In quadrilateral ABCD, we have AB = BC = CD = DA, AC = 14, and BD = 48. Find the perimeter of ABCD.

A quadrilateral with all four sides equal is a rhombus, whose area is half the product of the diagonals AC and BD.

Yeah, but we're not trying to find the area?

Sorry.

In that case, we make use of the fact that rhombuses have diagonals which intersect at right-angles.
Thus each side is the hypotenuse of a right triangle with legs equal to the half-diagonals.

Thus if s=length of each side of the rhombus, then
s²=(AC/2)²+(BD/2)²
using Pythagoras theorem.
Calculate
s=√(7²+24²)
=25

The perimeter is 4s.

To find the perimeter of quadrilateral ABCD, we need to determine the lengths of all four sides.

Given that AB = BC = CD = DA, let's assign a variable to represent the length of one side. Let's call it x. So, AB = BC = CD = DA = x.

Since AC is a diagonal, it bisects the quadrilateral into two congruent triangles. Let's draw the diagonal AC:

A _______ B
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____|_______|____
| |
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D ‾‾‾‾‾‾‾ C

Now we have two congruent right triangles, DAC and BAC, with hypotenuse AC and legs AD and DC.

Using the Pythagorean theorem, we can relate the lengths of AC, AD, and DC:

AC² = AD² + DC²

Since AD and DC are equal in length, we can say:

AC² = 2 * AD²

Substituting AC with its value of 14:

14² = 2 * AD²

196 = 2 * AD²

Dividing both sides by 2:

98 = AD²

Taking the square root of both sides:

√98 = AD

Simplifying:

√(49 * 2) = AD

√(7² * 2) = AD

√(7²) * √2 = AD

7√2 = AD

Since AB = BC = CD = DA = x, and AD = 7√2, we can equate them:

x = 7√2

We also know that BD = 48, which is equal to two sides (BC + CD) of the quadrilateral:

BD = BC + CD

48 = x + x

48 = 2x

Dividing both sides by 2:

24 = x

So, the length of each side is x = 24. Hence, the perimeter of the quadrilateral ABCD is:

Perimeter = AB + BC + CD + DA = 24 + 24 + 24 + 24 = 96 units.

Stop cheating Sheenybeany!