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

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==Problem==
 
==Problem==
 
The rectangle shown has side lengths of 8 and 4.
 
The rectangle shown has side lengths of 8 and 4.
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{{Template:Image needed}}
 
The area of the shaded region is  
 
The area of the shaded region is  
  
 
<math> \text{ (A) }\ 32 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 64 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 4 </math>
 
<math> \text{ (A) }\ 32 \qquad\text{ (B) }\ 16 \qquad\text{ (C) }\ 64 \qquad\text{ (D) }\ 12 \qquad\text{ (E) }\ 4 </math>
 
==Solution 1==
 
==Solution 1==
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Let x be the base of one of the shaded triangles that isn't equal to 4 necessarily.
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Let <math>x</math> be the base of one of the shaded triangles that isn't equal to 4 necessarily, as shown here:
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{{Template:Image needed}}
  
 
We know that the shaded triangles are two [[right triangle]]s because the interior angles of a [[rectangle]] are right angles.
 
We know that the shaded triangles are two [[right triangle]]s because the interior angles of a [[rectangle]] are right angles.

Latest revision as of 12:17, 22 April 2025

Problem

The rectangle shown has side lengths of 8 and 4.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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, as shown here:

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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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