Future- Posted June 3, 2007 Report Share Posted June 3, 2007 need this done by 5:00 P.m. today if possible. The side length of equilateral triangle I is10 feet and the side length of equilateral triangle II is 24 feet. What is the side length of equilateral triangle III, in feet, whose area is the sum of the areas of equilateral triangles I and II? anyone help? ty Link to comment Share on other sites More sharing options...
blade85 Posted June 3, 2007 Report Share Posted June 3, 2007 This is pretty easy, just find the area of triangle 1 and triangle 2 add them then work backwards to get the side length..... Link to comment Share on other sites More sharing options...
Future- Posted June 3, 2007 Author Report Share Posted June 3, 2007 This is pretty easy, just find the area of triangle 1 and triangle 2add them then work backwards to get the side length..... yah, i get confused. dunno how to do area of equil. triangles and i forgot how to do radicals which i think is involved? Link to comment Share on other sites More sharing options...
Con Artist Posted June 3, 2007 Report Share Posted June 3, 2007 use pythagorean theorem to find the area of the triangle. draw a line from the center of a side to the opposite angle and you can easily find the areas of each right triangle. Link to comment Share on other sites More sharing options...
Seto Posted June 3, 2007 Report Share Posted June 3, 2007 this is probably really wrong... but yea here goes lmao ..i havent taken math for about 5 months .... first off find the area of the first two triangles... a of triangle = b x h / 2 find the height of the triangle.. if its equaliteral its base is gunna be 1 side..so now u have to find the height ...if its equaliteral you can use pythagreous theorm ...split draw a line down the middle to form a right angle triangle, u know the base of the right angle triangle is = half the side lenght, and the hypothneus is = side length ..and u have to solve for the height 10^2 = 5^2 + x^2 100=25+x^2 75=x^2 sqrroot75 = 8.66 this will give you the height of the triangle to find the area plug into the equation so it will be 8.66 x 10 / 2 = 43.3 = aTriangleI now to do triangle II 24^2=12^2+x^2 576=144+x^2 432=x^2 sqrroot 432 = 20.78 plug into equation.. 20.78 x 12 / 2 = 124.68 = aTriangleII now u have the areas of the two triangles u can figure out the area if the third triangle by adding the two 124.68 + 43.3 = 167.98 = aTriangleIII now u know that you can plug into the equation A = b x h / 2 so 167.98 = b x h / 2 multiply by to to get rid of the 2 on that side.. now ur left with 335.96 = b x h ----------------------------------Im pretty sure up to here is right... goodluck ...lol you have to be left with only 1 variable so you know h = sqrroot(b^2-1/2b^2) so now ur left with 335.96 = b(sqrroot(b^2-1/2b^2) the square root will cancel out the squares so you're left with 335.96=b(b-1/2b) you expand 335.96=b^2-1/2b^2 you multiply everything by 2 to get rid of the 1/2 so its going to be 671.92=2b^2-b^2 671.92=b^2 sqrroot(671.92)=b and b=25.92 = 26feet b=base = 1 side in a equaliteral triangle so 26feet = ur anwser its probably wrong.. lemme check... this is wrong.. if u do the formula you will get 146.25 as the area of a triangle with 26feet on each side Link to comment Share on other sites More sharing options...
Bape. Posted June 3, 2007 Report Share Posted June 3, 2007 i thought u were getting a 3.89 gpa and that u weren't trying? anyways. find the height of each triangle by using the pythagorean theorem switching around the formukla to c2-b2=a2 don't forget since ur trying to find the height that means ur cutting the base in half so b would be 12 for triangle 2 and 5 for triangle 1 and the forumla would be: (24)2-(12)2=a2 do that with the other triangle after u find the heights find the area and then add them. Link to comment Share on other sites More sharing options...
Future- Posted June 3, 2007 Author Report Share Posted June 3, 2007 thanks guys.. shdw - one of my friends needed help.. i did this 2 years ago Link to comment Share on other sites More sharing options...
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