Mistake Master
The hypotenuse sits alone; the ratios are fixed
One theorem and two memorized triangles cover this whole topic. The arithmetic is squares and roots. The points slip away when the hypotenuse takes a leg's seat and the squares add instead of subtract, when the special-triangle factors cross wires, and when the work stops at $c^2$ or at a side the question never asked for. Seat the hypotenuse first.
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What this topic is about
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Right triangles run on one theorem and two memorized shapes. The points ride on three skills: putting the hypotenuse in the right seat of $a^2 + b^2 = c^2$, keeping the special-triangle ratios straight, and finishing the problem, taking the square root and answering the quantity that was asked.
§2
The hypotenuse sits alone
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The side opposite the right angle, always the longest, is $c$, and it sits alone on its side of the theorem. Which sides are known decides whether the squares add or subtract.
- Two legs known: ADD their squares, then root, to get the hypotenuse.
- Hypotenuse and one leg known: SUBTRACT squares, then root, to get the other leg.
- In word problems, the slanted thing (ladder, ramp, wire) is usually the hypotenuse.
Worked example. A $25$-foot ladder leans against a wall with its base $7$ feet out. How high does it reach?
The ladder is the slanted side: $c = 25$. The height is a leg.
$$\sqrt{25^2 - 7^2} = \sqrt{576} = 24 \text{ feet}.$$ Adding the squares treats the ladder as a leg and makes the wall taller than the ladder.
§3
The two special triangles
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Two angle patterns come with fixed side ratios, and the SAT expects them recalled, not re-derived.
- $45$-$45$-$90$: legs equal, hypotenuse $= \text{leg} \cdot \sqrt{2}$.
- $30$-$60$-$90$: hypotenuse $= 2 \cdot$ SHORT leg; long leg $= \text{short leg} \cdot \sqrt{3}$.
- Going from hypotenuse back to a leg DIVIDES by the factor.
Worked example. A $45$-$45$-$90$ triangle has a hypotenuse of $14$. Find each leg.
Leg $= \dfrac{14}{\sqrt{2}} = 7\sqrt{2}$.
$7\sqrt{2} \approx 9.9$, shorter than the hypotenuse, as a leg must be. Multiplying instead gives $14\sqrt{2} \approx 19.8$, longer than the hypotenuse, which no leg can be.
§4
Finish the problem
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The theorem's natural output is $c^2$, not $c$, and many questions want something past the sides: a perimeter, an area, a diagonal's use. Stopping early is this topic's cheapest lost point.
- After $c^2$, the square root still has to act.
- Perimeter adds ALL THREE sides; area is half the product of the LEGS.
- Familiar triples ($3$-$4$-$5$, $5$-$12$-$13$, $8$-$15$-$17$) scale by any factor.
Worked example. A right triangle has legs $9$ and $12$. Find its perimeter.
$\sqrt{81 + 144} = \sqrt{225} = 15$, the $3$-$4$-$5$ triple scaled by $3$.
$$9 + 12 + 15 = 36.$$ Stopping at $15$ answers a different question; so does the area, $54$.
§5
Squares, rectangles, and diagonals
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A square's diagonal makes a $45$-$45$-$90$ triangle with two sides; a rectangle's diagonal makes a right triangle with its length and width. The theorem travels into every figure with a corner.
Seat the hypotenuse first and let it decide add-or-subtract, recall the two special ratios exactly, and after the root, keep going until the asked quantity, side, perimeter, or area, is in hand.
§6
Three patterns that cost real points
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Three patterns recur on right-triangle questions. They are the same ones the diagnostic routes on.
The hypotenuse takes a leg's seat.
Squares get added when the hypotenuse is known, or the sides get combined without squaring at all, producing a leg longer than the hypotenuse.
Fix. Find the side opposite the right angle first; that is $c$, alone on its side. Known hypotenuse means subtract; two known legs mean add.
The special-triangle factors cross wires.
The $45$-$45$-$90$ triangle gets the doubling rule, the $\sqrt{2}$ multiplies when it should divide, or the doubling starts from the long leg.
Fix. Anchor both patterns to their short side: leg times $\sqrt{2}$ to the $45$-$45$-$90$ hypotenuse; SHORT leg doubled to the $30$-$60$-$90$ hypotenuse.
The work stops one move early, or on the wrong quantity.
$c^2$ gets reported instead of $c$, or the hypotenuse answers a perimeter question, or the perimeter answers an area question.
Fix. After the Pythagorean step, reread the ask: root to a length, then continue to the perimeter or area if that is what was named.
Ten quick checks across the patterns: add-or-subtract with the theorem, both special triangles in both directions, diagonals, and finishing to perimeter or area. Pick or type your answer, then check. Progress is saved.