Rectangle PQRS is inscribed in rectangle ABCD, as shown. If DR=3, RP=13, and PA=8, compute the area of rectangle ABCD.
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2007-01-11 10:13:28 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
132 is the answer.
11*12 is 132
the total height of the Rectangle is 8+3
then you know two sides of a triangle, that contains an unknown side, which is the width of the Rectangle ABCD (3+3=6. 11-6=5) So you have a right triangle that is made between PR (13, the hypotenuse), the height - 6 (3+3=6, 11-6=5, one leg) and you are left with the last leg, which is also the width of ABCD.
a^2 + b^2 = c^2
25 + b^2 = 13^2
b^2 = 169-25
b^2 = 144
b = 12, which is the width of the rectangle.
and 12*11 = 132
2007-01-11 10:28:12 · answer #1 · answered by Ritic 1 · 0⤊ 0⤋
8-3 = 5
Pythagorean triple: 5, 12, 13
Therefore, the area = 12(11) = 132
2007-01-11 18:28:36 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
First compute DA using Pythagoras. you get DA=12 which shows that ABCD IS NOT A SQUARE but has area 132.
2007-01-11 18:24:33 · answer #3 · answered by gianlino 7 · 0⤊ 0⤋
Segments RD, DS, QB and BP are equal, and so are CR, CQ, SA, and PA. Therefore, each side of ABCD is equal to 8 + 3, or 11. The area is therefore (11)(11) which is equal to 121.
2007-01-11 18:21:59 · answer #4 · answered by Someone 3 · 0⤊ 1⤋
121 i think
2007-01-11 18:18:41 · answer #5 · answered by B9O9R9I9C9U9A 3 · 0⤊ 1⤋