# RHS

RHS

In triangles, you must have studied about congruency of triangles.

Congruency of triangles means that:

• All corresponding angles are equal.
• All corresponding sides are equal.

However, in order to be sure that the two triangles are congruent, we do not necessarily need to have information about all sides and all angles. We use certain rules to prove the congruency of triangles.

Look at the triangles given below. Can we place these triangles on each other without any gaps or overlaps?

In this mini-lesson, we will explore about RHS congruency criterion by learning about its definition and proof with the help some solved examples and a few interactive questions for you to test your understanding.

Let's begin!

## Lesson Plan

 1 What is RHS Congruence Rule in Triangles? 2 Thinking Out of the Box! 3 Important Notes on RHS 4 Solved Examples on RHS 5 Interactive Questions on RHS

## What Is RHS Congruence Rule in Triangles?

RHS criterion of congruence stands for Right Angle-Hypotenuse-Side (full form of RHS congruence).

RHS congruence theorem states that, if the hypotenuse and side of one right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, the two triangles are congruent.

This rule is only applicable in right-angled triangles.

An important point to note here is that when we keep hypotenuse and any one of the other 2 sides of two right triangles equal, we are automatically getting three similar sides, as all three sides in a right triangle are related to each other and that relation is popularly known as Pythagoras theorem.

\begin{align} \text{hypotenuse}^2=\text{base}^2+\text{perpendicular}^2 \end{align}

So, in two right triangles, if the length of base and hypotenuse are 3 units and 5 units respectively, then perpendicular of both the triangles is of length 4 units.

Let's see how.

\begin{align} \text{hypotenuse}^2=\text{base}^2+\text{perpendicular}^2 \end{align}

\begin{align} 5^2=3^2+\text{perpendicular}^2 \end{align}

\begin{align} 25=9+\text{perpendicular}^2 \end{align}

\begin{align} 25-9=\text{perpendicular}^2 \end{align}

\begin{align} \text{perpendicular}^2=16 \end{align}

\begin{align} \text{perpendicular}=4\ \text{units} \end{align}

Think Tank
1. Under the RHS congruence criterion, we consider two sides and an angle in two right triangles. Under SAS criterion also, we consider two sides and an angle. So, think and tell why do we need to have RHS congruency rule as a separate criterion to prove congruency of triangles?

## What Is the Proof of RHS Congruence Theorem?

Let us do an activity to understand the proof of RHS congruence theorem.

### Proof

Try to draw two triangles $$\triangle ABC$$ and $$\triangle PQR$$ with any one of the angles as $$90^o$$.

Are these triangles congruent?

No.

Now, let's try to keep hypotenuse side equal in both the triangles along with one $$90^o$$ angle.

Let's try to make the hypotenuse side of $$\triangle PQR$$ equals to 10 units.

If we change the hypotenuse of a triangle, other side-lengths will also be changed to maintain the Pythagoras relation between the sides, i.e. $$\text{hypotenuse}^2=\text{base}^2+\text{perpendicular}^2$$

Can you see congruent triangles now?

Still no?

Now, let's keep one more side equal in both the triangles and observe the result.

Are we getting congruent triangles now?

Yes, in the above image, $$\triangle ABC \cong \triangle PQR$$.

We can place both the triangles on each other without any gaps and overlaps.

Hence, $$\triangle ABC \cong \triangle PQR$$ using RHS congruency rule.

### Example and Solution

State and proof whether the given triangles are congruent or not.

In the given triangles, $$\triangle ZXY$$ and $$\triangle PQR$$

$\text{side}\ XZ=\text{side}\ PQ$

$\text{side}\ YZ=\text{side}\ PR$

$\angle ZXY=\angle PQR=90^o$

$$\therefore \triangle ZXY \cong \triangle PQR$$, by RHS congruence criterion.

Important Notes
1. RHS congruency criterion is applicable only in right-angled triangles.
2. Under RHS rule, we consider only the hypotenuse and one corresponding side of the given two right triangles to prove the congruency of triangles.
3. Two congruent triangles are always equal in area.
4. When we place two congruent right-angled triangles on one another, there are no gaps and overlaps. They completely fit each other.

Now, look at some RHS criteria examples for a deeper understanding.

## Solved Examples

 Example 1

In the given isosceles triangle $$\triangle PQR$$, prove that the altitude $$PO$$ bisects the base of the triangle $$QR$$.

Solution

In the given triangle $$\triangle PQR$$, there are two small right-angled triangles formed and those are $$\triangle POQ$$ and $$\triangle POR$$,

Altitude $$PO$$ bisects $$QR$$ when $$OQ=OR$$.

So, let us prove that $$\triangle POQ \cong \triangle POR$$.

$$PQ=PR$$ (given as equal)

$$PO=PO$$ (common)

$$\angle POQ=\angle POR=90^o$$

So, by RHS congruence criterion,

$$\triangle POQ \cong \triangle POR$$

$$OQ=OR$$ (by CPCT)

$$\therefore$$ Altitude of triangle $$\triangle PQR$$ bisects the base $$QR$$ of the triangle.

 Hence Proved
 Example 2

In the given triangle, $$\triangle ABD$$, if $$AC$$ bisects side $$BD$$ and $$CE=CF$$, prove that the area of triangles $$\triangle BCE$$ and $$\triangle DCF$$ are equal.

Solution

We know that area of two congruent triangles is always equal.

So, to prove that the triangles $$\triangle BCE$$ and $$\triangle DCF$$ are equal, we just need to prove that they are congruent triangles.

$$\triangle BCE$$ and $$\triangle DCF$$ are right triangles, in which,

\begin{align} CB=CD\end{align} (as AC bisects BD)

\begin{align}CE=CF\end{align}

\begin{align}\angle CEB=\angle CFD=90^o\end{align}

$$\therefore \triangle BCE \cong \triangle DCF$$ (by RHS congruence criterion)

Hence, $$\triangle BCE$$ and $$\triangle DCF$$ are equal in area.

 Hence Proved

## Interactive Questions on RHS

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

## Let's Summarize

The mini-lesson targeted the fascinating concept of RHS. The math journey around RHS started with what a student already knew and went on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## FAQs on RHS

### 1. What are the Rules of Congruency?

There are 5 main rules of congruency of triangles and those are:

• SSS (side side side) rule
• ASA (angle side angle) rule
• SAS (side angle side) rule
• AAS (angle angle side) rule
• RHS (right-angle hypotenuse side) rule

### 2. What is RHS criterion in triangles?

If the hypotenuse and side of one right-angled triangle are equal to the hypotenuse and the corresponding side of another right-angled triangle, the two triangles are congruent.

### 3. How do you use RHS rule?

Under RHS rule, we show that in two right triangles, the length of the hypotenuse is equal along with the length of another corresponding side of the triangles. If we can prove this, that means the given triangles are congruent, otherwise not.

### 4. What is the difference between SAS and RHS?

In SAS rule, we consider the angle between two sides that we are equal, but in RHS, the placement of angle does not matter as we take hypotenuse and any one of the other two corresponding sides.

### 5. What is the full form of RHS congruence rule?

Full form of RHS is right-angle hypotenuse side.

### 6. Where do we use RHS congruence rule?

RHS rule is used to find the length of the missing side in right triangles, to find the area of triangles, to design buildings and towers, etc.

### 7. How to proof RHS congruence criterion?

RHS congruence criterion can be proved by keeping one side and hypotenuse of two right-angled triangles equal.