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# Geometry Problem

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Triangle AHI is equilateral. We know BC, DE, and FG are all parallel to HI and AB = BD = DF = FH. What is the ratio of the area of the trapezoid FGIH to the area of triangle AHI? Express your answer as a common fraction.

Guest Feb 13, 2018
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#1
+541
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I will call the distance from B to C x.

Since all of the points are equal distance from eachother, $$FG = 3x \operatorname{and} HI = 4x$$.

Area of a trapezoid = $$^{1}\!\!/\!_{2}(b_1+b_2)\times h$$, where h = height, b1 = length of bottom, and b2 = length of top.

Area of FGIH = $$7x(\tfrac{\sqrt{3}}{2}x)=\tfrac{7\sqrt{3}}{2}x^{2}$$

Area of ABC = $$x(\tfrac{\sqrt{3}}{2}x)=\tfrac{\sqrt{3}}{2}x^{2}$$

Area ratio of FGIH to ABC = $$\tfrac{7\sqrt{3}}{2}x^{2}:\tfrac{\sqrt{3}}{2}x^{2}$$ --> $$7:1$$

helperid1839321  Feb 13, 2018
#2
+85673
+1

Thaks, helperid...here's an alternative solution

Without a loss of generailty we can let  AB  = 1

So   AHI  =   (1/2)(4)*2 sin (60)  =  4√3       (1)

Area  of  AFG  =  (1/2)(3)^2 sin (60) = (9/4)√3    (2)

So area of trapezoid  FGIH  =   √3 [ 4 - 9/4]  = √ 3 [ 7/4]

So  ratio of area of trapezoid FGIH  to AHI  =

√3 {7 /4]  / [  4√3 ]   =   7 / 16

EDIT TO CORRECT AN IDIOTIC ERROR  !!!!

CPhill  Feb 13, 2018
edited by CPhill  Feb 13, 2018
#3
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Triangle AHI is equilateral. We know BC, DE, and FG are all parallel to HI and AB = BD = DF = FH.
What is the ratio of the area of the trapezoid FGIH to the area of triangle AHI?

$$\text{Let FG = s } \\ \text{Let HI = c } \\ \text{Area of the triangle AHI = A_{AHI} } \\ \text{Area of the triangle AGF = A_{AGF} } \\ \text{Area of the trapezoid FGIH = A_{FGIH} } \\ \text{Let AK = h (height of the triangle_{AGF}) } \\ \text{Let AL = H (height of the triangle_{AHI}) } \\ \text{Let AF = \frac34 c }$$

1.

$$\begin{array}{|rcll|} \hline A_{AHI} &=& A_{AGF} + A_{FGIH} \\\\ \dfrac{cH}{2} &=& \dfrac{sh}{2} + \left( \dfrac{s+c}{2}\right)(H-h) \quad & | \quad \times 2 \\\\ cH &=& sh + (s+c)(H-h) \\\\ \not{cH} &=& \not{sh} + sH-\not{sh}+\not{cH}-ch \\\\ \mathbf{ch} &\mathbf{=}& \mathbf{ sH } \qquad (1) \\\\ &\text{or}& \\\\ \mathbf{\dfrac{s}{c}} &\mathbf{=}& \mathbf{\dfrac{h}{H}} \qquad (2) \\ \hline \end{array}$$

2.

$$\begin{array}{|rcll|} \hline \text{ratio} &=& \dfrac{ A_{FGIH} } {A_{AHI}} \\\\ &=& \dfrac{ \left( \dfrac{s+c}{2}\right)(H-h) } {\dfrac{cH}{2}} \\\\ &=& \dfrac{(s+c)(H-h)}{cH} \\\\ &=& \dfrac{sH-sh+cH-ch}{cH} \quad & | \quad ch=sH \qquad (1) \\\\ &=& \dfrac{cH-sh}{cH} \\\\ &=& 1-\dfrac{sh}{cH} \quad & | \quad \dfrac{s}{c} = \dfrac{h}{H} \qquad (2) \\\\ &=& 1-\dfrac{h^2}{H^2} \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\dfrac{h}{H}\right)^2} \\\\ && \boxed{\mathbf{3.}\\ \dfrac{h}{\frac34 c} = \dfrac{H}{c} \\ h = \frac34 c \cdot \dfrac{H}{c} \\ h = \frac34 \cdot H \\ \mathbf{\dfrac{h}{H} = \frac34} } \\\\ \mathbf{\text{ratio}} & \mathbf{=} & \mathbf{ 1-\left(\frac34 \right)^2} \\\\ &=& 1- \frac{9}{16} \\\\ &=& \frac{16-9}{16} \\\\ &\mathbf{=}& \mathbf{\dfrac{7}{16}} \\ \hline \end{array}$$

So  the ratio  of areas is $$\mathbf{\dfrac{7}{16}}$$

heureka  Feb 13, 2018
edited by heureka  Feb 13, 2018
edited by heureka  Feb 13, 2018

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