All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Find the area (Posted on 2011-02-09)
The diagonals of the trapezoid ABCD intersect at P.
The area of the triangle ABP is 216 and the area of CDP is 150.

What is the area of the trapezoid ABCD?

 See The Solution Submitted by Ady TZIDON Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: Solution | Comment 3 of 4 |
(In reply to Solution by Bractals)

`To prove the following is easy.`
`"The diagonals of a trapezoid divide itinto four triangles. Two of the triangles( those having a side in common with thelateral  sides of the trapezoid ) haveequal areas."`
`Let WXYZ be an trapezoid with side WXparallel with side YZ and diagonals WYand XZ intersecting at point P.`
`   [WYX] = [WZX]`
`because they both have the same base andtheir altitudes are equal. Therefore,`
`   [XPY] = [WYX] - [WPX] = [WZX] - [WPX] = [ZPW].`
`To prove `
`"For any convex quadrilateral we have`
`   [WPX][YPZ] = [XPY][ZPW]."`
`Let WXYZ be any convex quadrilateral insteadof a trapezoid.`
`   [WPX] = (1/2)|PW||PX|*sin(WPX)    (1)     [XPY] = (1/2)|PX||PY|*sin(WPX)    (2)     [YPZ] = (1/2)|PY||PZ|*sin(WPX)    (3)     [ZPW] = (1/2)|PZ||PW|*sin(WPX)    (4) `
`because sin(XPY) = sin(YPZ) = sin(ZPW)                 = sin(WPX). Therefore,`
`   [WPX][YPZ] = (1/4)|PW||PX||PY||PZ|*                     [sin(WPX)]^2`
`   [XPY][ZPW] = (1/4)|PX||PY||PZ||PW|*                     [sin(WPX)]^2`
` `

 Posted by Bractals on 2011-02-10 18:33:34

 Search: Search body:
Forums (0)