+0

# triangle

0
7
1
+1123

Find the inradius of trianlge ABC.

Find the circumradius of triangle ABC.

Dec 18, 2023

#1
+259
+1

Calculate the semiperimeter (s) of the triangle:

s = (a + b + c) / 2 = (1 + sqrt(2) + sqrt(2)) / 2 = 1 + sqrt(2)

Use Heron's formula to find the area (K) of the triangle:

K = sqrt(s * (s - a) * (s - b) * (s - c))\

K = sqrt((1 + sqrt(2)) * (0 + sqrt(2)) * (1 + sqrt(2)) * (0 + sqrt(2)))

K= sqrt(2^(3 + 1/2)) = 2 * sqrt(2^(1/2)) = 2

Finally, calculate the inradius (r) using the formula:

r = K / s r = 2 / (1 + sqrt(2))

Use the Law of Cosines to find the angle A of the triangle:

cos(A) = (b^2 + c^2 - a^2) / (2bc

cos(A) = ((sqrt(2))^2 + (sqrt(2))^2 - 1^2) / (2 * sqrt(2) * sqrt(2))

cos(A) = 0 A = pi/2

Since angle A is a right angle, the circumradius (R) is simply half the hypotenuse:

R = a / 2 R = 1 / 2

Therefore, the inradius of triangle ABC is 2 / (1 + sqrt(2)) and the circumradius is 1 / 2.

Dec 18, 2023

#1
+259
+1

Calculate the semiperimeter (s) of the triangle:

s = (a + b + c) / 2 = (1 + sqrt(2) + sqrt(2)) / 2 = 1 + sqrt(2)

Use Heron's formula to find the area (K) of the triangle:

K = sqrt(s * (s - a) * (s - b) * (s - c))\

K = sqrt((1 + sqrt(2)) * (0 + sqrt(2)) * (1 + sqrt(2)) * (0 + sqrt(2)))

K= sqrt(2^(3 + 1/2)) = 2 * sqrt(2^(1/2)) = 2

Finally, calculate the inradius (r) using the formula:

r = K / s r = 2 / (1 + sqrt(2))

Use the Law of Cosines to find the angle A of the triangle:

cos(A) = (b^2 + c^2 - a^2) / (2bc

cos(A) = ((sqrt(2))^2 + (sqrt(2))^2 - 1^2) / (2 * sqrt(2) * sqrt(2))

cos(A) = 0 A = pi/2

Since angle A is a right angle, the circumradius (R) is simply half the hypotenuse:

R = a / 2 R = 1 / 2

Therefore, the inradius of triangle ABC is 2 / (1 + sqrt(2)) and the circumradius is 1 / 2.

BuiIderBoi Dec 18, 2023