anonymous
  • anonymous
plse help me to solve this sum
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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Jim766
  • Jim766
|dw:1370548669952:dw| the two triangles are right triangles, use Pythagorean theorem
Jim766
  • Jim766
they are asking you to find the area, not the sum..... find the area for each triangle find the area for the rectangle in the middle and add the three
Jim766
  • Jim766
do something similar for the 2nd part create a parallelogram and 1 triangle

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Jim766
  • Jim766
|dw:1370548971613:dw|
Jim766
  • Jim766
make sense?
anonymous
  • anonymous
yes , let me solve again thanks
Jim766
  • Jim766
opps the first one is screwed up.... those right triangles cant have the same base if they have different hypotenuse their bases together has to be 15, but the longer hypotenuse has to have a longer base
Jim766
  • Jim766
|dw:1370550901931:dw| use this for the 1st one the height for each triangle should be the same
anonymous
  • anonymous
why not 7.5
Jim766
  • Jim766
the lines drawn down are perpendicular to the bottom base. if one hypotenuse is 14 and the other hypotenuse is 13 and the heights are the same the base of each triangle must be different
Jim766
  • Jim766
if you use 7.5 you will get a different height for each triangle, which is not possible if that is a rectangle in the middle. Trapezoid the bases have to be parallel
anonymous
  • anonymous
what would be the height of traingles
Jim766
  • Jim766
c^2 - a^2 = b^2 14^2-8.4^2 = 11.2 13^2 -6.6^2 = 11.2
Jim766
  • Jim766
does that make sense
Jim766
  • Jim766
I get area of 196
Jim766
  • Jim766
in the 2nd one your triangle will have a height of 11.2, and a base of 15 parallelgram will have width of 10 and length of 14
anonymous
  • anonymous
|dw:1370551491584:dw| \[area of trapezoid=\frac{ 25+10 }{ 2 }h=17.5h\] also area=10*h+\[\frac{ 1 }{ 2 }*15*h\] \[area=\frac{ 20+15 }{2 }h=17.5h\] area in both cases is same.
anonymous
  • anonymous
|dw:1370552411543:dw| \[h ^{2}=13^{2}-x ^{2},\] \[also h ^{2}=14^{2}-\left( 15-x \right)^{2}\] \[169-x ^{2}=196-\left( x ^{2} -30x+225\right)\] \[169-x ^{2}=196-x ^{2}+30x-225\] 169-196+225=30x 30x=198 \[x=\frac{ 198 }{ 30 }=6.6\] now you can find the h and area

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