what is the area of the figure:

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carlosego carlosego · Answer 1 · 2019-05-09T22:01:19+02:00

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

[tex]A = \frac {b * h} {2}[/tex]

Where:

b: It's the base

h: It's the height

We have to:

[tex]cos (45) = \frac {b} {24}\\b = 24 * cos (45)\\b = \frac {\sqrt {2}} {2} * 24\\b = 12 \sqrt {2}[/tex]

The atura will be given by:

[tex]sin (45) = \frac {h} {24}\\h = 24 * sin (45)\\h = \frac {\sqrt {2}} {2} * 24\\h = 12 \sqrt {2}[/tex]

So, the area is:

[tex]A = \frac {12 \sqrt {2} * 12 \sqrt {2}} {2}\\A=\frac{(12\sqrt{2})^2}{2}\\A = 144[/tex]

Answer:

144

lublana lublana · Answer 2 · 2019-05-09T22:08:31+02:00

Answer:

Area=144

Step-by-step explanation:

In right triangle ABC,

[tex]\sin\left(45^o\right)=\frac{BC}{AB}[/tex]

[tex]\frac{1}{\sqrt{2}}=\frac{BC}{24}[/tex]

[tex]\frac{1}{\sqrt{2}}=\frac{Height}{24}[/tex]

[tex]\frac{24}{\sqrt{2}}=Height[/tex]

[tex]Height=\frac{24}{\sqrt{2}}[/tex]

similarly

[tex]\cos\left(45^o\right)=\frac{AC}{AB}[/tex]

[tex]\frac{1}{\sqrt{2}}=\frac{AC}{24}[/tex]

[tex]\frac{1}{\sqrt{2}}=\frac{Base}{24}[/tex]

[tex]\frac{24}{\sqrt{2}}=Base[/tex]

[tex]Base=\frac{24}{\sqrt{2}}[/tex]

Then area of right triangle [tex]=\frac{1}{2}\left(Base\right)\left(Height\right)[/tex]

[tex]=\frac{1}{2}\left(\frac{24}{\sqrt{2}}\right)\left(\frac{24}{\sqrt{2}}\right)[/tex]

[tex]=\frac{576}{4}=144[/tex]

Hence Area=144