The square below represents one whole.

Mathematics

1 answer:

grandymaker [24]2 years ago

6 0

Answer:

Fraction: 81/100

Decimal: 0.81

Percentage: 81%

Step-by-step explanation:

The question is complete but the diagram mentioned can be found in

brainly.com/question/12188410

First step is to find the area of the whole square and then the area of the part that was shaded.

From the diagram, there are 100 squares in the whole square. 81 of those squares are shaded.

Area of the whole square = 10 by 10

= 10×10 = 100unit²

Area of shaded region = (10 by 8) + 1

= 80+1 = 81unit²

shaded area expressed as a fraction = 81unit²/100unit²

= 81/100

Shaded area expressed as a decimal = 81/100

You take the number 81 and carry the point by two decimal place from the right

= 0.81

Shaded area expressed as a percent of the whole = (81/100) × 100

= (81×100)/100

= 81%

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marta [7]

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2 years ago

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2. Given a quadrilateral with vertices (−1, 3), (1, 5), (5, 1), and (3,−1):

zlopas [31]

<h2>Explanation:</h2>

In every rectangle, the two diagonals have the same length. If a quadrilateral's diagonals have the same length, that doesn't mean it has to be a rectangle, but if a parallelogram's diagonals have the same length, then it's definitely a rectangle.

So first of all, let's prove this is a parallelogram. The basic definition of a parallelogram is that it is a quadrilateral where both pairs of opposite sides are parallel.

So let's name the vertices as:

$A(-1,3) \\ \\ B(1,5) \\ \\ C(5,1) \\ \\ D(3,-1)$

First pair of opposite sides:

<u>Slope:</u>

$\text{For AB}: \\ \\ m=\frac{5-3}{1-(-1)}=1 \\ \\ \\ \text{For CD}: \\ \\ m=\frac{1-(-1)}{5-3}=1 \\ \\ \\ \text{So AB and CD are parallel}$

Second pair of opposite sides:

<u>Slope:</u>

$\text{For BC}: \\ \\ m=\frac{1-5}{5-1}=-1 \\ \\ \\ \text{For AD}: \\ \\ m=\frac{-1-3}{3-(-1)}=-1 \\ \\ \\ \text{So BC and AD are parallel}$

So in fact this is a parallelogram. The other thing we need to prove is that the diagonals measure the same. Using distance formula:

$d=\sqrt{(y_{2}-y_{1})^2+(x_{2}-x_{1})^2} \\ \\ \\ Diagonal \ BD: \\ \\ d=\sqrt{(5-(-1))^2+(1-3)^2}=2\sqrt{10} \\ \\ \\ Diagonal \ AC: \\ \\ d=\sqrt{(3-1)^2+(-5-1)^2}=2\sqrt{10} \\ \\ \\$