Question

The figure shows a right triangle ABC and three squares. Find AB.

Answer 1

The right triangle ABC has squares connected to its sides with areas 169, 144, and
`\(x^2 + y^2\)`. Solving, the side lengths of squares 1 and 2 are 13 and 12 units, respectively. The height AB is 5 units.

Without the specific side lengths, let's denote the sides of the squares and the triangle as follows:

- a: side length of square 1

- b: side length of square 2

- c: side length of square 3 (the side connected to AB, the height of the triangle)

- x: one leg of the right triangle (triangle ABC)

- y: the other leg of the right triangle

Now, the relationships among these quantities are given by the areas of the squares:

1.
`\( a^2 = 169 \)` (square 1 is connected to the hypotenuse BC)

2.
`\( b^2 = 144 \)` (square 2 is connected to the base AC)

3.
`\( c^2 = x^2 + y^2 \)` (square 3 is connected to the height AB)

Since
`\( a^2 = 169 \), \( a = 13 \)` (ignoring the negative root since it represents a length).

Similarly, b = 12 because
`\( b^2 = 144 \)`.

Now, the Pythagorean theorem for the right triangle ABC gives us:

`\[ c^2 = x^2 + y^2 \]`

Substitute the known values:

`\[ 13^2 = x^2 + y^2 \]`

Solving for x or y, we get x = 5.

Therefore, the height AB is y = 5 units.