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Dafna11 [192]
3 years ago
10

5. Two vertices of a rectangle are (8,-5) and (8,7). If the area of the rectangle is 72 square units, name the possible location

of the other two vertices. 6. A triangle with two vertices located at (5,-8) and (5, 4) has an area of 48 square units Determine one possible​
Mathematics
1 answer:
Phantasy [73]3 years ago
4 0

Answer:

<h3>#5</h3>

<u>Given vertices:</u>

  • (8, -5) and (8, 7)

These have same x-coordinate, so when connected form a vertical segment.

<u>The length of the segment is:</u>

  • 7 - (-5) = 12 units

The area of the rectangle is 72 square units, so the horizontal segment has the length of:

  • 72/12 = 6 units

<u>Possible location of the remaining vertices (to the left from the given):</u>

  • (8 - 6, -5) = (2, -5)

and

  • (8 - 6, 7) = (2, 7)
<h3>#6</h3>

<u>Similarly to previous exercise:</u>

  • (5, -8) and (5, 4) given with the area of 48 square units

<u>The distance between the given vertices:</u>

  • 4 - (-8) = 12 units

<u>The other side length is:</u>

  • 48/12 = 4 units

<u>Possible location of the other vertices (to the right from the given):</u>

  • (5 + 4, -8) = (9, -8)

and

  • (5 + 4, 4) = (9, 4)
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<h3><u>Solution:</u></h3>

Given that It cost 5 dollars for a child ticket and 8 dollars for a adult ticket

cost of each child ticket = 5 dollars

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Let "c" be the number of child tickets bought

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<em>Number of child tickets bought + number of adult tickets bought = 110</em>

c + a = 110 ----- eqn 1

<em><u>Also we can frame a equation as:</u></em>

Number of child tickets bought x cost of each child ticket + number of adult tickets bought x cost of each adult ticket = 820

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From eqn 1,

a = 110 - c  ------ eqn 3

Substitute eqn 3 in eqn 2

5c + 8(110 - c) = 820

5c + 880 - 8c = 820

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c = 20

Therefore from eqn 3,

a = 110 - 20 = 90

a = 90

Therefore number of child tickets bought is 20

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