The medians of a right triangle which are drawn from the vertices of the acute angles are 5 and √40. The value of the hypotenuse is: A. 10B. 2√40C. √13D. 2√13E. None of these

We are tasked with finding the hypotenuse of a right triangle where the medians drawn from the acute angles are given as 5 and √40. Here's the detailed solution:

### Step 1: Understanding the problem
In a right triangle, the medians are lines drawn from a vertex to the midpoint of the opposite side. The medians for the acute angles are related to the sides of the triangle by the following formula:

For a triangle with sides $a$, $b$, and $c$ (where $c$ is the hypotenuse), the median from any vertex to the opposite side is given by:
$$\text{Median}=\frac{1}{2}\sqrt{2b^2+2c^2-a^2},$$
where $a$, $b$, and $c$ correspond to different sides depending on the vertex considered.

### Step 2: Assign variables
Let the sides of the right triangle be $a$, $b$, and the hypotenuse $c$. The medians are given as:
1. Median from vertex opposite side $b$ = 5.
2. Median from vertex opposite side $a$ = √40.

Using the formula for the medians, we can set up equations for both medians.

#### Equation for the first median (opposite side $b$):
$$5=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}.$$
Squaring both sides:
$$25=\frac{1}{4}(2a^2+2c^2-b^2).$$
Simplifying:
$$\begin{array}{}\text{(1)}& 100=2a^2+2c^2-b^2.\end{array}$$

#### Equation for the second median (opposite side $a$):
$$\sqrt{40}=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}.$$
Squaring both sides:
$$40=\frac{1}{4}(2b^2+2c^2-a^2).$$
Simplifying:
$$\begin{array}{}\text{(2)}& 160=2b^2+2c^2-a^2.\end{array}$$

### Step 3: Solve the system of equations
From Equation (1):
$$100=2a^2+2c^2-b^2.$$
Rearrange to isolate $b^2$:
$$\begin{array}{}\text{(3)}& b^2=2a^2+2c^2-100.\end{array}$$

From Equation (2):
$$160=2b^2+2c^2-a^2.$$
Rearrange to isolate $a^2$:
$$\begin{array}{}\text{(4)}& a^2=2b^2+2c^2-160.\end{array}$$

### Step 4: Substitution
Substitute $b^2$ from Equation (3) into Equation (4):
$$a^2=2(2a^2+2c^2-100)+2c^2-160.$$
Simplify:
$$a^2=4a^2+4c^2-200+2c^2-160.$$
Combine like terms:
$$a^2=4a^2+6c^2-360.$$
Rearrange to isolate $a^2$:
$$\begin{array}{}\text{(5)}& 0=3a^2+6c^2-360.\end{array}$$

### Step 5: Solve for $c$ (hypotenuse)
Divide Equation (5) by 3:
$$\begin{array}{}\text{(6)}& 0=a^2+2c^2-120.\end{array}$$

From the Pythagorean theorem for a right triangle:
$$\begin{array}{}\text{(7)}& a^2+b^2=c^2.\end{array}$$

Substitute $b^2=2a^2+2c^2-100$ (from Equation 3) into Equation (7):
$$a^2+(2a^2+2c^2-100)=c^2.$$
Simplify:
$$a^2+2a^2+2c^2-100=c^2.$$
$$3a^2+2c^2-100=c^2.$$
Rearrange:
$$\begin{array}{}\text{(8)}& 3a^2+c^2-100=0.\end{array}$$

Now solve Equations (6) and (8) simultaneously:
From Equation (6): $a^2=120-2c^2.$
Substitute this into Equation (8):
$$3(120-2c^2)+c^2-100=0.$$
$$360-6c^2+c^2-100=0.$$
$$260-5c^2=0.$$
$$5c^2=260.$$
$$c^2=52.$$
$$c=\sqrt{52}=2\sqrt{13}.$$

### Final Answer:
The value of the hypotenuse is **2√13**. Hence, the correct option is D.