Question

What is the area of the figure:

Answer 1

Area=144

Explanation:

In right triangle ABC,

`\sin\left(45^o\right)=(BC)/(AB)`

`(1)/(√(2))=(BC)/(24)`

`(1)/(√(2))=(Height)/(24)`

`(24)/(√(2))=Height`

`Height=(24)/(√(2))`

similarly

`\cos\left(45^o\right)=(AC)/(AB)`

`(1)/(√(2))=(AC)/(24)`

`(1)/(√(2))=(Base)/(24)`

`(24)/(√(2))=Base`

`Base=(24)/(√(2))`

Then area of right triangle
`=(1)/(2)\left(Base\right)\left(Height\right)`

`=(1)/(2)\left((24)/(√(2))\right)\left((24)/(√(2))\right)`

`=(576)/(4)=144`

Hence Area=144

Answer 2

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=((12√(2))^2)/(2)\\A = 144`

Answer:

144