anonymous
  • anonymous
ABC is an isosceles triangle right angled at b and EDGH is a square where EB:BD is 2:1, the ratio of are of square to are of triangle are is a. 2:5 b.1:10 c.1:3 d.2:3
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
do u have any pics from ?
anonymous
  • anonymous
|dw:1378827923601:dw|
phi
  • phi
missing the critical part of the question what do you mean by the ratio of are is

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phi
  • phi
area of square to triangle ?
anonymous
  • anonymous
u can find the are of squareif u consider eb to be 2x bd to be x its is found to be 5x^2
phi
  • phi
I can see a complicated way to get the answer. |dw:1378830498991:dw|
phi
  • phi
the idea is to solve for x, and then the area of the triangle = (1/2) * (x+2)^2
phi
  • phi
we can find x using the Law of Sines. http://www.mathsisfun.com/algebra/trig-sine-law.html
phi
  • phi
we need to find the sin(C) and set up the ratio \[ \frac{x}{\sin C} = \frac{\sqrt{5}}{\sin 45} \]
anonymous
  • anonymous
thnx once again
anonymous
  • anonymous
how do you solve this 2 unknown one equation situation
phi
  • phi
using the "exterior angle= sum of two opposite angles" A+90 (the exterior angle) = 45 + C and C= A+45 from the info, sinA = 1/sqr(5) and cosA= 2/sqr(5) sin(C)= sin(A+45)= sinA cos45 + sin45 cos A = ((1/sqr(5) + 2/sqr(5)) * 1/sqr(2) sin(C)= 3 * 1/sqr(10)
phi
  • phi
using sin 45 = 1/sqr2 we have \[ \frac{x}{\frac{3}{\sqrt{10}}} = \frac{ \sqrt{5}}{ \frac{1}{\sqrt{2}}}\]
phi
  • phi
we get x=3 (how convenient) the area of the triangle is 1/2 * (x+2)^2 = 25/2 the area of the square is sqr5 * sqr5 = 5 ratio of square to triangle is 5/ (25/2)= 2/5
phi
  • phi
no

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