Daily Math · 2026-05-08

Question 1

A right triangle has legs of length 3 and 4. What is the radius of its inscribed circle?

Accepted Answer

For any triangle, the inradius is $r = A / s$, where $A$ is the area and $s$ is the semi-perimeter.

Hypotenuse: $\sqrt{3^2 + 4^2} = 5$. Area: $\dfrac{1}{2} \cdot 3 \cdot 4 = 6$. Semi-perimeter: $(3 + 4 + 5)/2 = 6$.

$r = 6 / 6 = 1$.

For right triangles specifically there's also a shortcut: $r = (a + b - c)/2 = (3 + 4 - 5)/2 = 1$.

Answer

$1$

Answer

$\dfrac{6}{5}$

Answer

$\dfrac{5}{2}$

Answer

$\sqrt{2}$

Question 2

You owe $\$10{,}000$ on a loan at $6\%$ annual interest, compounded annually. You pay $\$2{,}000$ at the end of each year. After how many years is the loan fully paid off (rounded up)?

Accepted Answer

Let $B_n$ be the balance after the $n$-th payment. $B_0 = 10000$, and $B_n = 1.06 \cdot B_{n-1} - 2000$.

$B_1 = 10600 - 2000 = 8600$
$B_2 = 9116 - 2000 = 7116$
$B_3 = 7542.96 - 2000 = 5542.96$
$B_4 = 5875.54 - 2000 = 3875.54$
$B_5 = 4108.07 - 2000 = 2108.07$
$B_6 = 2234.55 - 2000 = 234.55$

After year 6 there's only \$234.55 left, so a 6th payment of \$234.55 finishes it. Total: 6 years (the last payment is small).

Answer

5 years

Answer

6 years

Answer

7 years

Answer

10 years

Question 3

Simplify: $\quad \dfrac{x^2 - 9}{x^2 - 6x + 9}$ for $x 
e 3$.

Accepted Answer

Factor numerator and denominator:
$$\frac{x^2 - 9}{x^2 - 6x + 9} = \frac{(x-3)(x+3)}{(x-3)^2}$$

Cancel one factor of $(x - 3)$ (valid since $x 
e 3$):
$$= \frac{x + 3}{x - 3}$$

Answer

$\dfrac{x + 3}{x - 3}$

Answer

$\dfrac{x - 3}{x + 3}$

Answer

$x - 3$

Answer

$\dfrac{1}{x - 3}$

Problems & worked solutions

Problem 1 — Geometry

Problem 2 — Compound Interest

Problem 3 — Algebra