SOLUTION: In triangle WXY, WY=24m qns 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

Algebra ->  Pythagorean-theorem -> SOLUTION: In triangle WXY, WY=24m qns 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      Log On


   

Question 1120239: In triangle WXY, WY=24m qns 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 ankor@dixie-net.com(22740) About Me  (Show Source):
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In triangle WXY, WY=24m qns 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
:
Draw a right triangle and label it as described, it will be easy to understand
Hypotenuse = 18 + 14 = 32m
Find xy using the pythag:
xy = sqrt%2832%5E2+-+24%5E2%29
xy = 21.166m
Find YQ
21.166 - 9.8 = 11.366m is YQ
:
Find the area
A = 1%2F2*21.166 * 24
A = 253.992 sq/m is the area of WXY, but it asks for the area of XPY
:
Find PY
PY = sqrt%2821.166%5E2-14%5E2%29
PY = 15.875, the height of triangle XPY
A = 1%2F2*15.875*9.8
A = 77.785 sq/m