The sides of a triangle are 𝑝, 𝑝 + 1, 2𝑝 − 1 and its area is 2𝑝√10. What is the value of 𝑝?

To solve this problem, we can use Heron's formula for the area of a triangle, which states that the area of a triangle with sides of length a, b, and c is √[s(s - a)(s - b)(s - c)], where s is the semi-perimeter of the triangle, given by (a + b + c) / 2.

Given the sides of the triangle are p, p + 1, and 2p - 1, we can calculate s as follows:

s = (p + (p + 1) + (2p - 1)) / 2 s = (4p) / 2 s = 2p

We are also given that the area of the triangle is 2p√10. Substituting the values of a, b, c, and s into Heron's formula, we get:

2p√10 = √[2p(2p - p)(2p - (p + 1))(2p - (2p - 1))] 2p√10 = √[2p(p)(p - 1)(1)] 2p√10 = √[2p^3 - 2p^2] 2p√10 = p√(2p^2 - 2p)

Dividing both sides by p, we get:

2√10 = √(2p^2 - 2p)

Squaring both sides to get rid of the square root, we get:

40 = 2p^2 - 2p

Rearranging terms, we get:

2p^2 - 2p - 40 = 0

Dividing through by 2 to simplify, we get:

p^2 - p - 20 = 0

This is a quadratic equation in the form ax^2 + bx + c = 0, which can be solved using the quadratic formula x = [-b ± √(b^2 - 4ac)] / (2a). Applying this formula, we get:

p = [1 ± √((1)^2 - 4(1)(-20))] / (2 * 1) p = [1 ± √(1 + 80)] / 2 p = [1 ± √81] / 2 p = [1 ± 9] / 2

This gives us two possible solutions: p = 10/2 = 5 and p = -8/2 = -4. However, since the length of a side of a triangle cannot be negative, we discard the solution p = -4.

Therefore, the value of p is 5.