Isosceles triangles and are not congruent but have the same area and the same perimeter. The sides of have lengths , , and , while those of have lengths , , and . Which of the following numbers is closest to ?
The area of is and the perimeter is 18.
The area of is and the perimeter is .
Thus , so .
Thus , so .
We square and divide 36 from both sides to obtain , so . Since we know is a solution, we divide by to get the other solution. Thus, , so The answer is .
The area is , the semiperimeter is , and . Using Heron's formula, . Squaring both sides and simplifying, we have . Since we know is a solution, we divide by to get the other solution. Thus, , so The answer is .
Triangle , being isosceles, has an area of and a perimeter of . Triangle similarly has an area of and .
Now we apply our computational fortitude.
Plug in to obtainPlug in to obtainWe know that is a valid solution by . Factoring out , we obtainUtilizing the quadratic formula givesWe clearly must pick the positive solution. Note that , and so , which clearly gives an answer of , as desired.
Triangle T has perimeter so .
Using Heron's, we get .
We know that from above so we plug that in, and we also know that then .
We plug in 3 for in the LHS, and we get 54 which is too low. We plug in 4 for in the LHS, and we get 80 which is too high. We now know that b is some number between 3 and 4.
If , then we would round up to 4, but if , then we would round down to 3. So let us plug in 3.5 for b.
We get 67.375 which is too high, so we know that .
The answer is .
For this new triangle, say its legs have length and the base length . To see why I did this, draw the triangle on a Cartesian plane where the altitude is part of the y-axis! Then, we notice that and . It's better to let a side be some variable so we avoid having to add non-square roots and square-roots!!
Now, modify the square-root equation with ; you get , so . Divide by to get . Obviously, is a root as established by triangle ! So, use synthetic division to obtain , upon which , which is closest to (as opposed to ). That's enough to confirm that the answer has to be .
12A 01-02 12A 03-04 12A 05-06 12A 07-08
12A 09-10 12A 11-12 12A 13-14 12A 15-16
12A 17-17 12A 18-19 12A 20-20 12A 21-22
12A 23-24 12A 25-25