}{5}+3\to m=\frac{1}{5} \end{gathered}

These slopes obey the relation stated above, so the lines are perpendicular.

C.

\begin{gathered} y=\frac{1}{3}x+1\to m=\frac{1}{3} \\  \\ y-1=-3(x-5) \\ y-1=-3x+15 \\ y=-3x+16\to m=-3 \end{gathered}

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D.

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These slopes obey the relation stated above, so the lines are perpendicular.

E.

\begin{gathered} y+2=\frac{1}{3}(x-6) \\ y+2=\frac{1}{3}x-2 \\ y=\frac{1}{3}x-4\to m=\frac{1}{3} \\  \\ y=3x+4\to m=3 \end{gathered}

These slopes don't obey the relation stated above, so the lines aren't perpendicular.

The correct options are A, B, C and D.

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