SOLUTION: In triangle WXY, WY=24m and angle Y=90degree. P lies on WX such that YP is perpendicular to WX, WP=18m and PX=14m. Q lies on YX such that QX=9.8m. Find I) the length of YQ ii) th

Algebra ->  Pythagorean-theorem -> SOLUTION: In triangle WXY, WY=24m and angle Y=90degree. P lies on WX such that YP is perpendicular to WX, WP=18m and PX=14m. Q lies on YX such that QX=9.8m. Find I) the length of YQ ii) th      Log On


   

Question 1120240: In triangle WXY, WY=24m and angle Y=90degree. P lies on WX such that YP is perpendicular to WX, WP=18m and PX=14m. Q lies on YX such that QX=9.8m. Find
I) the length of YQ
ii) the area of triangle XPY
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Triangles XPY and XYW are similar, because they are both right triangles that share angle X. So

PX/XY = XY/WX
(XY)^2 = PX*WX = 14*32 = 448
XY = sqrt(448) = 8*sqrt(7)

Then from the Pythagorean Theorem in triangle XPY, YP = 6*sqrt(7).

Answers...

(I) YQ = XY-XQ = 6*sqrt(7)-9.8
(II) the area of triangle XPY is (1/2)(PX)(PY) = (1/2)(14)(6*sqrt(7) = 42*sqrt(7)