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In square ABCD, E is the midpoint of BC, and F is the midpoint of CD. Let G be the intersection of AE and BF. Let M be the midpoint of AB, and let N be the intersection of AE and DM.

 

 

 

a) Find the ratios BG:MN, FG:MN, and DN:MN.

b) Compute the ratio of the area of GFDN to the area of GBMN.

 Sep 1, 2019
 #1
avatar+1129 
+2

Here I will give you hint:        EDIT: If you want you can go down to CPhill's answer for the answer

 

 

You can calculate the area of MAD, BEA, CFB.

 

 

You can calculate the area of BMDF by taking the area of the entire square and subtracting MAD and CFB from it.

 

 

 

 

 

You can do some coordinate geometry to find info about BGNM

 Sep 1, 2019
edited by CalculatorUser  Sep 1, 2019
 #2
avatar+103148 
+2

I like doing this in the following manner....construct a square with a side  = 4  [ any side length would work ]

Let A  = (0,0)  B  = (0,4)  C = (4,4)  and D  = (4,0)  E  = (2,4)  F  = (4,2)  M = (0, 2)

 

Here's a pic : 

 

 

Triangles BCF  and DAM are congruent right triangles

 

And the slope of BF  =  [ 4-2] / [ 0 - 4  ]=  2/-4  = -1/2

And the slope of EA =   [ 4-0]  / [2-0]  = 4/2  = 2

So....these segements have reciprocal slopes  so they are perpendicular

 

This means that triangles   BCF  and BGE  are similar by AA congruency

 

BC  = 4     CF  = 2     BE  = 2    BF  =  √ [ BF^2 + CF^2]  = √ [4^2 + 2^2] = √20  = 2√5

 

So.....by similar  triangles......  BG / BE   =  BC/ BF   →  BG/ 2  = 4/ [ 2√5 ]   →  BG  = 4/√5 

 

And GE  =   √[BE^2  - BG^2]  = √[2^2 - (16/5]  = √  4 - (16/5]  = √ [ 20 - 16] / √5  =  2/√5   

 

And the  area of triangle  BCF  =  (1/2) BC * CF  = 4

And since BG / BC  =  (  4 / [ √5] ) / 4   =  1/ √5.....the area  of triangle BGE  = 4 (1/√5)^2  =  4/5

 

Likewise  triangles MAD  and  MNA   are similar  and we can show that triangles BGE  and ANM are congruent

 

So....BG/MN  = BG / GE   =  [4/√5 ] / [2 /√5]    =   2

 

And FG  =  BF - BG   =     2√5  - 4/√5   =  [10 √5 - 4√5] / 5  =  6√5/5  =  6/√5

 

So....FG/ MN  =  6/√5 / [ 2/√5] =  6/2  =  3

 

And DN  = DM - MN  =   BF - GE =  2√5  - 2/√5  =  [ 10√5 - 2√5] / 5 = 8√5/5  = 8/√5

 

So DN / MN  = [8/√5]/ [2/√5]  = 8 /2  =  4 

 

 

Area  of BMDF  =  Area of the square  less areas  of BCF  and DAM  =  4^2  - 2 [4] = 16 - 8  = 8

 

Area of  GBMN  =  area of BEA - areas of BGE  and ANM  =  (1/2)BE * BA  - 2(4/5)  = 

(1/2)(2)(4) - 8/5   =

4- 8/5   =

[20 - 8] / 5  =

12/5

 

So the area of  GFDN  = area of BMDF  - area of GBMN  =   8 - 12/5 =  [40 - 12]/5 = 28/5

 

So  GFDN / GBMN  =  [ 28/5] / [ 12/5] =  28/12   =  7 / 3

 

 

cool cool cool

 Sep 1, 2019

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