The rectangle in the diagram has an area of 640 cm^2. Points B and F are the midpoints of sides AC and AE, respectively. What is the area of triangle BDF?
+1005 well we know that FED+ DCB count for 50% of the total area, since they each are 25%. Further, the ABF triangle is 1/8 of the total, so we have 5/8 unshaded. 640/8 = 80.
This is because if we put points...
Let's say point X is Midpoint ED, Y is midpoint CD
BCD is one half BCDX, and BCDX is one half whole rect, so 1/4
Same applies to FED.
If we put a point Z at midpoint of FY and BX, AFB is one half ABFZ, which is 1/4 whole rect, so 1/8.
3*80 = 240
if you don't understand anything feel free to ask.
+1005 well we know that FED+ DCB count for 50% of the total area, since they each are 25%. Further, the ABF triangle is 1/8 of the total, so we have 5/8 unshaded. 640/8 = 80.
This is because if we put points...
Let's say point X is Midpoint ED, Y is midpoint CD
BCD is one half BCDX, and BCDX is one half whole rect, so 1/4
Same applies to FED.
If we put a point Z at midpoint of FY and BX, AFB is one half ABFZ, which is 1/4 whole rect, so 1/8.
3*80 = 240
if you don't understand anything feel free to ask.
+1005 Thank you!
But TBH, I think yours might be a bit better, but both work, and the other can be used as a different strategy to check:
essentially we did the same thing, you just subtracted the unshaded, I calculated proportion of unshaded and subtracted that from 1.
+129899 Area of rectangle = 2(AB) * 2(AF)
Area of triangle ABF =(1/2)(AB)(AF) = (1/2)(AF)(AB)
Area of triangle BCD = (1/2) *2((AF) (AB) = (AF)(AB)
Area of triangle FED = (1/2) * 2(AB)(AF) = (AF)(AB)
So area of rectangle = 2(AF) 2(AB) = 4(AF)(AB)
So
640 = 4 (AF)(AB) divide through by 4
160 =(AF)(AB)
So
The area of the three right triangles = (AF)(AB) ( 1/2+ 1 + 1) = (5/2)(AF)(AB) = (5/2)(160) = 400
So....the area triangle BDF = 640 - 400 = 240
