9-12: find the area of the quadrilaterals with the given vertices!
9. parallelogram: a(-4, -6), b(6, -6), c(-1, 5), d(9, 5)
10. triangle: e(-1, 5), f(4, -5), g(4, 7)
11. parallelogram: h(-6, -4), i(0, -4), j(3, 4), k(-3, 4)
12. rhombus: l(-2, -1), m(3, -1), n(6, 3), o(1, 3)

Question

Accepted Answer

### Problem 9: Parallelogram with vertices $ A(-4, -6) $, $ B(6, -6) $, $ C(-1, 5) $, $ D(9, 5) $
# Explanation:
## Step 1: Identify base and height
For a parallelogram, area $ = \text{base} \times \text{height} $. The base can be the length of $ AB $. The $ y $-coordinates of $ A $ and $ B $ are the same ($-6$), so $ AB $ is horizontal. Length of $ AB $: $ |6 - (-4)| = 10 $. The height is the vertical distance between the lines $ AB $ (where $ y = -6 $) and $ CD $ (where $ y = 5 $). So height $ = |5 - (-6)| = 11 $.
## Step 2: Calculate area
Area $ = \text{base} \times \text{height} = 10 \times 11 = 110 $.
# Answer:
$ 110 $

### Problem 10: Triangle with vertices $ E(-1, 5) $, $ F(4, -5) $, $ G(4, 7) $
# Explanation:
## Step 1: Identify base and height
The $ x $-coordinates of $ F $ and $ G $ are the same ($ 4 $), so $ FG $ is vertical. Length of $ FG $: $ |7 - (-5)| = 12 $. The base is $ FG = 12 $. The height is the horizontal distance from $ E $ to the line $ FG $ (where $ x = 4 $). So height $ = |4 - (-1)| = 5 $.
## Step 2: Calculate area of triangle
Area of triangle $ = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 $.
# Answer:
$ 30 $

### Problem 11: Parallelogram with vertices $ H(-6, -4) $, $ I(0, -4) $, $ J(3, 4) $, $ K(-3, 4) $
# Explanation:
## Step 1: Identify base and height
The $ y $-coordinates of $ H $ and $ I $ are the same ($-4$), so $ HI $ is horizontal. Length of $ HI $: $ |0 - (-6)| = 6 $. The height is the vertical distance between $ y = -4 $ (line $ HI $) and $ y = 4 $ (line $ JK $). Height $ = |4 - (-4)| = 8 $.
## Step 2: Calculate area
Area $ = \text{base} \times \text{height} = 6 \times 8 = 48 $.
# Answer:
$ 48 $

### Problem 12: Rhombus with vertices $ L(-2, -1) $, $ M(3, -1) $, $ N(6, 3) $, $ O(1, 3) $
# Explanation:
## Step 1: Identify diagonals
The diagonals of a rhombus are perpendicular bisectors. The first diagonal: $ LM $ (horizontal, $ y = -1 $). Length of $ LM $: $ |3 - (-2)| = 5 $. The second diagonal: $ NO $ (horizontal, $ y = 3 $)? Wait, no, $ L(-2, -1) $, $ M(3, -1) $; $ N(6, 3) $, $ O(1, 3) $. Wait, actually, the diagonals are $ LN $ and $ MO $? No, better: the distance between $ y = -1 $ and $ y = 3 $ is vertical distance, but actually, the diagonals are the vertical and horizontal? Wait, $ LM $ is from $ (-2, -1) $ to $ (3, -1) $, length $ 5 $. $ NO $ is from $ (6, 3) $ to $ (1, 3) $, length $ 5 $? Wait, no, the other diagonal: from $ L(-2, -1) $ to $ N(6, 3) $? No, better: the rhombus has diagonals that are horizontal and vertical? Wait, $ LM $ is horizontal (length $ 5 $), and the vertical distance between $ y = -1 $ and $ y = 3 $ is $ 4 $? Wait, no, the diagonals are $ LM $ (length $ 5 $) and the vertical distance between the two horizontal lines? Wait, no, actually, the area of a rhombus is $ \frac{1}{2} \times d_1 \times d_2 $, where $ d_1 $ and $ d_2 $ are the lengths of the diagonals. Let's find the diagonals. The first diagonal: $ LM $ (from $ (-2, -1) $ to $ (3, -1) $): length $ |3 - (-2)| = 5 $. The second diagonal: $ LO $? No, $ M(3, -1) $ to $ O(1, 3) $? No, better: the vertical distance between the two horizontal sides (where $ y = -1 $ and $ y = 3 $) is $ 4 $, but wait, the other diagonal: from $ H(-6, -4) $ no, wait, vertices $ L(-2, -1) $, $ M(3, -1) $, $ N(6, 3) $, $ O(1, 3) $. So $ LM $ is horizontal (length $ 5 $), and $ NO $ is horizontal (length $ 5 $)? No, that can't be. Wait, actually, the diagonals are $ LN $ and $ MO $. Wait, $ L(-2, -1) $, $ N(6, 3) $: distance? No, better: the area of a rhombus is base times height. The base $ LM $ length is $ 5 $ (from $ x=-2 $ to $ x=3 $, $ y=-1 $). The height is the vertical distance between $ y=-1 $ and $ y=3 $, which is $ 4 $? Wait, no, the slope of $ LM $ is 0 (horizontal), so the height is the vertical distance between the two horizontal lines? Wait, no, the other side: from $ M(3, -1) $ to $ J(3, 4) $? No, wait, the vertices are $ L(-2, -1) $, $ M(3, -1) $, $ N(6, 3) $, $ O(1, 3) $. So $ LM $ is horizontal (length $ 5 $), and the vertical distance between $ y = -1 $ and $ y = 3 $ is $ 4 $, but actually, the height is the vertical distance? Wait, no, the area can also be calculated as the length of $ LM $ (5) times the vertical distance between $ y = -1 $ and $ y = 3 $, which is $ 4 $? Wait, no, that gives $ 5 \times 4 = 20 $, but that's wrong. Wait, let's use the formula for area of a parallelogram (rhombus is a parallelogram) with base $ LM $ (length 5) and height the vertical distance between $ y = -1 $ and $ y = 3 $, which is $ 4 $? No, wait, the $ y $-coordinates of $ M(3, -1) $ and $ J(3, 4) $? No, the vertices are $ L(-2, -1) $, $ M(3, -1) $, $ N(6, 3) $, $ O(1, 3) $. So the vector $ LM $ is $ (5, 0) $, and the vector $ LO $ is $ (3, 4) $? Wait, no, better to use the shoelace formula. Let's apply shoelace formula.
## Step 1: Shoelace formula
List the vertices in order: $ L(-2, -1) $, $ M(3, -1) $, $ N(6, 3) $, $ O(1, 3) $, back to $ L(-2, -1) $.
Shoelace sum: $ (-2)(-1) + 3(3) + 6(3) + 1(-1) = 2 + 9 + 18 - 1 = 28 $
Other sum: $ (-1)(3) + (-1)(6) + 3(1) + 3(-2) = -3 - 6 + 3 - 6 = -12 $
Area $ = \frac{1}{2} |28 - (-12)| = \frac{1}{2} |40| = 20 $. Wait, but earlier mistake. Wait, no, let's recalculate shoelace:
$ x_1 = -2, y_1 = -1 $
$ x_2 = 3, y_2 = -1 $
$ x_3 = 6, y_3 = 3 $
$ x_4 = 1, y_4 = 3 $
$ x_5 = -2, y_5 = -1 $
Sum1: $ x_1y_2 + x_2y_3 + x_3y_4 + x_4y_5 = (-2)(-1) + 3(3) + 6(3) + 1(-1) = 2 + 9 + 18 - 1 = 28 $
Sum2: $ y_1x_2 + y_2x_3 + y_3x_4 + y_4x_5 = (-1)(3) + (-1)(6) + 3(1) + 3(-2) = -3 - 6 + 3 - 6 = -12 $
Area $ = \frac{1}{2} |Sum1 - Sum2| = \frac{1}{2} |28 - (-12)| = \frac{1}{2} \times 40 = 20 $. Wait, but also, the diagonals: length of $ LM $ is $ 5 $, length of the other diagonal (from $ (-2, -1) $ to $ (6, 3) $)? No, better: the area of a rhombus is $ \frac{1}{2} \times d_1 \times d_2 $. Let's find $ d_1 $ (distance between $ L(-2, -1) $ and $ N(6, 3) $): $ \sqrt{(6 - (-2))^2 + (3 - (-1))^2} = \sqrt{64 + 16} = \sqrt{80} = 4\sqrt{5} $. $ d_2 $ (distance between $ M(3, -1) $ and $ O(1, 3) $): $ \sqrt{(1 - 3)^2 + (3 - (-1))^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} $. Then area $ = \frac{1}{2} \times 4\sqrt{5} \times 2\sqrt{5} = \frac{1}{2} \times 40 = 20 $. Correct.
# Answer:
$ 20 $

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QUESTION IMAGE

Question

Explanation:

Problem 9: Parallelogram with vertices \( A(-4, -6) \), \( B(6, -6) \), \( C(-1, 5) \), \( D(9, 5) \)

Step 1: Identify base and height

Step 2: Calculate area

Step 1: Identify base and height

Step 2: Calculate area of triangle

Step 1: Identify base and height

Step 2: Calculate area

Answer:

Problem 10: Triangle with vertices \( E(-1, 5) \), \( F(4, -5) \), \( G(4, 7) \)