Can some one please help me figure the angles out? sorry I can't copy and paste the figures...
The 17in x 17in square below is given. We have chosen 4 points M, N, P, and Q one on each of its sides as shown in the figure and joint them to obtain a new four sided figure in red.
Show that the figure in red is also a square (Which means has all four angles 90º and all four sides equal). What is the angle QMN or the angle MQP? Justify your answer Why 90º? Not all the 4 sided figures with equal 4 sides are squares!
How are the points M,N,P and Q chosen?
Are they at the same distance from the adjacent vertex?
yes it is a smaller box inside another box
I am labeling as follows starting at the lower left of the big box and going clockwise:
A M B N C P D Q
AM = AQ
so AMQ is isosceles
<A is right
so
<AMQ is 45
same system to get <BMN = 45
and congruent to AMQ
AB is straight (180)
therefore <NMQ = 180-45-45 = 90
MQ=MN because corresponding sides of congruent triangles
etc
To determine if the figure in red is a square, we need to show that it satisfies the criteria of a square, which is having all four angles equal to 90 degrees and all four sides equal in length.
To find the angle QMN or the angle MQP, we can start by examining the properties of the given square and the chosen points M, N, P, and Q.
Since the square has all its sides equal in length, we know that MQ, QP, PN, and NM are all equal. Let's assume that the length of each side of the square is d units.
Now, let's consider the triangle MQN. We have two equal sides, MQ and QN, and we want to find the angle QMN.
To find the angle QMN, we can use the Law of Cosines, which states that in a triangle with sides a, b, and c and angle C opposite to side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
Applying this formula to triangle MQN, we have:
MN^2 = MQ^2 + QN^2 - 2MQ * QN * cos(QMN)
Since MQ = QN (both are equal to d), we can rewrite the equation as:
MN^2 = d^2 + d^2 - 2d * d * cos(QMN)
Simplifying further:
MN^2 = 2d^2 - 2d^2 * cos(QMN)
MN^2 = 2d^2(1 - cos(QMN))
Now, let's consider the triangle MQP. We have two equal sides again, but this time, we want to find the angle MQP. Using the same process as before, we can write:
MN^2 = d^2 + d^2 - 2d * d * cos(MQP)
MN^2 = 2d^2 - 2d^2 * cos(MQP)
MN^2 = 2d^2(1 - cos(MQP))
Comparing the expressions we derived for triangle MQN and MQP:
2d^2(1 - cos(QMN)) = 2d^2(1 - cos(MQP))
Simplifying:
1 - cos(QMN) = 1 - cos(MQP)
cos(QMN) = cos(MQP)
Since the cosine function is equal for two angles if and only if the angles are congruent, we conclude that angle QMN is congruent to angle MQP.
Since QMN and MQP are both angles formed by lines within a square, they are both internal angles of the square and therefore must each measure 90 degrees.
Therefore, the angle QMN and the angle MQP both measure 90 degrees.
Conclusion:
We have shown that the figure in red, formed by connecting the points M, N, P, and Q, is a square. We determined this by demonstrating that the angles QMN and MQP each measure 90 degrees, which satisfies the requirement for a square where all four angles are 90 degrees.