if the perimeter of the isosceles right angled tringle is 8cm,what is its area?

Answer 1

The angles bust be 45 degrees each other than the right angle.
So the ratio's of the sides are 1:1:√2
(1+ 1+ √2)x=8
x=8/(2+√2)
x=4(2-√2)=8-4√2 per side adjacent to th 45 d angle

The side is both a base and height so
A=(1/2)(8-4√2)^2

A=16(3-2√2)=2.745 + cm^2

Answer 2

Let the hypotenuse of the triangle have length H and the two other sides have length A and B.

Now the triangle is isosceles, which means that two of its sides have equal length. But H is bigger than both A and B: so it must be A and B that are equal.

By Pythagoras' Theorem,

H^2 = A^2 + B^2
= 2 * A^2 (because A^2 = B^2)
H = sqrt(2) * A (taking square roots)

Now the perimeter is the sum of the lengths of the three sides, which equals 8 cm. So

A + B + H = 8 cm
A + A + sqrt(2) * A = 8 cm (because A = B)
A * (2 + sqrt(2)) = 8 cm

A = 8 / (2 + sqrt(2)) cm
= 4 * sqrt(2) / (sqrt(2) + 1) cm
= 4 * sqrt(2) * (sqrt(2) - 1) cm (multiplying top and bottom by (sqrt(2) -1)).

Now the area of a right-angled triangle is equal to 1/2 * A * B.

Area = 1/2 * A * B
= 1/2 * A^2 (because A = B)
= 1/2 * 16 * 2 * (sqrt(2) - 1)^2 cm^2
= 16 * (2 - 2sqrt(2) +1) cm^2
= 16 * (3 - 2sqrt(2)) cm^2

Answer 3

Since the triangle is isosceles, two of the legs are the same length. Call this length 'a'. Call the hypoteneus 'b'.
We know that the sum of the sides is 8,
So a+a+b = 8.
Since the triangle is a right triangle we can use the Pythagorean Theorem relating the two smaller sides to the hypoteneus:
a^2 + a^2 = b^2
Now we have two equations with two variables. Solve for either variable and substitue in the other equation.
One way of doing it (sqrt indicates "square root"):
2*a^2 = b^2,
so sqrt( 2*a^2 ) = sqrt( b^2 ),
sqrt(2)*a = b.
Substitute into the other equation:
a + a + sqrt(2)*a = 8.
Factor out the 'a':
a * ( 2 + sqrt(2) ) = 8.
Thus a = 8 / (2+sqrt(2))
Now you should easily be able to find 'b'!

Answer 4

let length of hypotenuse be L.
& that of the two equal sides ie, the base & the perpendicular be = Xcm.
As, the length of the hypotenuse is always greater than the base or the perpendicular, The hypotenuse can not be one of the equal sides.

However, By question,
L+2X = 8...........(1)
We know since the triangle is a right angled one, according to Pythagoras' Theorem,
L²=X²+X²...............(2)

From (1) we ahve,
L= 8-2X
or, L² = (8-2X)²
or, L² = 64 - 32X+4X²
Coimpairing with (2),
64 - 32X+4X² = 2X²
or, 64 - 32X+2X²=0
or, X²-16X + 32 = 0
or, (X-8)² - 32 =0
or, (X-8)² - (4√2)² =0
or, (X-8-4√2)(X-8+4√2)=0

so either X=8+4√2 or X= 8-4√2, But X can not be 8+4√2 because in that case the perimeter will be greater than 8 cm.
So.
X=8-4√2

Hence area of the triangle= 0.5 * Base * Perpendicular
=0.5 * ( 8-4√2)²
=0.5 (64 - 64√2+32) cm²
= 0.5 * (96 - 64√2) cm ²

=2.745166 cm²

Answer 5

Let s = one of the equal legs
Then hypotenuse must be s*sqrt(2)
So perimeter = s+s+s*sqrt(2) = 2s+s*sqrt(2) = s(2+ sqrt(2)
s(2+ sqrt(2) = 8
s= 8/[(2+sqrt(2)]
Area = s^2/2 = {8/[(2+sqrt(2)]}^2/2
A= {64/[4+4sqrt(2) +2]}/2 = 32/(6+4sqrt(2)) = 16/(3+2sqrt(2))
A = 16(3-2sqrt(2))/[(3+2sqrt(2))(3-2sqrt(2))]
A= [48 -32sqrt(2)] cm^2
A = approximately 2.75 cm^2

Answer 6

consider the side of a triangle as 'a'.
as it is isosceles 2 sides r equal.
hyotenuse=sqrt(2a^2)=sqrt(2)a=1.44a
perimeter=2a+b=8
3.44a=8
a=2.4(approx)
area=2.4*2.4/2
=2.88

Answer 7

its area=product of the legs
let one ofthe legs be x
thehypotenuse=rt(x^2+x^2)
=rt2x^2
perimeter=sum of the equal sides and the hypotenuse
x+x+rt(2x^2)=8
rt(2x^2)=8-2x
2x^2=(8-2x)^2
2x^2=64-32x+4x^2
2x^2-32x+64=0
x^2-16x+32=0
x=[16+/-rt(256-128)]/2
so one leg=2.64
area =(2.64)^2
=6.96 cm^2

Answer 8

mmm
x+x+sqrt(2)*x=8
or, x(2+sqrt(2))= 8
or x= 8/(2+sqrt(2))

area= 1/2*x^2
=1/2*(8/(2+sqrt(2)))^2
=2.74