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allsm [11]
4 years ago
10

The picture shows a red parallelogram split into 6 equal parts and a blue parallelogram split into 6 equal parts. Drag figures s

o that the total area is equal to the number given.

Mathematics
2 answers:
Luden [163]4 years ago
5 0

Step-by-step explanation:

The area of a parallelogram is base times height. You are given the base and the height of each parallelogram, so you can find their areas.

Red parallelogram:

area = base * height = 12 yd * 7 yd = 84 sq yd

Blue parallelogram:

area = base * height = 24 yd * 3 yd = 72 sq yd

Each parallelogram is made up of 6 parts whose areas are equal. To find the area of each piece of the red parallelogram, we divide 84 sq yd by 6.

Red parallelogram piece:

area = 84 sq yd/6 = 14 sq yd

Blue parallelogram piece:

area = 72 sq yd/6 = 12 sq yd

Now we know that each small red polygon has area 14 sq yd, and each blue polygon has area 12 sq yd.

Part A: 52 sq yd

52 = 14 + 14 + 12 + 12

Use 2 small red polygons and 2 small blue polygons.

Part B: 70 sq yd

70 = 5 * 14

Use 5 small red polygons.

Part C: 36 sq yd

36 = 3 * 12

Use 3 small blue polygons.

Gre4nikov [31]4 years ago
3 0

Answer:

Step-by-step explanation:

The area of a parallelogram is base times height. You are given the base and the height of each parallelogram, so you can find their areas.

Red parallelogram:

area = base * height = 12 yd * 7 yd = 84 sq yd

Blue parallelogram:

area = base * height = 24 yd * 3 yd = 72 sq yd

Each parallelogram is made up of 6 parts whose areas are equal. To find the area of each piece of the red parallelogram, we divide 84 sq yd by 6.

Red parallelogram piece:

area = 84 sq yd/6 = 14 sq yd

Blue parallelogram piece:

area = 72 sq yd/6 = 12 sq yd

Now we know that each small red polygon has area 14 sq yd, and each blue polygon has area 12 sq yd.

Part A: 52 sq yd

52 = 14 + 14 + 12 + 12

Use 2 small red polygons and 2 small blue polygons.

Part B: 70 sq yd

70 = 5 * 14

Use 5 small red polygons.

Part C: 36 sq yd

36 = 3 * 12

Use 3 small blue polygons.

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