ABCD is a parallelogram in which AE is perpendicular to CD. Also, AC = 5 cm, DE = 4 cm, and the area of ∆ AED = 6 $c m^{2}$ . The perimeter and area of ABCD are

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26 cm, 24 {cm}^{2}

23 cm, 25 {cm}^{2}

16 cm, 23 {cm}^{2}

26 cm, 14 {cm}^{2}

answer is A.

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Detailed Solution

Given, ABCD is a parallelogram in which AE is perpendicular to CD.
AC = 5 cm, DE = 4 cm.
The area of ∆ AED = 6

c m^{2}

.
We know that the area of a triangle is given as the product of half times its altitude and base.
So,

12 A E × E D = 6 4 A E = 12 A E = 3 cm

So, applying Pythagoras theorem on triangle AEC, we get,

A E 2 + E C 2 = A C 2 32 + E C 2 = 52 E C 2 = 16 E C = 4 cm

Next, applying the Pythagoras theorem on triangle AED, we get,

A D 2 = A E 2 + E D 2 A D 2 = 32 + 42 A D 2 = 25 A D = 5 cm

So, the base CD of the parallelogram,
= DE + EC
= 4 + 4
= 8 cm
Now, we know that in a parallelogram, the opposite side pairs are of same length, therefore, we have,
AB = CD = 8 cm and AD = BC = 5 cm
Now, we know that the perimeter of a parallelogram is the sum total of all its side lengths.
So, we have the perimeter of parallelogram ABCD as,