Solving the Right Triangle Ratio Problem

The question asks us to find the value of a specific ratio involving side lengths in a right-angled triangle ABC. We are given that ∠C = 90°, points P and Q are on sides AC and BC respectively, and the ratios AP : PC = 1 : 2 and BQ : QC = 1 : 2.

Understanding the Ratios and Side Lengths

Let's use the given ratios to express the lengths of the segments on sides AC and BC.

• For side AC: AP : PC = 1 : 2. Let AP = \(x\) units. Then PC = \(2x\) units. The total length of AC is AP + PC = \(x + 2x = 3x\) units.
• For side BC: BQ : QC = 1 : 2. Let BQ = \(y\) units. Then QC = \(2y\) units. The total length of BC is BQ + QC = \(y + 2y = 3y\) units.

So, we have AC = \(3x\) and BC = \(3y\), where \(x\) and \(y\) are positive values representing parts of the lengths.

Applying the Pythagorean Theorem

Since ΔABC is a right-angled triangle with ∠C = 90°, we can use the Pythagorean theorem to relate the sides.

1. Finding AB²:

In ΔABC, AB² = AC² + BC².

Substituting the lengths we found:

\(AB^2 = (3x)^2 + (3y)^2\)

\(AB^2 = 9x^2 + 9y^2\)

2. Finding AQ²:

Consider the triangle ACQ. This is a right-angled triangle with ∠C = 90°. AQ is the hypotenuse.

AQ² = AC² + CQ²

We know AC = \(3x\) and QC = \(2y\).

\(AQ^2 = (3x)^2 + (2y)^2\)

\(AQ^2 = 9x^2 + 4y^2\)

3. Finding BP²:

Consider the triangle BCP. This is a right-angled triangle with ∠C = 90°. BP is the hypotenuse.

BP² = BC² + CP²

We know BC = \(3y\) and PC = \(2x\).

\(BP^2 = (3y)^2 + (2x)^2\)

\(BP^2 = 9y^2 + 4x^2\)

Calculating the Expression

We need to find the value of \(\frac{AQ^2+BP^2}{AB^2}\).

First, let's find the sum AQ² + BP²:

\(AQ^2 + BP^2 = (9x^2 + 4y^2) + (9y^2 + 4x^2)\)

\(AQ^2 + BP^2 = 9x^2 + 4x^2 + 4y^2 + 9y^2\)

\(AQ^2 + BP^2 = 13x^2 + 13y^2\)

\(AQ^2 + BP^2 = 13(x^2 + y^2)\)

Now, substitute the expressions for AQ² + BP² and AB² into the required ratio:

\(\frac{AQ^2+BP^2}{AB^2} = \frac{13(x^2 + y^2)}{9x^2 + 9y^2}\)

\(\frac{AQ^2+BP^2}{AB^2} = \frac{13(x^2 + y^2)}{9(x^2 + y^2)}\)

Assuming \(x^2 + y^2 \neq 0\) (which must be true for a non-degenerate triangle), we can cancel out the \((x^2 + y^2)\) term.

\(\frac{AQ^2+BP^2}{AB^2} = \frac{13}{9}\)

Thus, the value of the expression \(\frac{AQ^2+BP^2}{AB^2}\) is \(\frac{13}{9}\).

Let's compare this result with the given options:

• 4/9
• 8/3
• 4/3
• 13/9

Our calculated value matches the fourth option.

Revision Table: Key Concepts

Additional Information: Generalization

This type of problem is a classic application of the Pythagorean theorem with ratios. Notice that the final result \(\frac{13}{9}\) does not depend on the specific values of \(x\) or \(y\), meaning it doesn't depend on the exact dimensions of the triangle, only on the ratios given for P and Q.

In a right triangle ABC (at C), if P on AC divides it in ratio m:n (AP:PC = m:n) and Q on BC divides it in ratio p:q (BQ:QC = p:q), we can find a general formula for \(\frac{AQ^2+BP^2}{AB^2}\).

• Let AC = \((m+n)k_1\) and BC = \((p+q)k_2\).
• Then PC = \(nk_1\), AP = \(mk_1\), QC = \(qk_2\), BQ = \(pk_2\).
• \(AB^2 = AC^2 + BC^2 = ((m+n)k_1)^2 + ((p+q)k_2)^2\)
• \(AQ^2 = AC^2 + CQ^2 = ((m+n)k_1)^2 + (qk_2)^2\)
• \(BP^2 = BC^2 + CP^2 = ((p+q)k_2)^2 + (nk_1)^2\)

\(AQ^2 + BP^2 = ((m+n)k_1)^2 + (qk_2)^2 + ((p+q)k_2)^2 + (nk_1)^2\)

\(AQ^2 + BP^2 = k_1^2((m+n)^2 + n^2) + k_2^2((p+q)^2 + q^2)\)

The expression becomes quite complex and usually depends on the specific ratios and possibly the relative lengths \(k_1\) and \(k_2\), unless there's a special case. In our specific problem, the ratios were the same (1:2), and the sides were defined relative to these ratios (AC = 3x, BC = 3y), which allowed simplification.

Our problem corresponds to m=1, n=2, p=1, q=2. If we let \(AC = 3k\) and \(BC = 3l\), then \(x=k\) and \(y=l\). The result \(\frac{13(k^2+l^2)}{9(k^2+l^2)} = \frac{13}{9}\) shows independence from \(k\) and \(l\).