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# In rectangle ABCD, I is the midpoint of AD and E and F trisect BC.

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In rectangle ABCD, I is the midpoint of AD and E and F trisect BC.  BD intersects EI and FI and G and H.  Find the ratio of the area of triangle GHI to the area of rectangle ABCD.

Dec 13, 2020

#1
+1162
+2

In rectangle ABCD, I is the midpoint of AD and E and F trisect BC.  BD intersects EI and FI and G and H.  Find the ratio of the area of triangle GHI to the area of rectangle ABCD.

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Since a square is also a rectangle, we can use it to make life easier.

Let the square side be 3 units.

ID = 1.5         EF = 1           ∠EIF = 2* tan-1[(EF/2) / AB] = 18.92464442º

∠ADB = 45º           ∠EID = 90 + (∠EIF / 2) = 99.4623222º           ∠FID = 90 - (∠EIF / 2) = 80.5376778º

Now you have enough information to calculate the area of triangles GID and HID.

But if you're lazy like me, you can visit this website:      https://www.triangle-calculator.com/

Dec 13, 2020
#2
+186
+3

Here's a solution where the rectangle really is a rectangle and not necessarily a square.

Run horizontal lines GG' and HH' from G and H across to AB.

G' and H' will be used to write down the heights of the triangles GID and HID.

Triangles BEG and DIG are similar so $$\displaystyle \frac{EG}{GI}=\frac{BE}{ID}=\frac{1/3}{1/2}=\frac{2}{3}.$$

So BG' = (2/5)AB, and G'A = (3/5)AB,

so the area of the triangle DIG = (3/5)AB times (1/2)AD times (1/2) = (3/20) area of the rectangle.

Triangles BFH and DIH are similar so $$\displaystyle \frac{FH}{HI}=\frac{BF}{ID}=\frac{2/3}{1/2}=\frac{4}{3}.$$

So BH' = (4/7)AB and H'A = (3/7)AB,

so the area of the triangle DIH = (3/7)AB times (1/2)AD times (1/2) = (3/28) area of the rectangle.

Area of the triangle GHI is then (3/20) - (3/28) = (3/70) area of the rectangle.

Dec 13, 2020
#3
+1162
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Great solution, Tiggsy!!!

jugoslav  Dec 13, 2020
#4
+186
+1

Thanks, much appreciated.

Tiggsy  Dec 14, 2020