Pearson Geometry Common Core, 2011
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Pearson Geometry Common Core, 2011 View details
1. Areas of Parallelograms and Triangles
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Exercise 17 Page 620

Practice makes perfect
a To find the area of the paved surface, we will add the areas of the driving region and the four parking spaces.

We see that the driving region is a rectangle with a base of 50feet and a height 15 feet, and that each of the parking spaces is a parallelogram with a base of 10 feet and height 16feet. Let's use the area formula, A= b h, to find the areas of each of these sections.

Area, A bh A=bh
Driving region, A_d 50 * 15 A_d=750
Parking space, A_p 10 * 16 A_p=160

Now, we can find the total area of the paved surface, A. Remember that there are four parking spaces. A = A_d+ 4 * A_p ⇓ A = 750 + 4 * 160=1390 The paved surface has an area of 1390 ft^2.

b To find the area A of the paved surface, we can first find the area A_E of the entire parking lot, which is a 31 feet by 50 feet rectangle. Then, we can subtract the area A_t of the two flower spaces from the entire area. A = A_E - 2* A_t
c Let's first find the total area A_E of the entire parking lot. It is a rectangle with the dimensions 50 feet by 31 feet, so its total area is 1550 ft^2.
A_E= b h ⇓ A_E= 50 * 31 =1550 Now, we need to find the areas of the two triangular flower spaces. Let's focus on the left triangle first. Since the parking spaces are congruent, the total base length of the four parallelograms is 40 feet. This means that the base of the triangle is 10 feet.

Therefore, the area A_t of the left triangle is 80 ft^2. A_t = 1/2 b h ⇓ A_t = 1/2 10 * 16 = 80 Given that the triangles are congruent, the total area for the flower spaces, 2 A_t, is 160 ft^2. By subtracting this area from the entire area A_E, we can find the area A of just the paved surface. A = A_E- 2 * A_t ⇓ A =1550 - 160=1390 The paved surface has an area of 1390 ft^2. We ended with the same result as in Part A, so our method is correct.