Q: Quadrilateral ABCD is a rhombus with perimeter 52 meters. The length of diagonal AC is 24 meters. What is the area in square meters of rhombus ABCD?
![[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]](https://latex.artofproblemsolving.com/e/5/4/e54b35c6e9f6a1eaa16a3138ed06ebc73122be63.png)
If the perimeter of ABCD is 52, and ABCD is a rhombus, that means that the length of each side is 52 ÷ 4 = 13.
If the length of a diagonal is 24, that means that 1/2 of it -- the distance between the intersection of the diagonals and a vertex (A or C in this case) -- is 24 ÷ 2 = 12.
We subtract the square of 12 from the square of the side length, 13, and we get the square of 1/2 of the other diagonal. 13² - 12² = 25. √25 = 5, so the length of the other diagonal is 5 ⋅ 2 = 10.
When you multiply the diagonals together and divide that by 2, you get the area. That means the area is 10 ⋅ 24 ÷ 2 = 120, making the answer D.
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