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ABCD is a parallelogram.
AD = 10, AB = 12. DE = 8W here DE is perpendicular to AB.
To calculate the perimeter.
To calculate area .
Since ABCD is a parallelogram, AB = CD = 12 and
AD = BC = 10.
So the perimeter AB+BC+CD+DA = 2(AB+AD) = 2(12+10) = 44 units.
The area of the parallelogram ABCB = AB*perpedicular DE = 12*8 =96 sq units.
In this problem we are given a that the parallelogram ABCD has AD=10, AB=12 and DE=8 where DE is perpendicular to AB.
Now take the triangle ADE, using the Pythagorean Theorem
AD^2= AE^2 + DE^2
=> AE^2 = 10^2 -8^2 = 100- 64 = 36
=> AE = 6
So EB = 6
Now lets draw a line BF that is perpendicular to DC and F lies on DC. We see that the ADE and CBF are congruent.
So we have two the parallelogram in three peices, 2 triangles and a rectangle. The area of the rectangle is 8*6 = 48 and that of the triangles is 2* (1/2) 6 * 8 = 48.
So the area of the parallelogram is 48+ 48 = 96. And the perimeter is 10*2+ 12*2 = 20 + 24 = 44.
The required area and perimeter are 96 and 44 resp.
In a parallelogram, the opposite sides are parallel and equal.
AB = CD = 12
BC = AD = 10
The perimeter of a geometric shape is the sum of the lengths of the sides of that shape.
P = AB+BC+CD+AD
Since AB = CD and BC = AD, we could re-write the perimeter as:
P = 2(AB+BC)
P = 2(12+10)
P = 2*22
P = 44 units
The area of the parallelogram could be written as a sum of 2 triangles and a rectangle.
A = 2*A(AED) + A(DEBF)
To calculate the area of AED, we need to calculate the cathetus AE. We'll use the Pythagorean theroem:
AE^2 = AD^2 - DE^2
AE^2 = 100-64
AE^2 = 36
AE = 6
Area of AED = AE*ED/2
Area of AED = 6*8/2
Area of AED = 24 square units
To calculate the area of the rectangle DEBF, we'll calculate first the width EB.
EB = AB-AE
EB = 12-6
EB = 6
Area of DEBF = EB*DE
Area of DEBF = 6*8
Area of DEBF = 48 square units
The area of ABCD is:
A(ABCD) = 2*A(AED) + A(DEBF)
A(ABCD) = 2*24 + 48
A(ABCD) = 2*48
A(ABCD) = 96 square units.
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