Mon, 04 May 2026 14:59:47 +0000

A right triangle has one leg 8 cm longer than the other. If the hypotenuse is 20 cm, what is the area of the triangle in square centimeters? That question is more than just a math problem—it’s a gateway into understanding foundational geometry with real-world relevance. From classroom learning to DIY home projects and professional design, right triangles underpin countless applications, making this calculation part of a broader toolkit people use daily. In the U.S. market, interest in practical math is rising, especially among learners, educators, and professionals seeking clarity on geometry without confusion. This triangle’s dimensions reveal a balance between abstraction and application. With one leg measuring 8 cm longer than its counterpart, and a hypotenuse fixed at 20 cm, solving for the legs unlocks insights into proportional relationships. Many users search for this because they want precise results—whether to teach geometry, design a structural component, or confidence-building practice. The area calculation lies at the core, offering not just numbers, but a deeper grasp of spatial reasoning. Let’s explore how to solve this problem clearly and accurately.
First, define the triangle sides. Let the shorter leg be \( x \), so the longer leg is \( x + 8 \). Using the Pythagorean theorem—\( x^2 + (x + 8)^2 = 20^2 \)—we establish the equation that connects the legs to the hypotenuse. Expanding and simplifying gives a quadratic equation: \( 2x^2 + 16x + 64 = 400 \), then \( 2x^2 + 16x - 336 = 0 \), and dividing by 2 results in \( x^2 + 8x - 168 = 0 \). This quadratic equation is straightforward to solve using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = 8 \), \( c = -168 \). Calculating:
\( x = \frac{-8 \pm \sqrt{64 + 672}}{2} = \frac{-8 \pm \sqrt{736}}{2} \).
Simplifying \( \sqrt{736} = \sqrt{16 \cdot 46} = 4\sqrt{46} \), so \( x = \frac{-8 \pm 4\sqrt{46}}{2} = -4 \pm 2\sqrt{46} \).
Only the positive root matters: \( x = -4 +