Question

select the correct answer.
what is the area, in square centimeters, of the triangle shown?
triangle with right angle at b, angles at a and c are 45°, hypotenuse ac = 24 cm
options:
a. 144
b. 216
c. 432
d. 288

Explanation:

Step1: Identify triangle type

The triangle has a right angle and two \(45^\circ\) angles, so it's an isosceles right triangle. Thus, \(AB = BC\).

Step2: Find legs using trigonometry

For an isosceles right triangle with hypotenuse \(c = 24\) cm, let the legs be \(a\) and \(b\) ( \(a = b\) ). Using \(\sin(45^\circ)=\frac{a}{c}\), we have \(a = c\sin(45^\circ)=24\times\frac{\sqrt{2}}{2}=12\sqrt{2}\) cm? Wait, no—wait, in an isosceles right triangle, the legs are equal, and hypotenuse \(c = a\sqrt{2}\), so \(a=\frac{c}{\sqrt{2}}=\frac{24}{\sqrt{2}} = 12\sqrt{2}\)? Wait, no, that's wrong. Wait, actually, in a right - isosceles triangle, if the hypotenuse is \(h\), then each leg \(l\) satisfies \(l=\frac{h}{\sqrt{2}}\), but also, the area of a right triangle is \(\frac{1}{2}\times\text{leg}_1\times\text{leg}_2\). Since it's isosceles, \(\text{leg}_1=\text{leg}_2\). Alternatively, we can use the formula for the area of a right - isosceles triangle: if hypotenuse is \(h\), area \(A=\frac{h^{2}}{4}\). Wait, let's check. Let the legs be \(x\). Then by Pythagoras, \(x^{2}+x^{2}=h^{2}\), so \(2x^{2}=h^{2}\), \(x^{2}=\frac{h^{2}}{2}\), and area \(A = \frac{1}{2}x\times x=\frac{x^{2}}{2}=\frac{h^{2}}{4}\). So \(h = 24\), then \(A=\frac{24^{2}}{4}=\frac{576}{4}=144\)? Wait, no, that's not right. Wait, no, I made a mistake. Wait, in a right - isosceles triangle, the legs are equal, and if the hypotenuse is \(c\), then each leg \(a=\frac{c}{\sqrt{2}}\). Then area \(A=\frac{1}{2}\times a\times a=\frac{1}{2}\times\frac{c^{2}}{2}=\frac{c^{2}}{4}\). Wait, \(c = 24\), so \(A=\frac{24^{2}}{4}=\frac{576}{4}=144\)? But let's do it another way. Let's use trigonometry correctly. In a right triangle, \(\sin(45^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), so if we take angle \(C = 45^{\circ}\), then the opposite side to \(C\) is \(AB\), and hypotenuse is \(AC = 24\). So \(AB=AC\times\sin(45^{\circ})=24\times\frac{\sqrt{2}}{2}=12\sqrt{2}\), and \(BC = AB = 12\sqrt{2}\) (since it's isosceles). Then area \(A=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times12\sqrt{2}\times12\sqrt{2}=\frac{1}{2}\times144\times2 = 144\) square centimeters. Wait, that's correct. Alternatively, we can think that in an isosceles right triangle, the legs are equal, and if we consider the two non - hypotenuse sides as the base and height. Let's re - derive. Let the legs be \(a\). Then by Pythagoras, \(a^{2}+a^{2}=24^{2}\), \(2a^{2}=576\), \(a^{2}=288\), \(a = 12\sqrt{2}\). Then area \(A=\frac{1}{2}\times a\times a=\frac{1}{2}\times288 = 144\) square centimeters.

Answer:

A. 144