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Difference between revisions of "2012 CEMC Gauss (Grade 8) Problems/Problem 10"

(Created page with "==Problem== The rectangle shown has side lengths of 8 and 4. The area of the shaded region is <math> \text{ (A) }\ 32 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 64 \qquad\t...")
 
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==Solution 1==
 
==Solution 1==
 
Let x be the base of one of the shaded triangles that isn't equal to 4 necessarily.  
 
Let x be the base of one of the shaded triangles that isn't equal to 4 necessarily.  
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 +
We know that the shaded triangles are two [[right triangle]]s because the interior angles of a [[rectangle]] are right angles.
  
 
Since the full side lengths is <math>8</math>, the length of the base of the other shaded triangle is <math>8 - x</math>. This means that the total area of both of the triangles is:
 
Since the full side lengths is <math>8</math>, the length of the base of the other shaded triangle is <math>8 - x</math>. This means that the total area of both of the triangles is:

Revision as of 11:08, 21 April 2025

Problem

The rectangle shown has side lengths of 8 and 4.

The area of the shaded region is

$\text{ (A) }\ 32 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 64 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 4$

Solution 1

Let x be the base of one of the shaded triangles that isn't equal to 4 necessarily.

We know that the shaded triangles are two right triangles because the interior angles of a rectangle are right angles.

Since the full side lengths is $8$, the length of the base of the other shaded triangle is $8 - x$. This means that the total area of both of the triangles is:

$\frac{4 \times x}{2} + \frac{4 \times (8 - x)}{2} = \frac{4 \times x + 4 \times (8 - x)}{2} = \frac{4 \times x + 4 \times 8 - 4 \times x}{2} = \frac{32}{2} = \boxed {\textbf {(C) } 16}$.

~anabel.disher

Solution 2

To find the area of the shaded triangles, we can subtract the total area of the rectangle from the area of the non-shaded triangle.

The total area of the rectangle is $4 \times 8 = 32$.

The height of the non-shaded triangle is $4$. Since its base is equal to $8$, its area must be $\frac{4 \times 8}{2} = \frac{32}{2} = 16$.

Thus, the area of the two shaded triangles is $32 - 16 = \boxed {\textbf {(C) } 16}$.

~anabel.disher

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