45-45-90 triangles have sides 1:1:sqrt(2) - Memorize this as it comes in very handy!
H = root 2
Therefore the sides which by proportion should be 1 will be
H/root 2 = Root 2 * H /2
Area of a right angle triangle = 1/2 * b * h
b and h are the same here...
Therfore area = 1/2 * (root 2 H/2)^2
= 1/2 * 2H^2/4 = H^2/4
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AREA of 45-45-90 traingle
Source: Beat The GMAT — Problem Solving |
Given that a triangle with 45-45-90 is a special right isosceles triangle, you know that the hypotenuse is equal to X*root2 and the two other sides are equal to X.
Since the hypotenuse is H, the other to sides are both equal to H/(root2).
The area of a triangle is (base * height)/2 so:
(H/(root2))*(H/(root2))/2 is equal to: (H^2)/4
Since the hypotenuse is H, the other to sides are both equal to H/(root2).
The area of a triangle is (base * height)/2 so:
(H/(root2))*(H/(root2))/2 is equal to: (H^2)/4
I don't understand how you get the "4" in the denominator in your answer.
I think if you square (H/root2) and divide this by "2" (since the area is calculated as 1/2 *(H/root2)*(H/root2))
you have the "2" in the denominator.
If you now square the numerator you get ((H*H)/2) this still leaves you with the original "2" in the denominator.
Since we have two fractions combined now, the "2" from the denominator goes up and cancels out with the two from ((H*H)/2) which leaves you with H*H.
Correct me if I am wrong.
I think if you square (H/root2) and divide this by "2" (since the area is calculated as 1/2 *(H/root2)*(H/root2))
you have the "2" in the denominator.
If you now square the numerator you get ((H*H)/2) this still leaves you with the original "2" in the denominator.
Since we have two fractions combined now, the "2" from the denominator goes up and cancels out with the two from ((H*H)/2) which leaves you with H*H.
Correct me if I am wrong.
Yes, I'm sorry but your reasoning is wrong..
Let's have a look at it:
1/2 *(H/root2)*(H/root2)).
Let's take the first piece:
(H/root2)*(H/root2)) is equal to (H^2/2).
This happens because H*H = H^2
and root2*root2= 2
Then, (H^2/2) must be divided by 2, so we can write it as:
(H^2/2)*(1/2) that is equal to (H^2*1)/(2*2)= (H^2)/4
Best regards
Let's have a look at it:
1/2 *(H/root2)*(H/root2)).
Let's take the first piece:
(H/root2)*(H/root2)) is equal to (H^2/2).
This happens because H*H = H^2
and root2*root2= 2
Then, (H^2/2) must be divided by 2, so we can write it as:
(H^2/2)*(1/2) that is equal to (H^2*1)/(2*2)= (H^2)/4
Best regards
(a/Sin A )= (b/Sin B) = (c/Sin C)
where a,b,c are sides of the trinagle and A,B,C are the angles opposite to corresponding sides.
Here H is hypotenuse and corrsponding opposite angle will be 90 degress(Sin 90 degrees =1)
other two sides are equal and side opposite to them is 45 degrees(base =height)
Area is 1/2*base*height =1/2 *base^2
So H = base*2^(1/2)
As a whole area = (H^2)/4
where a,b,c are sides of the trinagle and A,B,C are the angles opposite to corresponding sides.
Here H is hypotenuse and corrsponding opposite angle will be 90 degress(Sin 90 degrees =1)
other two sides are equal and side opposite to them is 45 degrees(base =height)
Area is 1/2*base*height =1/2 *base^2
So H = base*2^(1/2)
As a whole area = (H^2)/4
neophyte
