Question 1210155: ABCE is a Parallelogram. ADEF is a Rhombus. Find the area of the composite figure.
https://ibb.co/Jj2jKPD3 Found 4 solutions by Edwin McCravy, CPhill, mccravyedwin, ikleyn:Answer by Edwin McCravy(20055) (Show Source):
We calculate GE by using the Pythagorean theorem on right triangle FGE.
AG is also , so diagonal of the rhombus, AE =
FG = GD = 5 so the other diagonal of the rhombus, FD is 10.
The area of a rhombus is the product of the diagonals, or,
Area of rhombus ADEF is .
Since ABCF is a parallellogram,
ABCD is an isosceles trapezoid (or trapezium if you live in the UK).
We draw two green lines at D and C perpendicular to AB and CD. They partition
the trapezoid into two congruent right triangles, and a rectangle in the middle.
Since AH = GD = 5, the area of right triangle AHD is
The area of right triangle BIC is also
All that's left of isosceles trapezoid ABCD is the area of rectangle HICD which
is
Area of rectangle =
Area of right triangle BIC =
Area of isosceles trapezoid ABCD =
Area of rhombus ADEF =
Area of isosceles trapezoid ABCD =
Area of composite figure =
That's approximately 422.4 sqruare units.
Edwin
You can put this solution on YOUR website! Absolutely! Let's break down this problem step-by-step.
**1. Finding the area of Rhombus ADEF**
* **Properties of a Rhombus:**
* All sides are equal.
* Diagonals bisect each other at right angles.
* **Given:**
* EF = 17 (side of the rhombus)
* FO = 5 (half of diagonal FD)
* **Finding EO:**
* Since triangle EOF is a right-angled triangle, we can use the Pythagorean theorem: $EF^2 = EO^2 + FO^2$
* $17^2 = EO^2 + 5^2$
* $289 = EO^2 + 25$
* $EO^2 = 289 - 25 = 264$
* $EO = \sqrt{264} = 2\sqrt{66}$
* **Finding Diagonals:**
* FD = 2 * FO = 2 * 5 = 10
* AE = 2 * EO = 2 * $2\sqrt{66}$ = $4\sqrt{66}$
* **Area of Rhombus ADEF:**
* Area = (1/2) * (diagonal 1) * (diagonal 2)
* Area = (1/2) * 10 * $4\sqrt{66}$
* Area = $20\sqrt{66}$
**2. Finding the area of Parallelogram ABCE**
* **Properties of a Parallelogram:**
* Opposite sides are equal and parallel.
* **Given:**
* CD = 11
* EF = 17
* **Finding AE:**
* since ADEF is a rhombus, AE=EF. Therefore AE = 17.
* **Finding the Height of the Parallelogram:**
* The height of the parallelogram is the same as the length EO.
* Height = $2\sqrt{66}$
* **Finding the Area of the Parallelogram:**
* Area = base * height
* Area = AE * EO
* Area = 17 * $2\sqrt{66}$
* Area = $34\sqrt{66}$
**3. Finding the Area of the Composite Figure**
* **Total Area:**
* Area of composite figure = Area of Rhombus ADEF + Area of Parallelogram ABCE
* Total Area = $20\sqrt{66} + 34\sqrt{66}$
* Total Area = $54\sqrt{66}$
**Therefore, the area of the composite figure is $54\sqrt{66}$.**
You can put this solution on YOUR website! . ABCF is a Parallelogram. ADEF is a Rhombus. Find the area of the composite figure.
https://ibb.co/Jj2jKPD3
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I edited your post to turn your incorrect writing into a correct problem.
In rhombus ADEF, the half of the long diagonal is = = units long.
So, we can find the area of the parallelogram ABCF as the product of the base FC = 11+5+5 = 21 by the height AG = .
So, the area of the parallelogram ABCF is = .
To it, we should add the area of triangle FDE = = .
Thus, the total combined area is + = .
ANSWER. The total combined area is square units.
Solved.
As simple as a cucumber.
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Edwin, you do not need to revise your solution - it is correct.
