Let line AC be perpendicular to line CE. Connect A to the mid-point of CE which is D, and connect E to the mid-point of AC which is B. If AD and EB intersect at point F, BC=CD=15cm, then find the area of △DFE.
+14995 Let line AC be perpendicular to line CE. Connect A to the mid-point of CE which is D, and connect E to the mid-point of AC which is B. If AD and EB intersect at point F, BC=CD=15cm, then find the area of △DFE.
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\(\overline{BE}=\sqrt{(15^2+30^2)cm^2}=\sqrt{1125cm^2}=33.541cm\\ \overline{EF}=\frac{1}{2}\cdot \overline{BE}\)
\(\large \frac{h_{△DFE}}{15cm}=\frac{\frac{1}{2}\cdot \overline{BE} }{\overline{BE}}=\frac{1}{2}\)
\(h_{△DFE}=7.5cm\)
\(A_{△DFE}=\frac{1}{2}\cdot \overline{ED}\cdot h_{△DFE}=\frac{1}{2}\cdot 15cm\cdot 7.5cm\)
\(A_{△DFE}=56.25cm^2\)
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+1490 Let line AC be perpendicular to line CE. Connect A to the mid-point of CE which is D, and connect E to the mid-point of AC which is B. If AD and EB intersect at point F, BC=CD=15cm, then find the area of △DFE.
AD = BE = sqrt( 302 + 152 )
FD = 1/3AD
FE = 2/3BE
Now we have all 3 sides of a triangle DFE.
We can use Heron's formula to calculate the area of this triangle.
Area of △DFE = 75 cm2
