Unique Point P

We know that the square has side 25cm, $PA=24$ cm and $PB=7$ cm. We can conclude that the triangle $APB$ is a right angled triangle at $P$ . Let angle $PAB=m$ (in degrees). Then $\sin m =$ perpendicular/ hypotenuse $= 7/25$ . Looking at triangle $PAD$ , if we get angle $PAD$ , then by applying Cosine rule, we can directly get $PD^2$ .

As in the triangle $PAD$ , then $\cos PAD$ is the same as $\cos(90-m)$ . We can use the transformation trigonometric identity, $\cos(90-m)= \sin m$ . Therefore $\cos PAD$ is same as $\sin m$ , that is $7/25$ . Now applying cosine rule to the triangle $PAD$ , we get, $PD^2=PA^2+ AD^2 - 2.PA.AD.\cos(90-m) = 625+576-2\times 25\times 24\times 7/25 =865$ . Therefore , $PD^2= 865$

[Latex edits]

Note that 7^2 + 24^2 = 25^2, therefore the triangle ABP is a right triangle. APB = 90 deg. So sin(BAP) = 7/25, and thus cos(DAP) = 7/25 via the Property of Complementary Angles.

Therefore, we can use the Law of Cosines to determine the value of PD^2. PD^2 = 25^2 + 24^2 - 2(25)(24) cos DAP PD^2 = 625 + 576 - 336 PD^2 = 865

In $\Delta APD$ , $AD=25$ and $AP=24$ . If we let $\theta = m\angle DAP$ , then according to the Law of Cosines,

$PD^2 = 25^2 + 24^2 - 2(25)(24)\cos \theta = 1201-1200\cos \theta$ .

We can determine the value of $\cos \theta$ as follows.

Since $AP^2 + BP^2 = AB^2$ , the converse of the Pythagorean Theorem tells us that $\Delta ABP$ must be a right triangle, with $m\angle APB = 90^\circ$ . Therefore $m\angle PAB + m\angle ABP = 90^\circ$ . We also know, since $ABCD$ is a square, that $m\angle PAB + m\angle DAP = 90^\circ$ . Therefore, $m\angle ABP = m\angle DAP$ , the angle we referred to as $\theta$ . Looking at right $\Delta ABP$ again, we see that

$\cos \theta = \frac{BP}{AB} = \frac{7}{25}$ , and

$PD^2 = 1201-1200\cos \theta = 1201-1200(\frac{7}{25}) = 865$ .

25^2 = 24^2+7^2 ? so angle APB = 90° sin(angle BAP) = 7/25 So cos(angle DAP) = 7/25 , properties of complementary angles, since angle A = 90° By the cosine law: PD^2 = 25^2 + 24^ - 2(25)(24)cos DAP = 625 + 576 - 2(25)(24)(7/25) = 1201 - 336 = 865 Answer:865

Notice that 25^2 = 24^2 + 7^2.

Therefore, the angle APB must be 90 degrees.

Sin (angle BAP) = 7/25

Since angle DAB is a right angle, angle DAP is the complement of angle BAP and the cosine of angle DAP must be 7/25.

Use the Law of Cosines to solve for the value of PD^2.

(PD)^2 = 25^2 + 24^2 + (2)(25)(24)(cos (angle DAP)

Substitute in the value of angle DAP as 7/25

(PD)^2 = 625 + 576 - (2)(25)(24)(7/25) (PD)^2 = 1201 - 336 (PD)^2= 865

Observe that $\displaystyle {PA}^2 + {PB}^2 = {AB}^2$ . Therefore triangle $ABP$ is a right triangle, wich implies that $\angle APB = 90^{\circ}$ .

Let $\theta = \angle PBA$ .

Since $\angle APB = 90^{\circ} \Rightarrow \angle PAB + \theta = 90^{\circ}$ .

Since $\angle BAD = 90^{\circ} \Rightarrow \angle PAB + \angle PAD = \angle PAB + \theta \Rightarrow \angle PAD = \theta$ .

Using the cosine rule, we observe the following:

$7^2 + 25^2 =24^2 + 2(7)(25)\cos \theta \Rightarrow 2(25)\cos\theta = 14$ ... (1)

$24^2 + 25^2 = {PD}^2 + 2(24)(25)\cos\theta \Rightarrow {PD}^2 = 1201 - 2(24)(25)\cos\theta$ ... (2)

By substitution of (1) in (2) we obtain:

${PD}^2 = 1201 - 14(24) = 865$ .

APB is a right triangle (24^2 + 7^2 = 25^2) Forming triangle APD, where we already knew AP and AD. By cosine law, we can obtain PD if we know angle PAD. We knew that angle PAD is complementary with angle PAB. Therefore, cos (angle PAD) = sin (angle PAB) = 7/25. By cosine law, we can compute the value of (PD)^2.

Angle PAD = 90 degrees - Angle PAB = Angle PBA cos(angle PAD) = (1201-PD^2)/1200 cos(angle PBA) = 7/25

(1201-PD^2)/1200=7/25 =>1201-PD^2 = 336 PD^2 = 865

Let E be the foot of the altitude of PE to AB. Also let F be the foot of the altitude of PF to AD. Triangle BPA has perimeter 25+24+7=56. Using Heron formula we find that the area of triangle BPA is 84. That means that 84= AB * PE/2, so PE=168/25. Since ABCD is a square FD=25-(168/25). Using Pythagorean theorem we find that EA^2 = 24^2-(168/25)^2=EF. Now, using again Pythagorean theorem we get: PD^2= PF^2+ FD^2 = 24^2-(168/25)^2 + (25-(168/25))^2=865.

Let \angle\ PAB = \alpha\ and \angle\ PAD = \beta\

From triangle PAB,

a = PA = 24, b = AB = 25 and c = PB = 7

\cos\alpha = \frac{a^2 + b^2 - c^2}{2ab} (Cosine Rule)

\cos\alpha = \frac{24^2 + 25^2 - 7^2}{2 \times 24 \times 25}

\cos\alpha = \frac{576 + 625 - 49}{1200}

\cos\alpha = \frac{24}{25}

\cos\alpha = 0.96

\alpha = 16.26^\circ \beta = 90^\circ - \alpha

\beta = 90^\circ - 16.26^\circ

\beta =73.74^\circ

From triangle PAD

a = PA = 24, b = AD = 25 and c = PD

c^2 = a^2 + b^2 - 2ab\cos\beta

c^2 =24^2 + 25^2 - 2\times24\times25\times\cos73.74 ^ \circ

c^2 = 576 + 625 - 336

c^2 = 865

i.e., PD^2 = 865