What Is an Isosceles Right Triangle?
An isosceles right triangle is a triangle that is both isosceles (two equal sides) and right-angled (one 90° angle). The two equal sides are the legs that form the right angle, and the side opposite the right angle is the hypotenuse.
Because the two legs are equal, the two angles opposite them must also be equal. Those two angles share the 90° that is left after the right angle, so each is 45°. That is why this shape is universally called the 45-45-90 triangle: its angles are always 45°, 45°, and 90°, no matter how large or small the triangle is.
Is an Isosceles Right Triangle Always a 45-45-90 Triangle?
Yes, and the reasoning is short. The three angles of any triangle add to 180°. One angle is the right angle, 90°, which leaves 90° for the other two. Since the two legs are equal, the base angles opposite them are equal, so they split that remaining 90° evenly into 45° each. There is no other possibility, so every isosceles right triangle is a 45-45-90 triangle, and every 45-45-90 triangle is an isosceles right triangle. The two names describe the same shape from two angles, one from its sides, one from its angles.
Properties of an Isosceles Right Triangle
The equal-legs-plus-right-angle combination locks in a tidy set of properties:
Two equal legs, one hypotenuse. The legs are the equal sides; the hypotenuse is longer than either and is opposite the 90° angle.
Angles fixed at 45°, 45°, 90°. Always, regardless of size.
Side ratio is always $1 : 1 : \sqrt{2}$. If each leg is $x$, the hypotenuse is $x\sqrt{2}$.
The altitude to the hypotenuse equals half the hypotenuse. It also bisects the hypotenuse and the right angle at once, landing exactly at the midpoint.
It is exactly half of a square. Two copies joined along the hypotenuse rebuild the original square.
The Side Ratio 1 : 1 : √2 (and Where √2 Comes From)
The hypotenuse formula is not a rule handed down from nowhere, it falls straight out of the Pythagorean theorem. With both legs equal to $x$ and the hypotenuse $h$:
$$h^2 = x^2 + x^2 = 2x^2,$$
and taking the positive square root of both sides:
$$h = \sqrt{2x^2} = x\sqrt{2}.$$
So the hypotenuse is always the leg multiplied by $\sqrt{2} \approx 1.414$. Read in reverse, the leg is the hypotenuse divided by $\sqrt{2}$:
$$x = \frac{h}{\sqrt{2}} = \frac{h\sqrt{2}}{2}.$$
That single $\sqrt{2}$ relationship is the whole engine of the 45-45-90 triangle. (The irrationality of $\sqrt{2}$ was itself one of the first great discoveries of Greek mathematics, the proof that it cannot be written as a fraction.)
Area and Perimeter of an Isosceles Right Triangle
Because the two legs are perpendicular and equal, the area formula simplifies more than for a general right triangle.
Area. With each leg equal to $x$, the legs serve as base and height:
$$\text{Area} = \frac{1}{2} \times x \times x = \frac{x^2}{2}.$$
If instead you know the hypotenuse $h$, substitute $x = h/\sqrt{2}$ to get $\text{Area} = \tfrac{h^2}{4}$.
Perimeter. Add the two equal legs and the hypotenuse:
$$\text{Perimeter} = x + x + x\sqrt{2} = 2x + x\sqrt{2} = x(2 + \sqrt{2}).$$
Examples of Isosceles Right Triangle
With the ratio and the formulas in place, here is the 45-45-90 triangle doing real work. The problems build from a direct hypotenuse calculation up to working backward from the area.
Example 1 - Each leg of an isosceles right triangle is 5 cm. Find the hypotenuse
The hypotenuse is the leg times $\sqrt{2}$:
$$h = 5\sqrt{2} \approx 7.07 \text{ cm}.$$
Final answer: $5\sqrt{2} \approx 7.07$ cm.
Example 2 - The hypotenuse of an isosceles right triangle is 10 cm. Find the length of each leg
A common first move is to divide the hypotenuse by 2, giving "5 cm" for each leg. Check that against the side ratio: if both legs were 5, the hypotenuse would be $5\sqrt{2} \approx 7.07$, not 10. So 5 is too small, the slip is dividing by 2 when the relationship is divide by $\sqrt{2}$.
Done correctly:
$$x = \frac{h}{\sqrt{2}} = \frac{10}{\sqrt{2}} = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.07 \text{ cm}.$$
Final answer: each leg is $5\sqrt{2} \approx 7.07$ cm.
Example 3 - Find the area of an isosceles right triangle whose legs are 8 cm each
$$\text{Area} = \frac{x^2}{2} = \frac{8^2}{2} = \frac{64}{2} = 32 \text{ cm}^2.$$
Final answer: 32 cm².
Example 4 - Find the perimeter of an isosceles right triangle with legs of 6 cm
$$\text{Perimeter} = 2x + x\sqrt{2} = 2(6) + 6\sqrt{2} = 12 + 6\sqrt{2} \approx 20.49 \text{ cm}.$$
Final answer: $12 + 6\sqrt{2} \approx 20.49$ cm.
Example 5 - The hypotenuse of an isosceles right triangle is 8 cm. Find its area and perimeter
First the leg: $x = 8/\sqrt{2} = 4\sqrt{2} \approx 5.66$ cm. Then:
$$\text{Area} = \frac{x^2}{2} = \frac{(4\sqrt{2})^2}{2} = \frac{32}{2} = 16 \text{ cm}^2,$$
$$\text{Perimeter} = 2x + h = 8\sqrt{2} + 8 \approx 19.31 \text{ cm}.$$
Final answer: area 16 cm²; perimeter $8 + 8\sqrt{2} \approx 19.31$ cm.
Example 6 - An isosceles right triangle has an area of 50 cm². Find the length of each leg and the hypotenuse
Work backward from the area formula:
$$\frac{x^2}{2} = 50 ;\Rightarrow; x^2 = 100 ;\Rightarrow; x = 10 \text{ cm}.$$
Then the hypotenuse is $h = 10\sqrt{2} \approx 14.14$ cm.
Final answer: each leg is 10 cm; hypotenuse $10\sqrt{2} \approx 14.14$ cm.
Why the Isosceles Right Triangle Matters
This shape earns its keep because its proportions are fixed, which makes it the reference triangle wherever a clean 45° angle is needed.
The 45° set square. Every drafting set includes a 45-45-90 set square, used to draw exact 45° and 90° lines without measuring an angle. Its reliability comes entirely from the fixed 1:1:√2 ratio.
Trigonometry's first exact values. The 45-45-90 triangle is where $\sin 45° = \cos 45° = \tfrac{1}{\sqrt{2}}$ and $\tan 45° = 1$ come from, three of the special-angle values memorised in every trig course. They are read straight off this triangle.
Diagonals of squares. Any square's diagonal is its side times $\sqrt{2}$, the same relationship, because the diagonal cuts the square into two of these triangles. This is how a tile-setter computes a diagonal run or a screen's corner-to-corner length.
Roof and truss design. A symmetric gable roof with a 45° pitch is built on two of these triangles back to back, which is why the rafter length is always the half-span times $\sqrt{2}$.
For a Grade 9 student, the 45-45-90 triangle is the bridge from the Pythagorean theorem into trigonometry, learn its ratio cold and the special angles in trig arrive already half-understood.
Common Errors When Working With Isosceles Right Triangles
Mistake 1: Halving the hypotenuse to get a leg
Where it slips in: Given the hypotenuse, the student divides by 2 instead of by $\sqrt{2}$ to find a leg.
Don't do this: Write leg $= h/2$, for example calling each leg of a 10 cm hypotenuse "5 cm".
The correct way: Divide by $\sqrt{2}$: leg $= h/\sqrt{2} = h\sqrt{2}/2$. A fast check: the leg should be about 0.707 of the hypotenuse, clearly more than half. The memorizer who recalls "divide by 2 for isosceles" is mixing this up with the altitude rule.
Mistake 2: Multiplying the hypotenuse by √2 instead of the leg
Where it slips in: The student reverses the ratio and multiplies the hypotenuse by $\sqrt{2}$ to get a leg.
Don't do this: Write leg $= h\sqrt{2}$, which makes the leg longer than the hypotenuse.
The correct way: The hypotenuse is the longest side, so $h = x\sqrt{2}$ goes leg → hypotenuse (multiply), and hypotenuse → leg divides by $\sqrt{2}$. If a "leg" comes out longer than the hypotenuse, the ratio was applied backward.
Mistake 3: Using the hypotenuse as the height in the area formula
Where it slips in: The student computes area as $\tfrac{1}{2} \times x \times h$, mixing a leg with the hypotenuse.
Don't do this: Treat the hypotenuse as the height of the triangle.
The correct way: The two equal legs are the base and height, so area $= \tfrac{x^2}{2}$. The hypotenuse is never the height here; the altitude to the hypotenuse is a different, shorter segment.
Key Takeaways
An isosceles right triangle has two equal legs and one 90° angle, making it the 45-45-90 triangle with angles 45°, 45°, 90°.
Its side ratio is always $1 : 1 : \sqrt{2}$: the hypotenuse equals a leg times $\sqrt{2}$.
Area is $\tfrac{x^2}{2}$ (or $\tfrac{h^2}{4}$ from the hypotenuse); perimeter is $x(2 + \sqrt{2})$.
To find a leg from the hypotenuse, divide by $\sqrt{2}$, never by 2.
The triangle is exactly half a square and is the source of the special-angle values for 45° in trigonometry.
Practice These Problems to Solidify Your Understanding
Each leg of an isosceles right triangle is 9 cm. Find the hypotenuse and the area.
The hypotenuse of an isosceles right triangle is 14 cm. Find each leg.
An isosceles right triangle has an area of 18 cm². Find each leg and the hypotenuse.
Answer to Question 1: hypotenuse $= 9\sqrt{2} \approx 12.73$ cm, area $= 40.5$ cm². Answer to Question 2: each leg $= 7\sqrt{2} \approx 9.9$ cm. Answer to Question 3: each leg $= 6$ cm, hypotenuse $= 6\sqrt{2} \approx 8.49$ cm. If Question 2 gave 7 cm, check that you divided by $\sqrt{2}$, not 2 (see Mistake 1).
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