There is a property that the right angled triangle has the largest area.
Now, we are given that two sides are equal. If two sides are equal then the angles opposite them will also be equal.
So we get a 45-45-90 right angled triangle.
Now, the base and perpendicular = 2.
So, the area = \(\frac{1}{2} \times 2 \times 2 = 2\)
2 is the largest area we can get.
Now, if the third side = 0, then essentially we get a line of length 2. and the area of triangle becomes 0.
So, the minimum area has to be greater than 0. and maximum area is 2.
The values of area possible lie between 0 and 2.
\(0 < Area
< 2\)
OA, A,B,C,Dkaterjigeorge wrote:
Carcass wrote:
In triangle ABC,AB = AC= 2. Which of the following could be the area of the triangle ABC ?
Indicate all possible areas❑ 0.5
❑ 1.0
❑ 1.5
❑ 2.0
❑ 2.5
❑ 3.0
I have no idea where im going wrong, please correct me.
so I draw a height from A to the base, call it h
call BC=b
we know that 0<b<4
so I took 2 extremes:
1) b=0.1
by Pythagoras, h= sqrt(2^2 - (0.1/2)^2) = 2
so minimum area is 0.09 (approx. equal to 0)
2) b=3.9
h = sqrt (2^2 - (3.9/2)^2) = 0.44
maximum area= 3.9*0.44/2= 0.9
I knowwww that something must be wrong can someone help out pleaseee
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