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Difficulty:
35%
(medium)
Question Stats:
69% (02:22) correct
31%
(03:29)
wrong
based on 49
sessions
History
Date
Time
Result
Not Attempted Yet
Correct Option E : 1000
Posted from my mobile device
Posted from my mobile device
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Posts: 1,324
Line QS is the diagonal of the rectangle.
The diagonal of the rectangle is a line of symmetry. The other half of the rectangle is a reflection of the cut up half through the diagonal line.
This means that the diagonal divides the area of the rectangle into 2 congruent triangles with equal area.
Furthermore, the 5 lines drawn from point Q each cut up the bottom-length into equal partitions. If we use each one of those equal partitions as the base of one triangle ——> we can say point Q is a shared vertex for each 1 of the 5 triangles created.
Rule: there exists only one perpendicular distance/height from a point/vertex to a straight line.
If we take the base as each one of the equal, divided lengths —-> then each triangle will have the same perpendicular height drawn from vertex Q to the shared side of the rectangle
Therefore, since each triangle has an equal length base and the same perpendicular height, each triangle area will be equal.
In summary, the 5 lines drawn from vertex Q divide the bottom-right half of the rectangle into 5 triangles with Equal Area.
Let each of these triangle’s area = A
There are 5 triangles created ——> thus the Area of the rectangle to the bottom right of the diagonal is 5A and the area of the whole rectangle is twice this amount or 10A
We are told that the SUM of the areas of (QWT) + (QWV) is 300 less than the SUM of the areas of (QWR) + (QSU)
Triangle QWT: made up of 3 triangles of equal area ——-> 3A
Triangle QWV: is just 1 triangle ——-> 1A
Sum is = 4A
Triangle QWR: made up of 4 triangles —-> 4A
Triangle QSU: made up of 3 triangles
Sure = 7A
Thus:
7A - 4A = 300
3A = 300
A = 100
above, we deduced that the area of the entire rectangle would be 10A
Substituting the value of A = 100 into 10A:
10A = 10(100) =
1,000
E
Posted from my mobile device
The diagonal of the rectangle is a line of symmetry. The other half of the rectangle is a reflection of the cut up half through the diagonal line.
This means that the diagonal divides the area of the rectangle into 2 congruent triangles with equal area.
Furthermore, the 5 lines drawn from point Q each cut up the bottom-length into equal partitions. If we use each one of those equal partitions as the base of one triangle ——> we can say point Q is a shared vertex for each 1 of the 5 triangles created.
Rule: there exists only one perpendicular distance/height from a point/vertex to a straight line.
If we take the base as each one of the equal, divided lengths —-> then each triangle will have the same perpendicular height drawn from vertex Q to the shared side of the rectangle
Therefore, since each triangle has an equal length base and the same perpendicular height, each triangle area will be equal.
In summary, the 5 lines drawn from vertex Q divide the bottom-right half of the rectangle into 5 triangles with Equal Area.
Let each of these triangle’s area = A
There are 5 triangles created ——> thus the Area of the rectangle to the bottom right of the diagonal is 5A and the area of the whole rectangle is twice this amount or 10A
We are told that the SUM of the areas of (QWT) + (QWV) is 300 less than the SUM of the areas of (QWR) + (QSU)
Triangle QWT: made up of 3 triangles of equal area ——-> 3A
Triangle QWV: is just 1 triangle ——-> 1A
Sum is = 4A
Triangle QWR: made up of 4 triangles —-> 4A
Triangle QSU: made up of 3 triangles
Sure = 7A
Thus:
7A - 4A = 300
3A = 300
A = 100
above, we deduced that the area of the entire rectangle would be 10A
Substituting the value of A = 100 into 10A:
10A = 10(100) =
1,000
E
Posted from my mobile device
Let,
RT = TU = UV = VW = WS = x
QR = y
Length of rectangle = RT+TU+UV+VW+WS = 5x
Width of rectangle = QR = y
Area of rectangle = 5xy
For each triangle the height is QR = y
Length of line segment WT = 3x
Length of line segment WV = x
Length of line segment WR = 5x
Length of line segment SU = 3x
Area of △QWT = 1/2*WT*QR = 1/2*3x*y
Area of △QWV = 1/2*WV*QR = 1/2*x*y
Area of △QWR = 1/2*WR*QR = 1/2*5x*y
Area of △QSU = 1/2*SU*QR = 1/2*3x*y
Given,
Area of △QWT + Area of △QWV = Area of △QWR + Area of △QSU - 300
1/2*3xy + 1/2*xy = 1/2*5xy + 1/2*3xy - 300
4xy = 7xy - 600
3xy = 600
xy = 200
∴ Area of rectangle = 5xy = 5*200 = \(1000 \)(E)
RT = TU = UV = VW = WS = x
QR = y
Length of rectangle = RT+TU+UV+VW+WS = 5x
Width of rectangle = QR = y
Area of rectangle = 5xy
For each triangle the height is QR = y
Length of line segment WT = 3x
Length of line segment WV = x
Length of line segment WR = 5x
Length of line segment SU = 3x
Area of △QWT = 1/2*WT*QR = 1/2*3x*y
Area of △QWV = 1/2*WV*QR = 1/2*x*y
Area of △QWR = 1/2*WR*QR = 1/2*5x*y
Area of △QSU = 1/2*SU*QR = 1/2*3x*y
Given,
Area of △QWT + Area of △QWV = Area of △QWR + Area of △QSU - 300
1/2*3xy + 1/2*xy = 1/2*5xy + 1/2*3xy - 300
4xy = 7xy - 600
3xy = 600
xy = 200
∴ Area of rectangle = 5xy = 5*200 = \(1000 \)(E)
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