Question

What is the area of the figure:

Answer 1

For this case we have that the area of the triangle is given by:

$A = \frac {b * h} {2}$

Where:

b: It's the base

h: It's the height

We have to:

$cos (45) = \frac {b} {24}\\b = 24 * cos (45)\\b = \frac {\sqrt {2}} {2} * 24\\b = 12 \sqrt {2}$

The atura will be given by:

$sin (45) = \frac {h} {24}\\h = 24 * sin (45)\\h = \frac {\sqrt {2}} {2} * 24\\h = 12 \sqrt {2}$

So, the area is:

$A = \frac {12 \sqrt {2} * 12 \sqrt {2}} {2}\\A=\frac{(12\sqrt{2})^2}{2}\\A = 144$

Answer:

144

Answer 2

Answer:

Area=144

Step-by-step explanation:

In right triangle ABC,

$\sin\left(45^o\right)=\frac{BC}{AB}$

$\frac{1}{\sqrt{2}}=\frac{BC}{24}$

$\frac{1}{\sqrt{2}}=\frac{Height}{24}$

$\frac{24}{\sqrt{2}}=Height$

$Height=\frac{24}{\sqrt{2}}$

similarly

$\cos\left(45^o\right)=\frac{AC}{AB}$

$\frac{1}{\sqrt{2}}=\frac{AC}{24}$

$\frac{1}{\sqrt{2}}=\frac{Base}{24}$

$\frac{24}{\sqrt{2}}=Base$

$Base=\frac{24}{\sqrt{2}}$

Then area of right triangle $=\frac{1}{2}\left(Base\right)\left(Height\right)$

$=\frac{1}{2}\left(\frac{24}{\sqrt{2}}\right)\left(\frac{24}{\sqrt{2}}\right)$

$=\frac{576}{4}=144$

Hence Area=144