Hello everyone,
I stuck in this problem: https://www.gregmat.com/problems/problem/B-is-the-perpendicular-bisector/
Please help me by explaining the strategy of this problem. Thanks in advance.
Hello everyone,
I stuck in this problem: https://www.gregmat.com/problems/problem/B-is-the-perpendicular-bisector/
Please help me by explaining the strategy of this problem. Thanks in advance.
link is not showing, can you upload picture?
Its kind of an elaborate method but one way to solve is
Let AB=x then BC=x since C is a bisector of BD CD=x
Let GB=y so now we have 2 similar triangles ABG and ACF comparing those two similar triangles
AC/AB=CF/y, since AC=2 AB CF=2y similarily compare triangle ABG and ADE
After comparing we get to know ED=3y
Now its easy, Area of triangle EFD =(0.5)(base)(height) here base = 3y and height =x hence area is 1.5xy
Now the quadrilateral CDFG,
It is the sum of the area of 2 triangles CGF and CDE so following the above method to calculate the area you get area of CGF = xy and area of CDF = xy hence the area of the quadrilateral is 2xy
quantity A=2xy
quantity B=1.5xy since XY are positive numbers A >B
Hope this helps
Thanks a lot!
B is the perpendicular bisector of AC, and C is the perpendicular bisector of BD. ∠CDE is 90 degrees
Quantity A
The area of quadrilateral CDFG
Quantity B
The area of triangle FED
How to solve this?
Let AB = x , BG = y
AB = BC = CD = x since they are bisectors
Triangle ABG is similar to ACF and ADE
Thus we get CF = 2y , DE = 3y
Area of tri ADE = 0.5 x AD x DE = 0.5 x 3x x 3y = 4.5xy
Area of tri ACG = 0.5 x AC x BG = 0.5 x 2x x y = xy
Area of tri ADF = 0.5 x AD x CF = 0.5 x 3x x 2y = 3xy
Area of tri FED = ADE - ADF = (4.5-3)xy = 1.5xy
Area of quad CDFG = ADE - FED - ACG = (4.5 - 1.5 - 1)xy = 2xy
Comparing FED < CDFG
Thus ans should be A
It would be easier if one would take x and y as 1 but this would be a more generic solution.
Thanks for the response. The answer is quantity A
Right I made a minor mistake
Took wrong height in case of BG
Edited it now
Thanks for the solution. Understood now
Hi,
Could anyone help me solve this problem please? Would much appreciate your help!
---- copying the text of the question-----
B is the perpendicular bisector of ACAC, and CC is the perpendicular bisector of BDBD. \angle CDE∠CDE is 9090 degrees.
Quantity A
The area of quadrilateral CDFGCDFG
Quantity B
The area of triangle FEDFED
Area of quad CDFG = area CGF + area CDF = \frac{1}{2} * BC * FC + \frac{1}{2} * CD * FC
As CD = FC, Area (CDFG) = \frac{1}{2} * BC * FC + \frac{1}{2} * BC * FC = BC * FC
Area FED = \frac{1}{2} * CD * ED = \frac{1}{2} * BC * ED
So you are comparing
BC * FC and \frac{1}{2} * BC * ED
Eliminating BC:
FC and \frac{ED}{2}
FC = 2GB
\frac{ED}{2} = \frac{3GB}{2}
As 2GB > 1.5 GB
Quantity A is greater
Thank you so much!!! It was extremely helpful. Really appreciate your time and kind explanation