Question

find the area of △cde. round the area to the nearest whole number, if necessary. a(-5, 4), b(0, 3), f(-2, 1), c(4, -1), e(2, -3), d(4, -5) the area is 16 × square units.

Explanation:

Step1: Identify coordinates of C, D, E

C(4, -1), D(4, -5), E(2, -3)

Step2: Find base length (CD)

Since C and D have same x - coordinate, \( CD=\vert - 1-(-5)\vert=\vert4\vert = 4 \)

Step3: Find height (horizontal distance from E to CD)

CD is vertical line \( x = 4 \), horizontal distance from E(2, - 3) to \( x = 4 \) is \( \vert4 - 2\vert=2 \)

Step4: Calculate area of triangle

Area of triangle \(=\frac{1}{2}\times base\times height=\frac{1}{2}\times4\times2 = 4 \)? Wait, no, wait. Wait, maybe I made a mistake. Wait, let's use the formula for area of triangle with coordinates. The formula is \( A=\frac{1}{2}\vert x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)\vert \)

Let \( (x_1,y_1)=(4,-1) \), \( (x_2,y_2)=(4,-5) \), \( (x_3,y_3)=(2,-3) \)

\( A=\frac{1}{2}\vert4(-5 - (-3))+4(-3 - (-1))+2(-1 - (-5))\vert \)

\(=\frac{1}{2}\vert4(-2)+4(-2)+2(4)\vert \)

\(=\frac{1}{2}\vert - 8-8 + 8\vert=\frac{1}{2}\vert - 8\vert = 4 \)? No, that can't be. Wait, maybe the base is horizontal? Wait, no, let's plot the points. C(4,-1), D(4,-5) is vertical line segment. E(2,-3). So the triangle has vertices at (4,-1), (4,-5), (2,-3). So the base CD is length 4 (from y=-1 to y=-5, x=4). The height is the horizontal distance from E to the line x = 4, which is 4 - 2 = 2. Then area is \( \frac{1}{2}\times4\times2 = 4 \)? But that seems too small. Wait, maybe I mixed up the points. Wait, maybe the base is DE or CE? Wait, no, let's recalculate using the shoelace formula correctly.

Shoelace formula: For points \( (x_1,y_1) \), \( (x_2,y_2) \), \( (x_3,y_3) \)

Area \(=\frac{1}{2}\vert x_1y_2+x_2y_3+x_3y_1 - x_2y_1 - x_3y_2 - x_1y_3\vert \)

Plugging in C(4,-1), D(4,-5), E(2,-3):

\( x_1 = 4,y_1=-1 \); \( x_2 = 4,y_2=-5 \); \( x_3 = 2,y_3=-3 \)

\( x_1y_2=4\times(-5)=-20 \)

\( x_2y_3=4\times(-3)=-12 \)

\( x_3y_1=2\times(-1)=-2 \)

Sum: -20 - 12 - 2 = -34

\( x_2y_1=4\times(-1)=-4 \)

\( x_3y_2=2\times(-5)=-10 \)

\( x_1y_3=4\times(-3)=-12 \)

Sum: -4 - 10 - 12 = -26

Subtract: -34 - (-26) = -8

Absolute value and half: \( \frac{1}{2}\vert - 8\vert = 4 \). Wait, but maybe the points are different? Wait, maybe I misread the coordinates. Let me check again. C(4,-1), D(4,-5), E(2,-3). Yes. So the area is 4? But the initial answer was 16, which is wrong. Wait, maybe I made a mistake in coordinates. Wait, maybe D is (4,-5)? C is (4,-1), so vertical distance is 4. E is (2,-3). So the base is CD = 4, height is horizontal distance from E to CD (x=4) is 2. So area is 4. But maybe the triangle is different. Wait, maybe the user made a mistake in the initial answer. Wait, let's check again.

Wait, another way: The base can be the horizontal distance? No, CD is vertical. Alternatively, use the formula for area of a triangle with base and height. The length of CD is \( \sqrt{(4 - 4)^2+(-5+1)^2}=\sqrt{0 + 16}=4 \). The length of CE: \( \sqrt{(4 - 2)^2+(-1 + 3)^2}=\sqrt{4 + 4}=\sqrt{8} \). The length of DE: \( \sqrt{(4 - 2)^2+(-5 + 3)^2}=\sqrt{4 + 4}=\sqrt{8} \). So it's an isoceles triangle with base 4 and equal sides \( \sqrt{8} \). The height of the triangle (from E to CD) can be calculated as \( \sqrt{(\sqrt{8})^2 - 2^2}=\sqrt{8 - 4}=\sqrt{4}=2 \). So area is \( \frac{1}{2}\times4\times2 = 4 \). So the correct area is 4.

Wait, but maybe I misread the coordinates. Let me check the graph again. C is at (4,-1), D is at (4,-5), E is at (2,-3). Yes. So the area is 4.

Answer:

4