ABCD is a parallelogram such that AB is parallel to DC and DA parallel to CB. The length of side AB is 20cm. E is a point between A and B such that the length of AE is 3cm. F is a point between points D and C. Find the length of DF such that the segment EF divide the parallelogram in two regions with equal areas.

(Geometry)

প্রশ্নে বলা হচ্ছে, ABCD একটি সামান্তরিক। AB || DC এবং DA || CB. AB = 20cm. A এবং B এর মধ্যে E এমন একটি বিন্দু যেখানে AE = 3cm. D এবং C এর মধ্যবর্তী বিন্দু F. এখন EF  সামান্তরিকটিকে দুইটি অঞ্চলে বিভক্ত কলে DF এর দৈর্ঘ্য বের করুন।

Let, A, be the area of the trapezoid AEFD.

$Hence, A_1 = \frac{1}{2} h \left(AE + DF\right) = \frac{1}{2} h \left(3 + DF\right),$ h is the height of the parallelogram.

Now, let A₂ be the area of the trapezoid EBCF.

$Hence, A_2=\frac{1}{2}h(EB + FC )$

He also have, EB = 20- AE = 17, FC= 20- DF.

We now substitute EB and F in $A_2 = \frac{1}{2}h \left(EB + FC\right)$

$\Rightarrow A_{2 } = \frac{1}{2} h \left(17 + 20 - DF\right) = \frac{1}{2} \left(37 - DF\right)$

For EF to divide the parallelogram into two regions of equal area, we need to have area A, and area

$A_2 equal \frac{1}{2} h \left(3 + DF\right) = \frac{1}{2} h \left(37-DF\right)$ [Multiply both sides by 2 and divide by h]

$$\begin{gathered} \Rightarrow 3 + DF = 37 - DF \\ \Rightarrow 2DF = 37-3 = 34 \\ \therefore DF = \frac{34}{2} = 17 cm \end{gathered}$$