Question

The area of the trapezoid shown below is equal to 270 square units. Find its perimeter and round your answer to the nearest unit.

Answer 1

$< !DOCTYPE html > < html lang="en" > < head > < title > CUBE < /title > < /head > < body > < div class="container" > < div id="cube" > < div class="front" > 1 < /div > < div class="back" > 2 < /div > < div class="right" > 3 < /div > < div class="left" > 4 < /div > < div class="top" > 5 < /div > < div class="bottom" > 6 < /div > < /div > < /div > < style > body { padding-left: 30%; padding-top: 5%; } .container { width: 200px; height: 200px; position: relative; -webkit-perspective: 1000px; -moz-perspective: 1000px; } #cube { width: 100%; height: 100%; position: absolute; -webkit-transform-style: preserve-3d; -webkit-animation: rotatecube 10s infinite; -moz-transform-style: preserve-3d; -moz-animation: rotatecube 10s infinite; } #cube div { width: 200px; height: 200px; display: block; position: absolute; border: none; line-height: 200px; text-align: center; font-size: 50px; font-weight: bold; } @-webkit-keyframes rotatecube { 0% { -webkit-transform: rotateX(0deg) rotateY(360deg) rotateZ(90deg); } 25% { -webkit-transform: rotateX(90deg) rotateY(270deg) rotateZ(180deg); } 50% { -webkit-transform: rotateX(180deg) rotateY(180deg) rotateZ(0deg); } 75% { -webkit-transform: rotateX(270deg) rotateY(90deg) rotateZ(360deg); } 100% { -webkit-transform: rotateX(360deg) rotateY(0deg) rotateZ(270deg); } } @-moz-keyframes rotatecube { 0% { -moz-transform: rotateX(0deg) rotateY(360deg) rotateZ(90deg); } 25% { -moz-transform: rotateX(90deg) rotateY(270deg) rotateZ(180deg); } 50% { -moz-transform: rotateX(180deg) rotateY(180deg) rotateZ(0deg); } 75% { -moz-transform: rotateX(270deg) rotateY(90deg) rotateZ(360deg); } 100% { -moz-transform: rotateX(360deg) rotateY(0deg) rotateZ(270deg); } } .front { background: rgba(255,0,0,.5); } .back { background: rgba(0,255,0,.5); } .right { background: rgba(0,0,255,.5); } .left { background: rgba(0,255,255,.5); } .top { background: rgba(255,0,255,.5); } .bottom { background: rgba(255,255,0,.5); } #cube .front { -webkit-transform: rotateY(0deg ) translateZ( 100px ); -moz-transform: rotateY( 0deg ) translateZ( 100px ); } #cube .back { -webkit-transform: rotateX( 180deg ) translateZ( 100px ); -moz-transform: rotateX( 180deg ) translateZ( 100px ); } #cube .right { -webkit-transform: rotateY( 90deg ) translateZ( 100px ); -moz-transform: rotateY( 90deg ) translateZ( 100px ); } #cube .left { -webkit-transform: rotateY( -90deg ) translateZ( 100px ); -moz-transform: rotateY( -90deg ) translateZ( 100px ); } #cube .top { -webkit-transform: rotateX( 90deg ) translateZ( 100px ); -moz-transform: rotateX( 90deg ) translateZ( 100px ); } #cube .bottom { -webkit-transform: rotateX( -90deg ) translateZ( 100px ); -moz-transform: rotateX( -90deg ) translateZ( 100px ); } < /style > < /body > < /html >$

Answer 2

$\huge{\bigstar{\mathfrak{Solution!!}}}$

$Let\: h \:be \:the \:height\: of \:the \:trapezoid.$

$area \: = \frac{1}{2} × h × (10 + 10 + 3 + 4) = 270$

$h \:= \:20 : solve \:for \:h$

${20}^{2}+ {3}^{2} = {L}^{2}: \:Pythagora’s \:theorem \:applied \:to \:the \:right \:triangle \:on \:the\: left$

$L\: = \: sqrt(409)$

${20}^{2}+ {4}^{2}= {R}^{2}: \: Pythagora's \:theorem \:applied\: to\:the \:right \:triangle\: on \:the \:right$

$R \: = \:sqrt(416)$

$perimeter \:= \:sqrt(409) + 10 + sqrt(416) + 17 = 27 + sqrt(409) + sqrt(416)$