---
title: "First product of inertia of right-angled triangle"
subject: "Strength of Materials"
chapter: 2
key: "semester_3/strength_of_materials:product-of-inertia-right-triangle"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/product-of-inertia-right-triangle
site: BCT Engineering
---

# First product of inertia of right-angled triangle

Strength of Materials · Chapter 2: Geometrical Properties of Sections

- Asked in: 2068 Chaitra, 4 marks
- Video: [20. Product of Inertia | Most Important Problem#2 | Complete Concept](https://www.youtube.com/watch?v=2Ir_peuXKBA) — MKS TUTORIALS by Manoj Sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/product-of-inertia-right-triangle
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-2

## Question

Find from the first principle product of inertia for a right angled triangle with base 'b' and height 'h' along XX and YY axes. (base and height are collinear with XX and YY axes respectively).

## Short answer

By integration of a vertical strip, the product of inertia about the axes along the base and height (at the right-angle corner) is I_xy = b²h²/24. About centroidal axes it becomes -b²h²/72.

## Answer

Take a right-angled triangle with the right-angle corner at the origin, base $b$ along the $x$-axis and height $h$ along the $y$-axis. The hypotenuse has equation $y = h\left(1 - \dfrac{x}{b}\right)$.

```text
 y
 h|\
  | \
  |  \
  |   \
  O----- b   x
```

#### Derivation from first principles
Take a vertical strip of width $dx$ at distance $x$, of height $y_1 = h\left(1 - \dfrac{x}{b}\right)$. Its area is $dA = y_1\,dx$. The product of inertia of the strip about the $x$ and $y$ axes is obtained by applying the parallel-axis theorem to the strip (its own $I_{xy}$ is zero, as it is symmetrical about its centre line parallel to $y$):

$$
dI_{xy} = (dA)\,x\cdot\frac{y_1}{2} = x\,\frac{y_1}{2}\,y_1\,dx = \frac{x\,y_1^2}{2}\,dx
$$

Hence
$$
I_{xy} = \int_0^b \frac{x\,h^2}{2}\left(1 - \frac{x}{b}\right)^2dx
= \frac{h^2}{2}\int_0^b\left(x - \frac{2x^2}{b} + \frac{x^3}{b^2}\right)dx
$$
$$
= \frac{h^2}{2}\left[\frac{b^2}{2} - \frac{2b^2}{3} + \frac{b^2}{4}\right]
= \frac{h^2}{2}\cdot\frac{b^2}{12}
$$

$$
\boxed{I_{xy} = \frac{b^2h^2}{24}}
$$

(positive, since the triangle lies in the first quadrant). About the centroidal axes, $\bar I_{xy} = \dfrac{b^2h^2}{24} - \dfrac{bh}{2}\cdot\dfrac{b}{3}\cdot\dfrac{h}{3} = -\dfrac{b^2h^2}{72}$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In the first-principle derivation, what is the product of inertia of the vertical strip about the x and y axes based on?
   A. dA times x^2
   B. dA times x times y1/2, using the strip's centroid at height y1/2
   C. dA times y1^2 only
   D. The strip's own Ixy, which is non-zero
   Answer: B. The strip's own Ixy is zero, so the parallel-axis theorem gives dIxy = dA * x * (y1/2). (Key point: Strip choice)

2. What is the equation of the hypotenuse and the integration limits?
   A. y = h(1 - x/b), x from 0 to b
   B. y = hx/b, x from 0 to h
   C. y = h(1 - x/h), x from 0 to h
   D. y = bx/h, x from 0 to b
   Answer: A. With the right angle at the origin the hypotenuse passes through (0,h) and (b,0), so y = h(1 - x/b) over 0 to b. (Key point: Limits and line)

3. What is Ixy about the axes along the legs, and about the centroidal axes?
   A. b^2h^2/12 and b^2h^2/36
   B. b^2h^2/24 and b^2h^2/72
   C. bh^3/12 and bh^3/36
   D. b^2h^2/24 and -b^2h^2/72
   Answer: D. Integrating gives b^2h^2/24; subtracting A*xbar*ybar = (bh/2)(b/3)(h/3) gives -b^2h^2/72. (Key point: Result)

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