Question

In the right ∆ABC, the hypotenuse AB = 17 cm. M is the midpoint of the hypotenuse. Find the legs if PAMC=32 cm and PBMC=25 cm

Answer 1

The length of the legs is 8.64cm and 14.64cm respectively

Explanation:

I've added an attachment to aid my explanation.

At different intervals, I'll be making reference to it.

Given

`AB = 17`

`PAMC = 32`

`PBMC = 25`

From the attachment, we have:

`y + z = AB`

Since, M is the Midpoint

`y = z = \½AB`

Substitute 17 for AB

`y = z = \½ * 17`

`y = z = 8.5`

Also, from the attachment

`v + x + z = PAMC`

`v + x + y = 32`

Substitute 8.5 for y

`v + x + 8.5 = 32`

`v + x = 32 - 8.5`

`v + x = 23.5` --------- (1)

Also, from the attachment

`v + w + z = 25`

Substitute 8.5 for z

`v + w + 8.5 = 25`

`v + w = 25 - 8.5`

`v + w = 17.5` ----------- (2)

Subtract (2) from (1)

`v - v + x - w = 23.5 - 17.5`

`x - w = 6`

Make x the subject

`x = 6 + w`

Apply Pythagoras Theorem:

We have that:

`AB^2 = AC^2 + BC^2`

The above can be replaced with

`17^2 = x^2 + w^2` (see attachment)

`289 = x^2 + w^2`

Substitute 6 + w for x

`289 = (6 + w)^2 + w^2`

`289 = 36 + 12w + w^2 + w^2`

`289 - 36 = 12w + 2w^2`

`253 = 12w + 2w^2`

Reorder

`2w^2 + 12w - 253 = 0`

Solve using quadratic equation:

`w = (-b \± √(b^2 - 4ac))/(2a)`

Where

`a = 2`

`b = 12`

`c = -253`

`w = (-12 \± √(12^2 - 4 * 2 * -253))/(2 * 2)`

`w = (-12 \± √(144 + 2024))/(4)`

`w = (-12 \± √(2168))/(4)`

`w = (-12 \± 46.56)/(4)`

Split:

`w = (-12 + 46.56)/(4)` or
`w = (-12 - 46.56)/(4)`

`w = (34.56)/(4)` or
`w = (-58.56)/(4)`

`w = 8.64` or
`w = -14.64`

But length can't be negative

So:

`w = 8.64`

Recall that:
`x = 6 + w`

`x = 6 + 8.64`

`x = 14.64`

Hence, the length of the legs is 8.64cm and 14.64cm respectively