SOLUTION: ABCD is a rhombus with perimeter 120 units and area 720 square units. P is the midpoint of line segment AD. find the area and perimeter of triangle ACP

Algebra ->  Surface-area -> SOLUTION: ABCD is a rhombus with perimeter 120 units and area 720 square units. P is the midpoint of line segment AD. find the area and perimeter of triangle ACP      Log On


   



Question 1078568: ABCD is a rhombus with perimeter 120 units and area 720 square units. P is the midpoint of line segment AD. find the area and perimeter of triangle ACP
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It would be good to know the side length of this rhombus.
A rhombus has 4 sides of the same length,
so a rhombus with perimeter 120 units,
has a side length of AB=BC=CD=AD=120units%2F4=30units .

A rhombus id a parallelogram,
so if we take one of its sides as the base,
its area (in square units) would be
30%2Agreen%28h%29=720 ,
where h is the height in units.
So, green%28h%29=720%2F30=24 .

It would also be good to understand what we need to find out,
so we need a sketch.



Since red%28P%29 is the midpoint of AD,
obviously AP=PD .

AREA OF ACP:
Any way you look at it, the area of ACP is
1%2F4 of the area of ABCD,
so (in square units) it is %281%2F4%29%2A720=highlight%28180%29 .
1) Either you realize that
the area of ACD is 1%2F2 of the area of ABCD,
and when you split ACD in equal parts with median PC,
the areas of the halves (ACP and DCP) are 1%2F2 of the area od ACD.
2) Or you realize that the area of parallelogram ABCD is AD%2Agreen%28h%29 ,
while the area of triangle ACD is,
with %281%2F2%29AP%2Agreen%28h%29=%281%2F2%29%28%281%2F2%29AC%29%2Agreen%28h%29 .

PERIMETER OF ACP:
We know AP=PD=%281%2F2%29AD=%281%2F2%29%2830units%29=15units .
We just need AC and PC .
The Pythagorean theorem, applied to right triangle DCX, tells us that
DX%5E2%2BCX%5E2=CD%5E2 ,
or with all lengths in length units,
DX%5E2%2B24%5E2=30%5E2 --> DX%5E2=18%5E2 --> DX=18 .
With that, with all length in length units, we find
AX=AD%2BDX=30%2B18=48 , and PX=PD%2BDX=15%2B18=33 .
The, applying the Pythagorean theorem to right triangles ACX and PCX,
with all lengths in length units,
AC%5E2=AX%5E2%2BCX%5E2=48%5E2%2B24%5E2=5%2A24%5E2 --> AC=24sqrt%285%29=about53.6656 , and
PC%5E2=PX%5E2%2BCX%5E2=33%5E2%2B24%5E2=1665=9%2A185 --> PC=3sqrt%28185%29=about40.8044%29 .
So the perimeter of ACP is
AP%2BAC%2BPC=highlight%2815%2B24sqrt%285%29%2B3sqrt%28185%29%29units=highlight%28about109.47%29 units.