A square is inscribed in a right angled triangle with legs p and q and has a common right angle with triangle. The diagonal of the square is given by

Question

Prepp Team · Accepted Answer

Finding the Diagonal of a Square Inscribed in a Right Triangle

This problem asks us to find the diagonal of a square that is inscribed in a right-angled triangle. The square shares the right angle with the triangle, and the legs of the triangle are given as $p$ and $q$.

Understanding the Setup

Let's consider the right-angled triangle as ABC, with the right angle at vertex A. Let the legs be AB of length $q$ and AC of length $p$. A square, let's call it ADEF, is placed such that vertex A of the square coincides with vertex A of the triangle. Side AD lies along AB, and side AE lies along AC. Vertex F of the square lies on the hypotenuse BC.

Let the side length of the square be $s$. Then, AD = AE = EF = FD = $s$.

Using Coordinate Geometry Approach

We can place the triangle on a coordinate plane. Let vertex A be at the origin $(0,0)$. Since AB is along the x-axis and AC is along the y-axis, vertex B will be at $(q, 0)$ and vertex C will be at $(0, p)$.

The hypotenuse BC is a line segment connecting $(q, 0)$ and $(0, p)$. The equation of the line passing through these two points can be found using the two-point form or the intercept form.

Using the intercept form $\frac{x}{a} + \frac{y}{b} = 1$, where $a$ is the x-intercept and $b$ is the y-intercept:

The x-intercept is $q$.
The y-intercept is $p$.

So, the equation of the line BC is $\frac{x}{q} + \frac{y}{p} = 1$.

Now, consider the square ADEF with side length $s$. Vertex A is at $(0,0)$. AD lies along the x-axis (on AB), so D is at $(s, 0)$. AE lies along the y-axis (on AC), so E is at $(0, s)$. Vertex F, which completes the square, will have coordinates $(s, s)$ because its distance from AD (on x-axis) is $s$ and its distance from AE (on y-axis) is $s$.

Finding the Side Length of the Square

Since vertex F $(s, s)$ lies on the hypotenuse BC, its coordinates must satisfy the equation of the line BC:

$\frac{s}{q} + \frac{s}{p} = 1$

Now, we solve this equation for $s$:

$s \left(\frac{1}{q} + \frac{1}{p}\right) = 1$

Combine the terms inside the parentheses by finding a common denominator:

$s \left(\frac{p + q}{pq}\right) = 1$

To isolate $s$, multiply both sides by $\frac{pq}{p + q}$:

$s = \frac{pq}{p + q}$

So, the side length of the inscribed square is $\frac{pq}{p + q}$.

Calculating the Diagonal

The question asks for the diagonal of the square. For any square with side length $s$, the length of the diagonal $d$ is given by $d = s\sqrt{2}$.

Substitute the value of $s$ we found:

$d = \left(\frac{pq}{p + q}\right) \sqrt{2}$

This can be written as:

$d = \frac{\sqrt{2}pq}{p + q}$

Therefore, the diagonal of the inscribed square is $\frac{\sqrt{2}pq}{p + q}$.

Reviewing the Options

Let's compare our result with the given options:

Option 1: $\frac{\ pq}{p+q}$ (This is the side length, not the diagonal)
Option 2: $\frac{\ pq}{2p+q}$ (Incorrect)
Option 3: $\frac{\sqrt 2\ pq}{p+q}$ (Matches our calculated diagonal)
Option 4: $\frac{2\ pq}{p+q}$ (Incorrect)

Our calculated diagonal matches Option 3.

Revision Table: Inscribed Square in Right Triangle

Concept	Description	Formula/Value
Triangle Legs	Lengths of the sides forming the right angle	$p$, $q$
Square Side (s)	Length of the side of the inscribed square sharing the right angle	$\frac{pq}{p+q}$
Square Diagonal (d)	Length of the diagonal of the inscribed square	$s\sqrt{2} = \frac{\sqrt{2}pq}{p+q}$
Hypotenuse Equation	Line equation for the hypotenuse BC	$\frac{x}{q} + \frac{y}{p} = 1$

Additional Information: Similar Triangles Method

Another way to solve this problem is using similar triangles. Let the side of the square be $s$. The vertex F on the hypotenuse creates two smaller right triangles, FDB and FEC, which are similar to the original triangle ABC.

Consider triangle FDB. FD = $s$. DB = $q-s$. Triangle FDB is similar to triangle ABC.

The ratio of corresponding sides is equal. The height of FDB is FD=$s$. The base is DB=$q-s$. The height of ABC is p and the base is q (or vice versa depending on orientation).

If we consider the altitude from A to BC, or use ratios of sides relative to the legs:

Alternatively, draw a line from F parallel to AC, meeting AB at G (which is D). Draw a line from F parallel to AB, meeting AC at E. This forms square ADEF.

Consider the triangle ABC and the line segment EF parallel to AB (if AC is base) or FD parallel to AC (if AB is base). Let's use the orientation where AC is along y and AB along x.

Triangle BDF is similar to triangle BAC. The ratio of corresponding sides must be equal.

$\frac{FD}{AC} = \frac{BD}{AB}$

$\frac{s}{p} = \frac{q-s}{q}$

Cross-multiply:

$sq = p(q-s)$

$sq = pq - ps$

$sq + ps = pq$

$s(q + p) = pq$

$s = \frac{pq}{p+q}$

This method also gives the same side length for the inscribed square, and the diagonal is still $s\sqrt{2} = \frac{\sqrt{2}pq}{p+q}$. Both methods confirm the result.

Prepp Team · Answer

$\frac{\ pq}{p+q}$

Prepp Team · Answer

$\frac{\ pq}{2p+q}$

Prepp Team · Answer

$\frac{2\ pq}{p+q}$

Finding the Diagonal of a Square Inscribed in a Right Triangle

Understanding the Setup

Using Coordinate Geometry Approach

Finding the Side Length of the Square

Calculating the Diagonal

Reviewing the Options

Revision Table: Inscribed Square in Right Triangle

Additional Information: Similar Triangles Method

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Concept	Description	Formula/Value
Triangle Legs	Lengths of the sides forming the right angle	\(p\), \(q\)
Square Side (s)	Length of the side of the inscribed square sharing the right angle	\(\frac{pq}{p+q}\)
Square Diagonal (d)	Length of the diagonal of the inscribed square	\(s\sqrt{2} = \frac{\sqrt{2}pq}{p+q}\)
Hypotenuse Equation	Line equation for the hypotenuse BC	\(\frac{x}{q} + \frac{y}{p} = 1\)