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26 Nov 2020, 01:15
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D
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In two equilateral parallelograms, A and B, the sum of the lengths of the diagonals of A is equal to the sum of the lengths of the diagonals of B. In A, the length of the longer diagonal is 20 more than the length of the shorter diagonal. In B, the length of the longer diagonal is 8 more than the length of the shorter diagonal. What is the area of B minus the area of A?
A. 336
B. 168
C. 84
D. 42
E. 0
A. 336
B. 168
C. 84
D. 42
E. 0
26 Nov 2020, 02:02
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CONCEPTS USED: Area of parallelogram, Equilateral parallelogram, Simplification, Substitution
Area of parallelogram= Base*Height
Rhombus is usually defined as an equilateral parallelogram.
Area of a Rhombus= 1/2*d1*d2 (d1,d2 are the diagonals of a rhombus)
SOLUTION:Let the diagonals of plgm A be XA & YA. Let XA be the longer diagonal.
Let the diagonals of plgm B be XB & YB. Let XB be the longer diagonal.
Thus, according to the question:
(i) XA+YA=XB+YB
(ii) XA=YA+20
(iii) XB=YB+8
The diagonals intersect at 90 degrees as the parallelograms are equilateral.
Area of the parallelogram(s):
-A's Area(A) = XA*YA/2 =YA*(YA+20)/2 =YA^2+20AY/2
-B's Area(B) = XB*YB/2= YB(YB+8)/2 = (YB^2 + 8YB)/2
(iv)Hence, B-A= (YB^2+8YB-(YA^2+20AY))/2
Substitute (ii) and (iii) in (i).
2YA+ 20 =2YB +8
On simplifying YA+10=YB+4 and squaring both the sides
YA^2+20YA+100 =YB^2+8YB+16 => YB^2+8YB-(YA^2+20YA)=84
Divide by two(By iv) we have the difference as 42.(D)
Hope this helps
Keep studying 
Devmitra Sen
Area of parallelogram= Base*Height
Rhombus is usually defined as an equilateral parallelogram.
Area of a Rhombus= 1/2*d1*d2 (d1,d2 are the diagonals of a rhombus)
SOLUTION:Let the diagonals of plgm A be XA & YA. Let XA be the longer diagonal.
Let the diagonals of plgm B be XB & YB. Let XB be the longer diagonal.
Thus, according to the question:
(i) XA+YA=XB+YB
(ii) XA=YA+20
(iii) XB=YB+8
The diagonals intersect at 90 degrees as the parallelograms are equilateral.
Area of the parallelogram(s):
-A's Area(A) = XA*YA/2 =YA*(YA+20)/2 =YA^2+20AY/2
-B's Area(B) = XB*YB/2= YB(YB+8)/2 = (YB^2 + 8YB)/2
(iv)Hence, B-A= (YB^2+8YB-(YA^2+20AY))/2
Substitute (ii) and (iii) in (i).
2YA+ 20 =2YB +8
On simplifying YA+10=YB+4 and squaring both the sides
YA^2+20YA+100 =YB^2+8YB+16 => YB^2+8YB-(YA^2+20YA)=84
Divide by two(By iv) we have the difference as 42.(D)
Hope this helps
Keep studying Devmitra Sen
26 Nov 2020, 06:16
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Bunuel
20+x+x=8+y+y, 12=2y-2x, 6=y-x
areaB-A: (20+x)x/2-(8+y)y/2
substitute x or y in the equation: answer 42
(D)
