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Here are a few recommended readings before getting started.
Try your knowledge on these topics.
The figure below looks like a parallelogram.
In certain situations, it is required to find the length of a line segment or a side of a polygon. To find those lengths, it is recommended to calculate the distance between the endpoints of the segment, or between two vertices of a polygon. If those points are plotted on a coordinate plane, the Distance Formula can be used.
Given two points A(x_1, y_1) and B(x_2, y_2) on a coordinate plane, their distance d is given by the following formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
Start by plotting A(x_1,y_1) and B(x_2,y_2) on the coordinate plane. Both points can be arbitrarily plotted in Quadrant I for simplicity. Note that the position of the points in the plane does not affect the proof. Assume that x_2 is greater than x_1 and that y_2 is greater than y_1.
a= x_2-x_1, b= y_2-y_1
Use the Distance Formula to calculate the distance between the points plotted in the coordinate plane. If needed, round the answer to two decimal places.
Ramsha and Davontay are learning to make picture frames so they want to determine whether the quadrilateral ABCD shown below is a rectangle.
Other than just using the Distance Formula, try the Converse of the Pythagorean Theorem to determine whether the angles of the quadrilateral are right angles.
Substitute ( - 3,0) & ( 4,- 1)
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
---|---|---|---|
Segment | Points | Substitute | Simplify |
AC | A( - 3, 0) & C( 4, - 1) | AC= sqrt(( 4-( - 3))^2+( - 1- 0)^2) | AC= 5sqrt(2) |
AB | A( - 3, 0) & B( - 2, 2) | AB= sqrt(( - 2-( - 3))^2+( 2- 0)^2) | AB= sqrt(5) |
BC | B( - 2, 2) & C( 4, - 1) | BC= sqrt(( 4-( - 2))^2+( - 1- 2)^2) | BC= 3sqrt(5) |
Substitute values
By following the same procedure, it can be shown that ∠ A, ∠ C, and ∠ D are also right angles.
Since each angle of ABCD is a right angle, the quadrilateral is a rectangle. Therefore, Ramsha is correct. She's on her way toward making some nice frames.
Start by finding the radius of the circle.
Substitute ( 0,0) & ( 0,2)
Substitute ( 0,0) & ( 1,sqrt(3))
Sometimes the length of a segment is not what is needed. Instead, the coordinates of the point that lies exactly in the middle of a segment are needed.
If the points are plotted on a coordinate plane, the Midpoint Formula can be used to find the coordinates of the midpoint.
The midpoint M between two points A(x_1, y_1) and B(x_2, y_2) on a coordinate plane can be determined by the following formula.
M(x_1 + x_2/2,y_1 + y_2/2 )
The formula above is called the Midpoint Formula.
LHS+x_m=RHS+x_m
LHS+x_1=RHS+x_1
Commutative Property of Addition
.LHS /2.=.RHS /2.
M(x_1 + x_2/2,y_1 + y_2/2 )
Use the Midpoint Formula to calculate the coordinates of the midpoint M between the points plotted on the coordinate plane.
Paulina has joined the bandwagon of making picture frames. She wants to prove that the diagonals of the parallelogram shown below bisect each other.
See solution.
Use the Midpoint Formula to show that the diagonals intersect at their midpoint.
Substitute ( - 2,2) & ( 3,- 3)
Add terms
a+(- b)=a-b
Put minus sign in front of fraction
a/b=a÷ b
M(x_1+x_2/2,y_1+y_2/2) | |||
---|---|---|---|
Diagonal | Endpoints | Substitute | Simplify |
BD | B( - 2, 2) & D( 3, - 3) | M(- 2+ 3/2,2+( - 3)/2) | M(0.5,- 0.5) ✓ |
AC | A( - 3, 0) & D( 4, - 1) | M(- 3+ 4/2,0+( - 1)/2) | M(0.5,- 0.5) ✓ |
The midpoint of both diagonals is the same as their point of intersection. Therefore, the diagonals bisect each other.
Paulina's best friend is obsessed with triangles. He has requested for Paulina to make him a triangular picture frame. Suppose Paulina can prove the Triangle Midsegment Theorem for the triangle below. She will better understand how to work with triangles and therefore make a better frame!
See solution.
Start by recalling the Triangle Midsegment Theorem.
Triangle Midsegment Theorem |
The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half its length. |
M(x_1+x_2/2,y_1+y_2/2) | |||
---|---|---|---|
Side | Endpoints | Substitute | Simplify |
AB | A( 0, 0) and B( 2, 2.5) | M_1(0+ 2/2,0+ 2.5/2) | M_1(1, 1.25) |
BC | B( 2, 2.5) and C( 6, 0) | M_2(2+ 6/2,2.5+ 0/2) | M_2(4, 1.25) |
It can be seen above that both M_1 and M_2 have the same y-coordinate 1.25. This means that M_1M_2 is a horizontal segment. Since AC is on the x-axis, it can be said that it is also a horizontal segment. Therefore, the midsegment is parallel to AC. M_1M_2 ∥ AC Finally, it needs to be proven that M_1M_2 is half AC. To do this, the Distance Formula will be used.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
---|---|---|---|
Segment | Endpoints | Substitute | Simplify |
M_1M_2 | M_1( 1, 1.25) and M_2( 4, 1.25) | M_1M_2=sqrt(( 4- 1)^2+( 1.25- 1.25)^2) | M_1M_2= 3 |
BC | A( 0, 0) and C( 6, 0) | AC=sqrt(( 6- 0)^2+( 0- 0)^2) | AC=6 |
Since 3 is half 6, the length of M_1M_2 is half the length of AC. M_1M_2=1/2AC By following the same procedure, it can be proven that the other two midsegments of △ ABC are parallel and half the length of the third side. The Triangle Midsegment Theorem has been proven. Paulina has just leveled up and her friend will get a great frame!
The topics covered in this lesson can now be applied to the challenge. The figure below looks like a parallelogram; without using measuring tools, how can it be proven that the quadrilateral is a parallelogram?
See solution.
Place the quadrilateral on a coordinate plane. Then, translate it so that it has a vertex at the origin and a consecutive vertex on the x-axis.
To prove that the sides have the same length, the Distance Formula will be used.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) | |||
---|---|---|---|
Segment | Points | Substitute | Simplify |
AD | A( 0, 0) & D( 3, 0) | AD= sqrt(( 3- 0)^2+( 0- 0)^2) | AD= 3 |
BC | B( 2, 2) & C( 5, 2) | BC= sqrt(( 5- 2)^2+( 2- 2)^2) | BC= 3 |
The length of the opposite sides AD and BC is 3. This means that AD and BC are congruent.
The quadrilateral has been proven to have a pair of congruent parallel sides. Therefore, it is a parallelogram.