Answer:
E-F and E-D
C-B and C-D
Step-by-step explanation:
The circle and triangle are as shown. The options are:
By drawing radius lines from the center of the circle to the tangent points B, D, and F, we can divide the triangle into 3 kites. Therefore, only segments that are legs of the same kite are congruent. So the answer must be E-F and E-D, and C-B and C-D.
By inspecting the integrand, the "obvious" choice for substitution would be
<em>u</em> = <em>y</em> + <em>x</em>
<em>v</em> = <em>y</em> - <em>x</em>
<em />
Solving for <em>x</em> and <em>y</em>, we would have
<em>x</em> = (<em>u</em> - <em>v</em>)/2
<em>y</em> = (<em>u</em> + <em>v</em>)/2
in which case the Jacobian and its determinant are
The trapezoid <em>R</em> has two of its edges on the lines <em>x</em> + <em>y</em> = 8 and <em>x</em> + <em>y</em> = 9, so right away, we have 8 ≤ <em>u</em> ≤ 9.
Then for <em>v</em>, we observe that when <em>x</em> = 0 (the lowest edge of <em>R</em>), <em>v</em> = <em>y</em> ; similarly, when <em>y</em> = 0 (the leftmost edge of <em>R</em>), <em>v</em> = -<em>x</em>. So
-<em>x</em> ≤ <em>v</em> ≤ <em>y</em>
-(<em>u</em> - <em>v</em>)/2 ≤ <em>v</em> ≤ (<em>u</em> + <em>v</em>)/2
-<em>u</em> + <em>v</em> ≤ 2<em>v</em> ≤ <em>u</em> + <em>v</em>
-<em>u</em> ≤ <em>v</em> ≤ <em>u</em>
<em />
So, the integral becomes
