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

Answer:

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

Step-by-step 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:

$w = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}$

Where

$a = 2$

$b = 12$

$c = -253$

$w = \frac{-12 \± \sqrt{12^2 - 4 * 2 * -253}}{2 * 2}$

$w = \frac{-12 \± \sqrt{144 + 2024}}{4}$

$w = \frac{-12 \± \sqrt{2168}}{4}$

$w = \frac{-12 \± 46.56}{4}$

Split:

$w = \frac{-12 + 46.56}{4}$ or $w = \frac{-12 - 46.56}{4}$

$w = \frac{34.56}{4}$ or $w = \frac{-58.56}{4}$

$w = 8.64$ 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