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

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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

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Thank you so much!!! It was extremely helpful. Really appreciate your time and kind explanation