Question

Seth is using the figure shown below to prove the Pythagorean Theorem using triangle similarity: In the given triangle PQR, angle P is 90° and segment PS is perpendicular to segment QR. The figure shows triangle PQR with right angle at P and segment PS. Point S is on side QR. Part A: Identify a pair of similar triangles. (2 points) Part B: Explain how you know the triangles from Part A are similar. (4 points) Part C: If RS = 4 and RQ = 16, find the length of segment RP. Show your work. (4 points)

Answer 1

Answer:

Part A: $\triangle RPQ \sim \triangle RSP$

Part B. All angles are same, so the triangles are similar.  

Part C. RP = 8

Step-by-step explanation:

We are given a right angled triangle $\triangle RPQ$ with $\angle P = 90^\circ$.

PS is perpendicular to the hypotenuse RQ of $\triangle RPQ$ and S lies on RQ.

Part A:

To identify the pair of similar triangles.

$\triangle RPQ \sim \triangle RSP$.

Part B:

To identify the type of similarity.

Kindly refer to the image attached in the answer area.

Let us consider the triangles $\triangle RPQ \ and\ \triangle RSP$.

$\angle RSP =\angle RPQ =90^\circ$

Also, $\angle R$ is common to both the triangles under consideration.

Now, we can see that two angles of two triangles are equal.

So, third angle of the two triangles will also be same.

i.e. All three angles of two triangles $\triangle RPQ \ and\ \triangle RSP$ are equal to each other.

So, by A-A-A (Angle - Angle - Angle) similarity, we can say that $\triangle RPQ \sim \triangle RSP$.

Part C:

RS = 4

RQ = 16, Find RP.

There is one property of similar triangles that:

The ratio of corresponding sides of two similar triangles is equal.

i.e.

$\dfrac{RS}{RP} = \dfrac{RP}{RQ}\\\Rightarrow RP ^2 = RS \times RQ\\\Rightarrow RP ^2 = 4 \times 16\\\Rightarrow RP ^2 = 64\\\Rightarrow \bold{RP = 8\ units}$