ABC is an isoscelea triangle with AB =AC and its perimeter is 64cm .the altitude from A to BC is 24cm.find the length of AC and BC?
Let AB=AC = x, and the base BC = 2y
Then you have
2x+2y = 64
y^2 + 24^2 = x^2
(32-x)^2 + 24^2 = x^2
x = 25
You might have guessed this right off, since one of the basic right triangles with integer sides is
7-24-25
Of course, a 3-4-5 triangle scaled up by 6 would be 18-24-40 but that gives too large a perimeter.
To find the lengths of AC and BC in an isosceles triangle ABC, we can use the Pythagorean theorem and the fact that the altitude from A to BC divides the base BC into two equal parts.
Let AC be x cm. Since AB = AC, we have AB = x cm as well.
By the Pythagorean theorem, we know that the square of the hypotenuse (BC) is equal to the sum of the squares of the other two sides (AB and AC). Therefore, we have:
BC^2 = AB^2 + AC^2
Since AB = x, the equation becomes:
BC^2 = x^2 + x^2
BC^2 = 2x^2
Taking the square root on both sides, we get:
BC = √(2x^2)
BC = √2x
The perimeter of the triangle is given as 64 cm, so:
AB + AC + BC = 64
x + x + √2x = 64
2x + √2x = 64
Now, we also know that the altitude from A to BC is 24 cm. This forms a right-angled triangle with sides of length 24 cm, BC, and half of AB (which is x/2). Using the Pythagorean theorem again, we can write:
(BC/2)^2 + 24^2 = BC^2
(√2x/2)^2 + 24^2 = (√2x)^2
(2x/2)^2 + 576 = 2x^2
x^2 + 576 = 2x^2
x^2 = 576
x = √576
x = 24
Substituting the value of x back into the equation 2x + √2x = 64, we have:
2(24) + √2(24) = 64
48 + √48 = 64
Simplifying, we get:
√48 = 64 - 48
√48 = 16
√16 * √3 = 16
4√3 = 16
Dividing both sides by 4, we get:
√3 = 4
Therefore, the length of AC is 24 cm, and the length of BC is 4√3 cm.
To find the lengths of AC and BC in an isosceles triangle, we can use the Pythagorean theorem. Let's consider the right-angled triangle formed by the altitude from A to BC:
In the right-angled triangle, the length of the hypotenuse is equal to the length of AC, and one of the other sides is equal to BC.
Using the Pythagorean theorem, we can find the length of AC:
AC^2 = AB^2 + BC^2
AC^2 = (BC/2)^2 + BC^2 [Since AB = AC in an isosceles triangle]
Simplifying the equation:
AC^2 = BC^2/4 + BC^2
AC^2 = (5/4)BC^2
Now, let's find the length of BC:
Perimeter of Triangle ABC = AB + AC + BC
64cm = AB + AC + BC
64cm = BC + BC/2 + BC [Using AB = AC and simplifying]
Combining the common terms:
64cm = (5/2)BC
Now we can solve for BC:
BC = (64cm * 2) / 5
BC = 256/5
BC = 51.2cm
Substituting this value back into the equation for AC^2:
AC^2 = (5/4) * (51.2cm)^2
AC^2 = (5/4) * 2621.44cm^2
AC^2 = 3276.8cm^2
AC = √3276.8cm
AC ≈ 57.2cm
Therefore, the approximate lengths of AC and BC are 57.2cm and 51.2cm, respectively.