Respuesta :

jimman
for this problem you have to look at the triangle as a whole. you see that the area of the rectangle can be represented as the area of the big triangle minus the two smaller triangles. we are given the vertical sides of the rectangle and opposite sides of rectangles are always equal in measure therefore side RQ=5. so we can find the area of triangle RQC by doing Pythagorean theorem
[tex]x = \sqrt{13 {}^{2} - 5 {}^{2} } [/tex]

when you plug that in you get x=12 for segment RC and plugging in the values into the area of a triangle (l*w)/2 you get the area of triangle RQC=30. Now we need the area of triangle PBQ the area can be determined as segment (BP*PQ)/2 Where BP=15 and to find PQ we know the opposite sides of a rectangle are equal and since we don't know segment AR and PQ but their opposite sides of a rectangle we can say that AR and PQ = X.
plugging this in to the area of a triangle formula we get 7.5X. now we need to find the area of triangle ABC which is [(AP+PB)*(AR+RC)]/2, we know that AP+PB=15+5 which is 20 so Now we need to find AR+RC AR=X and RC=12 so AR+RC= 12+X. Now when we plug it in to the area formula of a triangle we get 10x+120. Now we're almost done we just need the area of the rectangle, which is L*W so PA*AR which is 5*X so area of rectangle=5X. Now to get our answer we take area of big triangle minus area of smaller triangles. So it would look like this:
[tex]10x + 120 - (7.5x + 30) = 2.5x + 90[/tex]
Now we can do a little trick we know that the rectangles area is 5X we can write out the full expression as Area=5X, now we can find the area part from the big triangle minus smaller triangle operation performed above. this leaves us with just the expression for the rectangles area in the triangle: 2.5X+90 is also equal to the rectangles area, now we can set
[tex]2.5x + 90 = 5x[/tex]
then solve for X
[tex]x = 36[/tex]
AND FINALLY we plug X back into the good old rectangle area L*W which we know know is
[tex]5 \times 36 = area \: of \: rectangle[/tex]
[tex]180 = area \: of \: rectangle \: apqr[/tex]