How to find the area of a triangle in the following diagram?


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11 months ago by

Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of  $\bigtriangleup ABC$ABC?

 

add commentfollow this post modified 9 months ago by Tomi Owens   • written 11 months ago by Rohit U  

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11 months ago by

 From the information given,  $\bigtriangleup ABC\sim\bigtriangleup BED$ABCBED . The hypotenuse of right triangle BED is  $\sqrt{1^2+2^2}=\sqrt{5}$12+22=5 

 $\frac{AC}{BC}=\frac{BD}{ED}=\frac{2}{1}$ACBC =BDED =21   $AC=2BC$AC=2BC 

 $\frac{AC}{AB}=\frac{BD}{BE}$ACAB =BDBE  => $\frac{AC}{1}=\frac{2}{\sqrt{5}}$AC1 =25  => $AC=\frac{2}{\sqrt{5}}$AC=25  

Since AC=2BC, BC= $\frac{1}{\sqrt{5}}$15 , triangle ABC is right triangle os the area is just

 $\frac{1}{2}\times AC\times BC=\frac{1}{2}\times\frac{2}{\sqrt{5}}\times\frac{1}{\sqrt{5}}=\frac{1}{5}$12 ×AC×BC=12 ×25 ×15 =15  

add comment modified 11 months ago by David Robinson   • written 11 months ago by David Robinson  
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