Area of Tri.

[humakhan]

The lenth of the hypotenuse of an isoscles right triangle is 30meters. find the area of the triangle. round to the nearest tenth, if necessay.

i made the picture
then i did:

a^2 + b^2 = c^2
? = ? = 30^2
sqrt a^2 + b^2 = sqrt900

so then
sqrta+b = 30
2x = 30
x = 15

so a and b = 15

then offcourse u have to find the area....
1/2 bh
1/2 (15)(15)
1/2 (225)
112 1/2 so this is what i got as my answer.

 

[soroban]

Hello, humakhan!

You fumbled the square root part . . .

The length of the hypotenuse of an isoscles right triangle is 30 meters.
Find the area of the triangle. Round to the nearest tenth, if necessary.

You say you made a sketch.
Did you notice the word "isoceles"? . . . The two legs are the same length!

      *
      | \
      |   \
     a|     \30
      |       \
      |         \
      *-----------*
           a

The area of a triangle is: \(\displaystyle \,A\:=\:\frac{1}{2}\cdot(\text{base})(\text{height})\)

In this problem: \(\displaystyle \,\text{base }=\,a\,\) and \(\displaystyle \,\text{height }= \,a\)

\(\displaystyle \;\;\)Hence: \(\displaystyle \,A\:=\:\frac{1}{2}\cdot a^2\;\;\) [1]


From Pythagorus, we have: \(\displaystyle \,a^2\,+\,a^2\:=\:30^2\;\;\Rightarrow\;\;2a^2\,=\,900\;\;\Rightarrow\;\;a^2\,=\,450\)

\(\displaystyle \;\;\)Substitute into [1]: \(\displaystyle \,A\;=\;\frac{1}{2}\cdot(450)\;=\;225\;\;\) . . . There!

 

[TchrWill]

The length of the hypotenuse of an isosceles right triangle is 30meters. find the area of the triangle. round to the nearest tenth, if necessay.

If the triangle is isosceles, the two les are equal to 30(sqrt2)/2 =21.213

Therefore, the area is 21.213^2/2

 

[galactus]

\(\displaystyle \L\\\sqrt{a^{2}+b^{2}}=30\)

Since a=b

\(\displaystyle \L\\\sqrt{2a^{2}}=30\)

\(\displaystyle \L\\2a^{2}=900\)

\(\displaystyle \L\\a^{2}=450\)

\(\displaystyle \L\\a=b=\sqrt{450}=15\sqrt{2}\)

Area:

\(\displaystyle \L\\\frac{(15\sqrt{2})^{2}}{2}=225\)

Also, the area of an isosceles triangle is given by \(\displaystyle \L\\\frac{1}{2}a^{2}sin{\theta}\)

The angle between the two sides of equal length is \(\displaystyle \L\\\frac{{\pi}}{2}=90\ degrees\)

So, \(\displaystyle \L\\\frac{1}{2}(15\sqrt{2})^{2}sin{\frac{{\pi}}{2}}=225\)

 

[humakhan]

thanks. for your help so very much. i'll remember the square root next time