Can someone please help me without any links or guessing? I’m putting a lot of points into this one so if you could just help me
that would be very much appreciated!
Justin made a large kite consisting of one parallelogram and two identical triangles, as shown below. What is the area of Justin's kite?
A) 14 ft sq
B) 20 ft sq
C) 28ft sq
D) 40ft sq
Justin made a large kite consisting of one parallelogram and two identical triangles, as shown below. What is the area of Justin's kite?
A) 14 ft sq
B) 20 ft sq
C) 28ft sq
D) 40ft sq
1 answer:
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Answer:
* The equation of the median of the trapezoid is 10x + 6y = 39
Step-by-step explanation:
* Lets explain how to solve the problem
- The slope of the line whose end points are (x1 , y1) , (x2 , y2) is
- The mid point of the line whose end point are (x1 , y1) , (x2 , y2) is
- The standard form of the linear equation is Ax + BC = C, where
A , B , C are integers and A , B ≠ 0
- The median of a trapezoid is a segment that joins the midpoints of
the nonparallel sides
- It has two properties:
# It is parallel to both bases
# Its length equals half the sum of the base lengths
* Lets solve the problem
- The trapezoid has vertices R (-1 , 5) , S (! , 8) , T (7 , -2) , U (2 , 0)
- Lets find the slope of the 4 sides two find which of them are the
parallel bases and which of them are the non-parallel bases
# The side RS
∵
# The side ST
∵
# The side TU
∵
# The side UR
∵
∵ The slope of ST = the slop UR
∴ ST// UR
∴ The parallel bases are ST and UR
∴ The nonparallel sides are RS and TU
- Lets find the midpoint of RS and TU to find the equation of the
median of the trapezoid
∵ The median of a trapezoid is a segment that joins the midpoints of
the nonparallel sides
∵ The midpoint of RS =
∵ The median is parallel to both bases
∴ The slope of the median equal the slopes of the parallel bases = -5/3
∵ The form of the equation of a line is y = mx + c
∴ The equation of the median is y = -5/3 x + c
- To find c substitute x , y in the equation by the coordinates of the
midpoint of RS
∵ The mid point of Rs is (0 , 13/2)
∴ 13/2 = -5/3 (0) + c
∴ 13/2 = c
∴ The equation of the median is y = -5/3 x + 13/2
- Multiply the two sides by 6 to cancel the denominator
∴ The equation of the median is 6y = -10x + 39
- Add 10x to both sides
∴ The equation of the median is 10x + 6y = 39
* The equation of the median of the trapezoid is 10x + 6y = 39