Question

quiz - special triangles
8
x
45°
45°
8√2
a) 4√2 b) 16√2
c) 16√6 d) 32
a
b
c
d

Explanation:

Step1: Identify the right triangles

The figure has two right - angled isosceles triangles (since angles are \(45^{\circ}\), \(45^{\circ}\) and \(90^{\circ}\)). In a right - angled isosceles triangle, the legs are equal and the hypotenuse \(h\) is related to the leg length \(l\) by \(h = l\sqrt{2}\).

Step2: Analyze the lower right triangle

The lower right triangle has a hypotenuse? No, wait, the base of the trapezoid is \(8\sqrt{2}\) and the angle at the base is \(45^{\circ}\). The triangle formed at the bottom right is a right - angled isosceles triangle (right angle and \(45^{\circ}\) angle). Let the leg length (the non - hypotenuse sides) be \(y\). Using the formula for the hypotenuse of a right - angled isosceles triangle \(h=y\sqrt{2}\), here \(h = 8\sqrt{2}\), so \(y\sqrt{2}=8\sqrt{2}\), which gives \(y = 8\).

Step3: Analyze the upper left triangle

The upper left triangle is also a right - angled isosceles triangle (right angle and \(45^{\circ}\) angle). The leg of this triangle (which is equal to \(x\)) and the other leg (which is equal to the leg of the lower right triangle, \(y = 8\)) are equal. Wait, no, in the upper left triangle, the two legs are equal (since it's a right - angled isosceles triangle with a \(45^{\circ}\) angle). Wait, actually, the left - most vertical side and the top side \(x\) are legs of a right - angled isosceles triangle, and the lower right triangle has legs equal to the left - most vertical side. Let's re - express:

In a right - angled isosceles triangle, if the hypotenuse is \(H\) and the leg is \(L\), then \(H = L\sqrt{2}\). For the lower right triangle, the base is \(8\sqrt{2}\) (which is the hypotenuse of the lower right isosceles right triangle). So if we let the leg of the lower right triangle be \(a\), then \(8\sqrt{2}=a\sqrt{2}\), so \(a = 8\).

Now, the upper left triangle is also a right - angled isosceles triangle. The vertical side (which is equal to \(a = 8\)) and the top side \(x\) are the legs. Wait, no, the angle at the top is \(45^{\circ}\), so the upper left triangle has legs \(x\) and the vertical side, and since it's a right - angled isosceles triangle, \(x\) is equal to the vertical side. But the vertical side is equal to the leg of the lower right triangle, which is \(8\)? Wait, no, maybe I made a mistake. Wait, the entire figure: the bottom side is \(8\sqrt{2}\), the lower right triangle is a right - angled isosceles triangle with hypotenuse? No, the bottom side is the base, and the two non - parallel sides: the left vertical side and the right slant side. Wait, the triangle formed by the left vertical side, the bottom base, and the diagonal: no, the diagonal divides the trapezoid into two triangles. The right - hand triangle: right angle, \(45^{\circ}\) angle, so it's isosceles right - angled. So the legs of the right - hand triangle are equal. Let the legs be \(m\) (the vertical leg) and \(n\) (the horizontal leg). Then \(m = n\), and the hypotenuse (the side with length \(8\sqrt{2}\)) is related by \(m\sqrt{2}=8\sqrt{2}\), so \(m = 8\). Now, the left - hand triangle: right angle, \(45^{\circ}\) angle, so it's also isosceles right - angled. The legs of the left - hand triangle are \(x\) (the top side) and \(m = 8\) (the vertical side). So since it's isosceles right - angled, \(x=m = 8\)? Wait, no, the options don't have 8. Wait, maybe I messed up the hypotenuse and leg.

Wait, maybe the bottom side \(8\sqrt{2}\) is the leg of a larger triangle? No, let's start over.

The figure is a trapezoid with two right angles (left side) and two \(45^{\circ}\) angles (top and bottom right). The diagonal divides it into two right - angled isosceles triangles.

The lower triangle (with base \(8\sqrt{2}\)): it's a right - angled isosceles triangle (right angle at the left - bottom, \(45^{\circ}\) at the bottom right). So the two legs (the left - bottom vertical leg and the bottom horizontal leg) are equal. Let the length of each leg be \(a\). Then, by the Pythagorean theorem, \(a^{2}+a^{2}=(8\sqrt{2})^{2}\).

\(2a^{2}=64\times2\)

\(a^{2}=64\)

\(a = 8\)

Now, the upper triangle (with top side \(x\)): it's also a right - angled isosceles triangle (right angle at the top - left, \(45^{\circ}\) at the top). The two legs (the top - left horizontal leg \(x\) and the top - left vertical leg) are equal. The top - left vertical leg is equal to the left - bottom vertical leg, which is \(a = 8\). Wait, no, in the upper triangle, the angle at the top is \(45^{\circ}\), so the legs are \(x\) and the vertical side (which is 8). So since it's a right - angled isosceles triangle, \(x\) is equal to the vertical side? But that would make \(x = 8\), but 8 is not an option. Wait, maybe I got the hypotenuse and leg reversed.

Wait, maybe the bottom side \(8\sqrt{2}\) is the leg of the upper triangle? No, let's look at the options. The options are \(4\sqrt{2}\), \(16\sqrt{2}\), \(16\sqrt{6}\), 32. Wait, maybe the entire figure is a combination of two isosceles right triangles, and the base \(8\sqrt{2}\) is the leg of a triangle where the hypotenuse is \(x\sqrt{2}\)? No, wait, maybe the two triangles together form a larger isosceles right triangle?

Wait, another approach: In an isosceles right triangle, the ratio of leg to hypotenuse is \(1:\sqrt{2}\).

Looking at the figure, the bottom side is \(8\sqrt{2}\), and the top side is \(x\). The vertical sides are equal. Let the vertical side be \(h\).

For the lower right triangle (isosceles right triangle), hypotenuse? No, the lower right triangle has a right angle, so the two legs are \(h\) and (let's say) \(b\), and angle \(45^{\circ}\), so \(h = b\). The base of the trapezoid is \(x + b=8\sqrt{2}\)? No, the base of the trapezoid is \(8\sqrt{2}\), and the top is \(x\). The difference between the base and the top is \(8\sqrt{2}-x\), but since the two non - parallel sides are equal (because of the \(45^{\circ}\) angles), and each non - parallel side forms an isosceles right triangle with the vertical drop.

Wait, maybe the height of the trapezoid is \(h\). For the right - hand side, the triangle formed with the base (length \(8\sqrt{2}-x\)) and height \(h\) is an isosceles right triangle, so \(8\sqrt{2}-x=h\). For the left - hand side, the triangle formed with the top \(x\) and height \(h\) is an isosceles right triangle, so \(x = h\).

Substituting \(h=x\) into \(8\sqrt{2}-x=h\), we get \(8\sqrt{2}-x=x\)

\(8\sqrt{2}=2x\)

\(x = 4\sqrt{2}\)

Ah! That makes sense. So we set up the equation:

Since the right - hand side triangle (the one with base \(8\sqrt{2}-x\)) is isosceles right - angled, its base equals its height (\(h\)). The left - hand side triangle (the one with base \(x\)) is also isosceles right - angled, so its base (\(x\)) equals its height (\(h\)). So \(8\sqrt{2}-x=h\) and \(x = h\). Substituting \(h = x\) into the first equation:

\(8\sqrt{2}-x=x\)

\(8\sqrt{2}=2x\)

\(x=\frac{8\sqrt{2}}{2}=4\sqrt{2}\)

Answer:

A) \(4\sqrt{2}\)