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tiny-mole [99]
3 years ago
10

A quadrilateral has vertices at $(0,1)$, $(3,4)$, $(4,3)$ and $(3,0)$. Its perimeter can be expressed in the form $a\sqrt2+b\sqr

t{10}$ with $a$ and $b$ integers. What is the sum of $a$ and $b$?
Mathematics
1 answer:
seraphim [82]3 years ago
3 0

Answer:

a + b = 12

Step-by-step explanation:

Given

Quadrilateral;

Vertices of (0,1), (3,4) (4,3) and (3,0)

Perimeter = a\sqrt{2} + b\sqrt{10}

Required

a + b

Let the vertices be represented with A,B,C,D such as

A = (0,1); B = (3,4); C = (4,3) and D = (3,0)

To calculate the actual perimeter, we need to first calculate the distance between the points;

Such that:

AB represents distance between point A and B

BC represents distance between point B and C

CD represents distance between point C and D

DA represents distance between point D and A

Calculating AB

Here, we consider A = (0,1); B = (3,4);

Distance is calculated as;

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

(x_1,y_1) = A(0,1)

(x_2,y_2) = B(3,4)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

AB = \sqrt{(0 - 3)^2 + (1 - 4)^2}

AB = \sqrt{( - 3)^2 + (-3)^2}

AB = \sqrt{9+ 9}

AB = \sqrt{18}

AB = \sqrt{9*2}

AB = \sqrt{9}*\sqrt{2}

AB = 3\sqrt{2}

Calculating BC

Here, we consider B = (3,4); C = (4,3)

Here,

(x_1,y_1) = B (3,4)

(x_2,y_2) = C(4,3)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

BC = \sqrt{(3 - 4)^2 + (4 - 3)^2}

BC = \sqrt{(-1)^2 + (1)^2}

BC = \sqrt{1 + 1}

BC = \sqrt{2}

Calculating CD

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = C(4,3)

(x_2,y_2) = D (3,0)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

CD = \sqrt{(4 - 3)^2 + (3 - 0)^2}

CD = \sqrt{(1)^2 + (3)^2}

CD = \sqrt{1 + 9}

CD = \sqrt{10}

Lastly;

Calculating DA

Here, we consider C = (4,3); D = (3,0)

Here,

(x_1,y_1) = D (3,0)

(x_2,y_2) = A (0,1)

Substitute these values in the formula above

Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

DA = \sqrt{(3 - 0)^2 + (0 - 1)^2}

DA = \sqrt{(3)^2 + (- 1)^2}

DA = \sqrt{9 +  1}

DA = \sqrt{10}

The addition of the values of distances AB, BC, CD and DA gives the perimeter of the quadrilateral

Perimeter = 3\sqrt{2} + \sqrt{2} + \sqrt{10} + \sqrt{10}

Perimeter = 4\sqrt{2} + 2\sqrt{10}

Recall that

Perimeter = a\sqrt{2} + b\sqrt{10}

This implies that

a\sqrt{2} + b\sqrt{10} = 4\sqrt{2} + 2\sqrt{10}

By comparison

a\sqrt{2} = 4\sqrt{2}

Divide both sides by \sqrt{2}

a = 4

By comparison

b\sqrt{10} = 2\sqrt{10}

Divide both sides by \sqrt{10}

b = 2

Hence,

a + b = 2 + 10

a + b = 12

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3 0
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How do I do this problem?
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so a quadrilateral is a figure with 4 sides. So looking at the picture I see 5 rhombi and 7 squares. So all together thats 12 quadrilaterals  

So to find the ration you have to do:

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so in this case it would be

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which you can simplify

normally you'd simplify to 1:2 but that doesn't seem like an option so I'd go with the one closest to that ratio


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