# Can you answwr this question ? Q. $△$LMN is an isosceles right-angled triangle. If the square of the hypotenuse of $△$LMN is 32 $c{m}^{2}$, then what is the length of each of the two equal sides?

Please ignore the solution to the first solution I gave:
LN)2 = 32 cm (Hypotenuse 2)
And LM and MN are equal as is an isosceles right angled triangle.
Let us assume both LM and LN to be 'y'

According to Pythagoras Theorem, we can tell that:
(perpendicular2) + (base2) = (hypotenuse2)
So, here ,
(LM)2 + (MN)2 = (Hypotenuse2) = 32 cm 2
y2 + y2 = 32 cm2
=> 2(y2) = 32 cm 2
=> y2 = 32 cm 2 / 2 = 16 cm
=> y = ​√16 cm = 4 cm
As, LM and MN are equal so, the length of each of the 2 equal sides is 4 cm .

• 2
According to pythagoras theorem,
Let a side be a
then,
a2+a2=32
2a2=32
a2=16
a=4cm (Ans)
• 0
Dear Faiz,

(LN) 2 = 32 cm 2 (Hypotenuse)
And, LM = MN as it an isosceles right angled triangle
So, we can assume LM to 'y' and also MN to be 'y'

So, According to Pythagoras Theorem we can write:

(LM2) + (MN)2  = (LN)​2 ... (1)
We have taken both LM and LN to be 'y' as both are equal.
So, By substituting (1), we can tell that:
2 + y 2 = (LN)2
2 + y 2 = (32)2
2(y 2) =  1024 cm
(y 2) = 1024 cm     =  512 cm
2

y = √512 cm​
~ 22.62 cm

As we took y as LM and MN we can say that both are equal so, the lenght of each of 2 equal sides is ~ 22.62 cm

• -1
32 multiplied with 3 = 96
• 2
What are you looking for?