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spin [16.1K]
3 years ago
6

Given the rectangle abcd shown below has a total area of 72. E is in the midpoint of bc and f is the midpoint of dc. What is the

area of the inscribed triangle aef.

Mathematics
1 answer:
scoray [572]3 years ago
8 0

Refer to the attached image.

Given the rectangle ABCD of length 'l' and height 'h'.

Therefore, CD=AB = 'l' and BC = AD = 'h'

We have to determine the area of triangle AEF.

Area of triangle AEF = Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

Area of triangle ADF = \frac{1}{2}bh

= \frac{1}{2}(DF \times AD)

= \frac{1}{2}(\frac{l}{2} \times h)

=\frac{lh}{4}

Area of triangle ECF = \frac{1}{2}bh

= \frac{1}{2}(CF \times CE)

= \frac{1}{2}(\frac{l}{2} \times \frac{h}{2})

=\frac{lh}{8}

Area of triangle ABE = \frac{1}{2}bh

= \frac{1}{2}(AB \times BE)

= \frac{1}{2}(l \times \frac{h}{2})

=\frac{lh}{4}

Now, area of triangle AEF =

Area of rectangle ABCD - Area of triangle ADF - Area of triangle ECF - Area of triangle ABE

= 72 - (\frac{lh}{4} + \frac{lh}{8} + \frac{lh}{4})

= 72 - (\frac{2lh+lh+2lh}{8})

=72 - (\frac{5lh}{8})

=72 - (\frac{5 \times 72}{8})

=\frac{72 \times 8 - (5 \times 72)}{8}

= 27 units

Therefore, the area of triangle AEF is 27 units.

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