Why is the maximum area of a triangle with two fixed sides 7 and 10 equal to 35?
I know the area formula is one-half times base times height, but the problem gives two side lengths rather than a height. How can I justify that the maximum happens when those two sides are perpendicular? Is there also a quick SAT-style way to see it without calculus?
16 Answers
Use the side of length 10 as the base. The side of length 7 can contribute at most 7 units of perpendicular height, and it reaches that full height only when th…
Why can the height never be more than 7? That is the part I was missing.
Drop a perpendicular from the opposite vertex to the 10-unit base. That height is only the vertical component of the 7-unit side. Tilting the side makes the ver…
Can I explain it without using the sine formula? I understand base and height better than trig.
Yes. Picture a 7 by 10 rectangle. A diagonal cuts it into two triangles of area 35. If you slant the 7-unit side, the matching parallelogram becomes shorter, so…
If this is multiple choice, is there a safe shortcut using the answer choices?
First establish the upper bound: the height cannot exceed 7, so the area cannot exceed 35. Then show 35 is actually possible by making the sides perpendicular. …
What if the side of length 10 were the hypotenuse of a right triangle? Could that create a larger area?
No. If 10 is the hypotenuse, the other leg is shorter than 10, so using 7 as the base gives an area below one-half times 7 times 10. The maximum uses 7 and 10 a…
I keep mixing up a side length with the height. What should I draw first so I do not make that mistake?
Choose one side as the base, then draw a separate perpendicular from the opposite vertex to that base line. Label that segment h. The 7-unit side equals h only …
People keep saying this should take only a few seconds. Is the fastest-looking method always the best one on the SAT?
Use the shortest method you can still justify. Here, a quick sketch plus the height bound is both fast and solid. A memorized shortcut is risky if you cannot ex…
How do I recognize the same pattern when the side lengths are different?
For two fixed sides a and b, the largest possible area is ab divided by 2, reached when the included angle is 90 degrees. You can see it from height at most b, …
Mentor note: 35 square units is correct for the newly stated problem. The cleanest SAT-level justification is that the perpendicular height cannot exceed the ot…