stumped on Math Destroyer! please help

[NguyenDDS]

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I'm restudying for my DAT 🙂 And I have a question about a problem in Math Destroyer....

The question asks:

"The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square."


Can someone explain to me why the answer is 10? I read the solution in the back of the book, however, it doesn't make sense to me and is more complicated than it needs to be. In the solutions, it uses the distance formula between the two points and then uses a^2+b^2=c^2. Even if I used these formulas to find the solution, it only solves half of the square (one triangle) and thus should be doubled to 20.

Why can't I just solve for the sides of the square in the beginning by subtracting 6-2=4 and 4-2=2? And then multiplying 4x2 to get 8? Maybe I'm over thinking it...... 🙄

Thank you!!

 

[LetsGo2DSchool]

The square is not oriented evenly with the y and x axes. Use the distance formula to find the hypotenuse of the right triangle that the diagonal creates (= sqrt 20). From here you can see that it's a 45-45-90 right triangle meaning the length of one side of the triangle (sqrt 20/sqrt 2) is also equal to the length of one side of the square. Thus area will equal (sqrt 20/sqrt 2)^2 = 10.

 

[DentistDMD]

I'm restudying for my DAT 🙂 And I have a question about a problem in Math Destroyer....

The question asks:

"The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square."


Can someone explain to me why the answer is 10? I read the solution in the back of the book, however, it doesn't make sense to me and is more complicated than it needs to be. In the solutions, it uses the distance formula between the two points and then uses a^2+b^2=c^2. Even if I used these formulas to find the solution, it only solves half of the square (one triangle) and thus should be doubled to 20.

Why can't I just solve for the sides of the square in the beginning by subtracting 6-2=4 and 4-2=2? And then multiplying 4x2 to get 8? Maybe I'm over thinking it...... 🙄

Thank you!!

You wouldn't necessarily get the dimensions of a square if you calculate it that way for this type of probelm. Destroyer's solution ensures that you are getting the correct dimensions (sides of isosceles right triangle, which is half of a square.)

1) Since only right triangle satisfies a^2 + b^2 = c^2, you first find the distance
between the two points and substitute it for c.

2) Since the figure has to be a square, a and b must be equivalent in length. 2a^2 = 20
a^2 = 10
Since the area of a square is one side squared, 10 is the answer.

 

[es1113]

I'm restudying for my DAT 🙂 And I have a question about a problem in Math Destroyer....

The question asks:

"The endpoints of the diagonal of a square are located at (2,4) and (6,2). Find the area of the square."


Can someone explain to me why the answer is 10? I read the solution in the back of the book, however, it doesn't make sense to me and is more complicated than it needs to be. In the solutions, it uses the distance formula between the two points and then uses a^2+b^2=c^2. Even if I used these formulas to find the solution, it only solves half of the square (one triangle) and thus should be doubled to 20.

Why can't I just solve for the sides of the square in the beginning by subtracting 6-2=4 and 4-2=2? And then multiplying 4x2 to get 8? Maybe I'm over thinking it...... 🙄

Thank you!!

for an isoceles right triangle, the hypotenus is x(2^.5) and the two other sides are both X. The diagonal in this question would be route20 which is also equal to (2^.5)(10^.5) so you just divide by 2^.5 to get the value for the sides of teh square and then you square that and you get 10

 

[es1113]

dang im slow

 

[NguyenDDS]

Thank you!! It makes more sense now. I keep thinking the square is in line with x and y... 🙂

 

[dentist911]

just to make things go faster on the exam day just remember...

diagonal of a square = x √2
diagonal of a cube = x √3

where x = the side of the square/cube

 

[Tallon]

just to make things go faster on the exam day just remember...

diagonal of a square = x √2
diagonal of a cube = x √3

where x = the side of the square/cube

What?