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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.

 Jul 24, 2020
 #1
avatar+10085 
+1

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.

 

Hello Guest!

 

\(\overline{BE}=\sqrt{(15^2+30^2)cm^2}=\sqrt{1125cm^2}=33.541cm\\ \overline{EF}=\frac{1}{2}\cdot \overline{BE}\)

\(\large \frac{h_{△DFE}}{15cm}=\frac{\frac{1}{2}\cdot \overline{BE} }{\overline{BE}}=\frac{1}{2}\)

\(h_{△DFE}=7.5cm\)

\(A_{△DFE}=\frac{1}{2}\cdot \overline{ED}\cdot h_{△DFE}=\frac{1}{2}\cdot 15cm\cdot 7.5cm\)

\(A_{△DFE}=56.25cm^2\)

laugh  !

 Jul 24, 2020
edited by asinus  Jul 24, 2020
edited by asinus  Jul 24, 2020
 #3
avatar+1326 
+1

You have an error here:::

 

EF ≠ 1/2 * BE                    EF = 2/3 * BE

 

The centroid divides a median into two line segments, and their ratio is 2 : 1

Dragan  Jul 24, 2020
 #2
avatar+1326 
+1

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 cm 

 Jul 24, 2020
edited by Dragan  Jul 24, 2020

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