Question

If the area of a triangle ABC = 32cm2, AD is the median of AABC and E is the midpoint of AD, then the area of ADEF is equal to​

Answer 1

Answer:

Expert Verified Answer!

Hello Mate!

In ∆ABC, AD is a median hence,

ar(∆ADB) = ar(∆ADC) = ½ ar( ∆ ABC )

ar(∆ADB) = ½ × 32 cm²

ar(∆ADB) = 16 cm²

Now, in ∆ADB BE is median hence,

ar(∆BED) = ar(∆AEB) = ½ ar(∆ABD)

ar(∆BED) = ½ × 16 cm²

ar(∆BED) = 8 cm²

Hence ar(∆BED) = 8 cm².

Have great future ahead!

Answer 2

The area of ΔDEB is 8 cm².

Given:

Area of ΔABC = 32 cm²

                 BD = BC               {AD median divides the base into two equal parts.

                 AE = ED                { E is the midpoint of the median AD

To Find:

Area of ΔDEB.

Solution:

A median divides a triangle into two parts with equal areas.

So,

                    $ar(\triangle\bold{ADB}) = \cfrac{1}{2}*ar(\triangle\bold{ABC} )= ar(\triangle\bold{ADC})\\ar(\triangle\bold{ADB}) = \cfrac{1}{2} * 32\\ar(\triangle\bold{ADB}) = 16~cm^2$

And in the triangle ADB, BE is the median. So,

                     $ar(\triangle\bold{DEB}) = \cfrac{1}{2}*ar(\triangle\bold{ABD} )= ar(\triangle\bold{AEB})\\ar(\triangle\bold{DEB}) = \cfrac{1}{2} * 16\\ar(\triangle\bold{DEB}) = 8~cm^2$

Hence the area of ΔDEB is 8 cm².