Square ABCD has M as the midpoint of AB, N as the midpoint of BC, P as the midpoint of CD and Q as the midpoint of MP. If the area of AMNPDQ is 20, what is the area of ABCD?
make your sketch and join QN
in MBCP, you will see 4 equal congruent triangles,
label each one as x units
so outline AMNPDQ and you will see it is made up
of 4x
so 4x = 20
x = 5
all of ABCD is 8x
8x = (8(5) = 40 units
To find the area of the square ABCD, we need to determine the side length of the square. Let's start by examining the structure of the problem.
We know that M is the midpoint of AB, N is the midpoint of BC, P is the midpoint of CD, and Q is the midpoint of MP. Using this information, we can see that triangle AMN is similar to triangle CPQ. This is because they share the same angles and the ratio of their side lengths is 1:2, which is the same for all corresponding sides of similar triangles.
Since AMN and CPQ are similar triangles, we can deduce that the ratio of their areas is the square of the corresponding side ratio. The side ratio is 1:2, so the area ratio is 1^2:2^2 = 1:4.
Given that the area of AMNPDQ is 20, we can determine that the area of triangle AMN is 20/4 = 5.
Now, let's find the length of MN. Since M is the midpoint of AB and N is the midpoint of BC, we can conclude that MN is parallel to and half the length of AD. Similarly, MN is parallel to and half the length of CD since P is the midpoint of CD and Q is the midpoint of MP.
Therefore, MN is half the length of both AD and CD.
Since opposite sides of a square are equal, AD = CD, so MN = AD/2 = CD/2.
Now, we have the area of AMN (5) and the length of MN (AD/2 = CD/2). To find the area of the square ABCD, we need to determine the length of AD (or CD).
Since triangle AMN is a right triangle (formed by the diagonal MN of the square), we can use the Pythagorean theorem to find the length of AD (or CD) if we know the length of MN.
The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Therefore, in triangle AMN, we have:
MN^2 = AM^2 + AN^2
Since AM = MN/2 and AN = MN/2 (because M and N are midpoints), we can substitute these values into the equation:
MN^2 = (MN/2)^2 + (MN/2)^2
Simplifying further:
MN^2 = (MN^2)/4 + (MN^2)/4
Combining like terms:
MN^2 = (2/4) * MN^2 + (2/4) * MN^2
MN^2 = (4/4) * MN^2
MN^2 = MN^2
This equation is true for any positive value of MN.
Thus, we cannot find the exact value of MN without additional information or without solving for the lengths of AM and AN from the area of AMN. Therefore, with the given information, we cannot determine the length of AD (or CD) and hence cannot find the area of the square ABCD.