Choose the best decimal representation for 5 3/7
2 answers:
You might be interested in
Answer:
The area of the small square is 1 cm^2
Step-by-step explanation:
The large square consist in four identical rectangles and one small square.
Then the area of the small square will be equal to the difference between the area of the large square and the areas of the rectangles.
Because we have 4 equal rectangles, if R is the area of one rectangle, and S is the area of the large square, the area of the small square will be:
area = S - 4*R
We know that the area of the large square is 49 cm^2
Then:
S = 49cm^2
Remember that the area of a square of side length K is:
A = K^2
Then the side length of the large square is:
K^2 = 49 cm^2
K = √(49 cm^2) = 7cm
And we know that the diagonal of one rectangle is 5cm.
Remember that for a rectangle of length L and width W, the diagonal is:
D = √(L^2 + W^2)
Then:
D = √(L^2 + W^2) = 5cm
And for how we construct this figure, we must have that the length of the rectangle plus the width of the rectangle is equal to the side length of the large square, then:
L + W = 7cm
L = (7cm - W)
Replacing this in the diagonal equation, we get:
√((7cm - W)^2 + W^2) = 5cm
(7cm - W)^2 + W^2 = (5cm)^2 = 25cm^2
49cm^2 - 14cm*W + W^2 + W^2 = 25cm^2
2*W^2 - 14cm*W + 49cm^2 = 25cm^2
2*W^2 - 14cm*W + 49cm^2 - 25cm^2 = 0
2*W^2 - 14cm*W + 24cm^2 = 0
We can solve this for W using the Bhaskara's formula, the solutions are:
Then we have two solutions, and we only need one (because the length will have the other value)
We can take:
W = (14 cm + 2cm)/4 = 4cm
Then using the equation:
L + W = 7cm
L + 4cm = 7cm
L = 7cm - 4cm = 3cm
L = 3cm
Now remember that the area of one rectangle of length L and width W is:
R = L*W
Then the area of one of these rectangles is:
R = 4cm*3cm = 12cm^2
Now we can compute the area of the small square:
area = S - 4*R = 49cm^2 - 4*12cm^2 = 1cm^2
The area of the small square is 1 cm^2
Step 1
Find the perimeter of rectangle ABDE
we know that
the perimeter of rectangle is equal to
In this problem
substitute
Step 2
Find the perimeter of triangle BCD
we know that
the perimeter of triangle is equal to
In this problem we have
Applying the Pythagoras theorem
substitute
Find the ratio of the perimeter of rectangle ABDE to the perimeter of triangle BCD
we have
the perimeter of rectangle is equal to
the perimeter of the triangle is
so
the ratio is equal to
therefore
<u>the answer Part 1) is the option B</u>
Step 3
Find the area of polygon ABCDE
we know that
The area of polygon is equal to the sum of the area of rectangle plus the area of triangle
Area of rectangle is equal to
Area of the triangle is equal to
the height h of the triangle is equal to
substitute
The area of polygon is
therefore
<u>the answer part 2) is the option C</u>