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2 years ago

Find the area of the shaded region. Round your answer to the nearest tenth. Put the units on the answer for full credit

Mathematics

1 answer:

Finger [1]2 years ago

6 0

Answer:

Area of Shaded Region: 189 square centimeters

Step-by-step explanation:

~ To find the area of the shaded region, let us calculate the area of the rectangle added to the area of the two triangles, or the area of the whole triangle created from the two smaller triangles ~

1. Given dimensions of 15 by 9 centimeters, the area of the rectangle would be ⇒ 15 cm * 9 cm = 135 cm^2

2. It would be simpler if we were to consider the area of the larger triangle. The height in this case, by the Partition Postulate, would be 27 cm - 15 cm ⇒ 12 centimeters. Now provided that the shape with dimensions 15 cm by 9 cm is a rectangle, opposite sides are ≅, and thus ⇒ the base of this large triangle is 9 cm. That being said we can calculate the area of the triangle by ⇒ 1/2 * base * altitude ⇒ 1/2 * 9 cm * 12 cm ⇒ 6 cm * 9 cm ⇒ 54 cm^2

3. The area of the shaded region is thus 135 cm^2 + 54 cm^2 ⇒ 189 cm^2

Area of Shaded Region: 189 square centimeters

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Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.

So, if we sum the first N odd numbers, we have

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1$

The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have

$\displaystyle 2\sum_{i=1}^N i = 2\cdot\dfrac{N(N+1)}{2}=N(N+1)$

The second sum is simply the sum of N ones:

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So, the final result is

$\displaystyle \sum_{i=1}^N 2i-1 = 2\sum_{i=1}^N i - \sum_{i=1}^N 1 = N(N+1)-N = N^2+N-N = N^2$

which ends the proof.

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Answer:

Step-by-step explanation:

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