an isosceles triangle abc is inscribed in a square with a the area of 4 what is the perimeter of triangle abc?

Answer 1

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Let the triangle have one side of the square as its base.

The area of the square is 4
Area = (side)^2
Side = Square root (Area)
Side = Square root (4)
Side = 2

Thus the base of the isosceles triangle is 2.

This isosceles triangle can be split into 2 right angled triangles through the mid-point of the base.

Thus the other 2 sides of the triangle are the hypotenuses of the right angled triangles.

Hypotenuse = Square root ( side1^2 + side2^2 )
Hypotenuse = Square root ( (2/2)^2 + 2^2 )
Hypotenuse = Square root ( 1 + 4 )
Hypotenuse = √5

Other sides of the triangle = √5

Hence the perimeter of the triangle = 2 + √5 + √5
Perimeter = 2 ( 1 + √5 )

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Answer 2

There are several ways of reading this question,

"inscribed in a square with an area of 4"

I take that to mean the area of the square is 4 (square units), so the length of one side is 2 units

If you draw your isosceles triangle (at least two sides equal)inside of a square you notice something interesting, you can split the isosceles triangle in half giving two equal triangles each with with one angle of 90 degrees, which allows you to determine the length of one side. if a = base = 1 unit, and b = height = 2 units then length of 3rd side c = sqrt of (a^2 + b^2)

c = sqrt(1^2 + 2^2) = sqrt 5

Your isosceles has a base a = 2 units, and two sides = sqrt 5 units

Perimeter = a + b + c = 2 +sqrt 5 +sqrt 5

sqrt 5 = 2.236

Perimeter = 6.472 units

Answer 3

I'll assume that you mean the equal sides are greater than 2 units. Therefore, the height of the triangle is 2, and the base is 2. The isosceles sides are each sqrt(5) [c² = a² + b²], so the perimiter of the triangle would be 2 + 2sqrt(5).

There are a lot of isosceles triangles that can be inscribed in a square. You should provide a drawing of your problem via a web link.

Answer 4

component to a triangle is a million/2b*h. because AB=BC, BD is the top, h=40 8. So what's the bottom. because it is isosceles, D is the midpoint of BC. let's call the size of BC = x and the size of AB = y (= length of BC to boot) 2y + x =128 (perimeter of the triangle) or x = 128-2y (x/2)^2 + 40 8^2 = y^2 (Pythag. Thm.) Then change 128-2y in for x contained in the perfect equation above to unravel for y. Use this to discover x (from the fringe equation...it will be a lot less stressful to unravel for x), it is the bottom. section = (a million/2)*40 8* x or 24*x --------------------------------------... 2d problem: Draw the diagonal of the sq. on your drawing. note that that's a diameter chord of the circle so the diagonal of the sq. is 8. To get the size of a side, use pythag. thm. back: a^2 + b^2 = 8^2 yet with the aid of the indisputable fact that's a sq., a and b are the same so that you somewhat have 2*(a^2) = sixty 4 . be certain for a (the size of the fringe of the sq.), and multiply with information from 4 to get the fringe.

Answer 5

If square has an area of 4cm², each side will be of length 2cm.
Assume base of triangle lies on side of square ie length of base of isosceles triangle = 2cm.
Apex of triangle will then bisect opposite side of square.
Let length of equal sides of isosceles triangle be x cm. There are now 3 triangles inside the square.
In two of them the hypotenuse is x cm with sides1cm and 2 cm.
x² = 1² + 2² = 5
x = 5^(1/2) =2.24cm
perimeter of isosceles triangle is 2.24 + 2.24 + 2 cm = 6.48 cm

Answer 6

12

Answer 7

12

Answer 8

The length of one of the equal sides can be found using Pythagoras.

x^2 =2^2 + 1^2
x= sq root of 5 =2.24units.
Perimeter if triangle = 2.24+2.24+2=6.48 units.
The base of the triangle= length of square = root 4 = 2
the apex of the triangle is at the centre of the opposite side of the square.
Draw a diagram...it always makes things clearer

Answer 9

2 + 2 sqrt(5)

The base of triangle is two. And it is made of two right triangle with sides 1 and 2. Therefore hypothenuse is sqrt(5).

Answer 10

Area of 4 what ???