ABC is a triangle right angled at C with BC = a and AC = b. If p is the length of the perpendicular from C on AB, then which one of the following is correct?

Question

Prepp Team · Accepted Answer

Understanding the Right Triangle and Perpendicular

The question asks for the relationship between the sides of a right-angled triangle and the length of the perpendicular drawn from the right-angle vertex to the hypotenuse. We are given a right-angled triangle ABC, where the right angle is at C. The lengths of the sides BC and AC are given as 'a' and 'b', respectively. Let 'p' be the length of the perpendicular from vertex C to the hypotenuse AB.

Let's visualize the triangle and the perpendicular.

Triangle ABC is right-angled at C.
The legs of the right triangle are AC = b and BC = a.
The hypotenuse is AB.
A perpendicular CD is drawn from C to AB, and its length is CD = p.

Calculating the Hypotenuse Length

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. This is known as the Pythagorean theorem.

Applying the Pythagorean theorem to triangle ABC:

\[AB^2 = AC^2 + BC^2\]

Substituting the given values:

\[AB^2 = b^2 + a^2\]

So, the length of the hypotenuse AB is:

\[AB = \sqrt{a^2 + b^2}\]

Relating Area to Find the Formula

The area of a triangle can be calculated using different base and height combinations. In a right-angled triangle, we have two easy ways to calculate the area:

Using the two legs as base and height.
Using the hypotenuse as the base and the perpendicular from the right angle to the hypotenuse as the height.

Method 1: Using the Legs

The area of triangle ABC using legs AC and BC is:

\[\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{height}\]

Using BC as base and AC as height (or vice versa):

\[\text{Area}(\triangle ABC) = \frac{1}{2} \times BC \times AC = \frac{1}{2} \times a \times b = \frac{1}{2} ab\]

Method 2: Using the Hypotenuse and Perpendicular

The area of triangle ABC using the hypotenuse AB as base and the perpendicular CD (with length p) as height is:

\[\text{Area}(\triangle ABC) = \frac{1}{2} \times \text{base} \times \text{height}\]

Using AB as base and CD as height:

\[\text{Area}(\triangle ABC) = \frac{1}{2} \times AB \times CD = \frac{1}{2} \times \sqrt{a^2 + b^2} \times p\]

So,

\[\text{Area}(\triangle ABC) = \frac{1}{2} p \sqrt{a^2 + b^2}\]

Equating the Area Expressions

Since both methods calculate the area of the same triangle, the results must be equal:

\[\frac{1}{2} ab = \frac{1}{2} p \sqrt{a^2 + b^2}\]

Multiply both sides by 2 to simplify:

\[ab = p \sqrt{a^2 + b^2}\]

Squaring Both Sides to Find the Relationship

To get rid of the square root and find a relationship involving squares, we square both sides of the equation:

\[(ab)^2 = (p \sqrt{a^2 + b^2})^2\]

\[a^2 b^2 = p^2 (\sqrt{a^2 + b^2})^2\]

\[a^2 b^2 = p^2 (a^2 + b^2)\]

This equation represents the relationship between the legs (a, b) and the perpendicular from the right angle to the hypotenuse (p) in a right-angled triangle.

Comparing with the Options

Let's compare our derived relationship \(a^2 b^2 = p^2 (a^2 + b^2)\) with the given options:

Option 1: \(a^2 b^2 = p^2 (a^2 + b^2)\)
Option 2: \(a^2 b^2 = p^2 (a^2 - b^2)\)
Option 3: \(2a^2 b^2 = p^2 (a^2 + b^2)\)
Option 4: \(a^2 b^2 = 2p^2 (a^2 + b^2)\)

Our derived relationship matches Option 1.

Alternatively, we can rearrange the result as \(\frac{1}{p^2} = \frac{a^2 + b^2}{a^2 b^2} = \frac{a^2}{a^2 b^2} + \frac{b^2}{a^2 b^2} = \frac{1}{b^2} + \frac{1}{a^2}\). This form \(\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\) is also a common way to express this relationship and is derived directly from \(a^2 b^2 = p^2 (a^2 + b^2)\) by dividing both sides by \(p^2 a^2 b^2\).

Revision Table: Key Formulas

Concept	Formula	Applies to
Pythagorean Theorem	\(c^2 = a^2 + b^2\) (where c is hypotenuse)	Right-angled triangles
Area of Triangle (Base & Height)	\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)	Any triangle
Relationship of Perpendicular (p) to Hypotenuse with Legs (a, b)	\(a^2 b^2 = p^2 (a^2 + b^2)\) or \(\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}\)	Right-angled triangles (p is from right angle to hypotenuse)

Additional Information: Geometric Mean Theorem

The segments created by the altitude (perpendicular from the right angle) on the hypotenuse have specific relationships with the altitude and the legs. Let D be the foot of the perpendicular from C to AB. Then AB is divided into segments AD and DB. In right triangle ABC with altitude CD:

Altitude Rule: The altitude to the hypotenuse is the geometric mean of the two segments it divides the hypotenuse into. \(CD^2 = AD \times DB\) or \(p^2 = AD \times DB\).
Leg Rule 1: Each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. \(AC^2 = AB \times AD\) or \(b^2 = \sqrt{a^2+b^2} \times AD\).
Leg Rule 2: \(BC^2 = AB \times DB\) or \(a^2 = \sqrt{a^2+b^2} \times DB\).

While these rules provide alternative ways to relate the segments and sides, using the area method is often the most straightforward way to derive the formula \(a^2 b^2 = p^2 (a^2 + b^2)\).

Prepp Team · Answer

a 2b 2= p 2(a 2&ndash; b 2)

Prepp Team · Answer

2a 2b 2= p 2(a 2+ b 2)

Prepp Team · Answer

a 2b 2= 2p 2(a 2+ b 2)

Understanding the Right Triangle and Perpendicular

Calculating the Hypotenuse Length

Relating Area to Find the Formula

Method 1: Using the Legs

Method 2: Using the Hypotenuse and Perpendicular

Equating the Area Expressions

Squaring Both Sides to Find the Relationship

Comparing with the Options

Revision Table: Key Formulas

Additional Information: Geometric Mean Theorem

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