Mixing
What is the area of the rectangle?
$begingroup$
The diagram shows a rectangle and a line X parallel to its base. Two points A and B lie on X (A is outside and B is inside the rectangle). The sum of the areas of the two shaded triangles is 10cm. What is the area of the rectangle. I need to know the full solution and the answer of this question. Thanks in advance?
rectangles
edited Jan 20 at 6:04
Larry
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
$endgroup$
add a comment |
$begingroup$
The diagram shows a rectangle and a line X parallel to its base. Two points A and B lie on X (A is outside and B is inside the rectangle). The sum of the areas of the two shaded triangles is 10cm. What is the area of the rectangle. I need to know the full solution and the answer of this question. Thanks in advance?
rectangles
edited Jan 20 at 6:04
Larry
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
$endgroup$
1
$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16
add a comment |
$begingroup$
The diagram shows a rectangle and a line X parallel to its base. Two points A and B lie on X (A is outside and B is inside the rectangle). The sum of the areas of the two shaded triangles is 10cm. What is the area of the rectangle. I need to know the full solution and the answer of this question. Thanks in advance?
rectangles
edited Jan 20 at 6:04
Larry
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
$endgroup$
The diagram shows a rectangle and a line X parallel to its base. Two points A and B lie on X (A is outside and B is inside the rectangle). The sum of the areas of the two shaded triangles is 10cm. What is the area of the rectangle. I need to know the full solution and the answer of this question. Thanks in advance?
rectangles
rectangles
edited Jan 20 at 6:04
Larry
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
edited Jan 20 at 6:04
Larry
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
edited Jan 20 at 6:04
Larry
2,41331129
edited Jan 20 at 6:04
Larry
2,41331129
edited Jan 20 at 6:04
Larry
2,41331129
2,41331129
asked Jan 20 at 4:35
SurenSuren
11
asked Jan 20 at 4:35
SurenSuren
11
asked Jan 20 at 4:35
SurenSuren
11
11
1
$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16
add a comment |
1
$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16
1
1
$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16
$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The area of the rectangle is 20 cm^2.
Detailed answer:
Since the line segment AX is parallel to the base, the height of the triangles remain constant. So we can move point A to point B without changing the areas.
As for the area of the rectangle let's look at the top rectangle which has height h_1 and the bottom rectangle which has height h_2, and together these rectangle form the whole. They both have the same base b, and the rectangles dimensions are b and h_1 + h_2.
The top rectangle has area A_1 = 1/2 bh_1.
The bottom rectangle has area A_2 = 1/2 bh_2.
You already said that A_1 + A_2 = 10.
So A_1 + A_2 = 1/2 b(h_1 + h_2)
The area of our desired rectangle is b(h_1 + h_2).
So we have 10 = 1/2 b(h_1 + h_2)
Which means 20 = b(h_1 + h_2).
Short answer:
The top shaded triangle is 1/2 the top rectangle, the bottom shaded triangle is 1/2 the bottom rectangle, and together they take up half the big rectangle.
So double 10 cm^2 to get 20 cm^2.
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
The area of the rectangle is 20 cm^2.
Detailed answer:
Since the line segment AX is parallel to the base, the height of the triangles remain constant. So we can move point A to point B without changing the areas.
As for the area of the rectangle let's look at the top rectangle which has height h_1 and the bottom rectangle which has height h_2, and together these rectangle form the whole. They both have the same base b, and the rectangles dimensions are b and h_1 + h_2.
The top rectangle has area A_1 = 1/2 bh_1.
The bottom rectangle has area A_2 = 1/2 bh_2.
You already said that A_1 + A_2 = 10.
So A_1 + A_2 = 1/2 b(h_1 + h_2)
The area of our desired rectangle is b(h_1 + h_2).
So we have 10 = 1/2 b(h_1 + h_2)
Which means 20 = b(h_1 + h_2).
Short answer:
The top shaded triangle is 1/2 the top rectangle, the bottom shaded triangle is 1/2 the bottom rectangle, and together they take up half the big rectangle.
So double 10 cm^2 to get 20 cm^2.
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
$endgroup$
add a comment |
$begingroup$
The area of the rectangle is 20 cm^2.
Detailed answer:
Since the line segment AX is parallel to the base, the height of the triangles remain constant. So we can move point A to point B without changing the areas.
As for the area of the rectangle let's look at the top rectangle which has height h_1 and the bottom rectangle which has height h_2, and together these rectangle form the whole. They both have the same base b, and the rectangles dimensions are b and h_1 + h_2.
The top rectangle has area A_1 = 1/2 bh_1.
The bottom rectangle has area A_2 = 1/2 bh_2.
You already said that A_1 + A_2 = 10.
So A_1 + A_2 = 1/2 b(h_1 + h_2)
The area of our desired rectangle is b(h_1 + h_2).
So we have 10 = 1/2 b(h_1 + h_2)
Which means 20 = b(h_1 + h_2).
Short answer:
The top shaded triangle is 1/2 the top rectangle, the bottom shaded triangle is 1/2 the bottom rectangle, and together they take up half the big rectangle.
So double 10 cm^2 to get 20 cm^2.
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
$endgroup$
add a comment |
$begingroup$
The area of the rectangle is 20 cm^2.
Detailed answer:
Since the line segment AX is parallel to the base, the height of the triangles remain constant. So we can move point A to point B without changing the areas.
As for the area of the rectangle let's look at the top rectangle which has height h_1 and the bottom rectangle which has height h_2, and together these rectangle form the whole. They both have the same base b, and the rectangles dimensions are b and h_1 + h_2.
The top rectangle has area A_1 = 1/2 bh_1.
The bottom rectangle has area A_2 = 1/2 bh_2.
You already said that A_1 + A_2 = 10.
So A_1 + A_2 = 1/2 b(h_1 + h_2)
The area of our desired rectangle is b(h_1 + h_2).
So we have 10 = 1/2 b(h_1 + h_2)
Which means 20 = b(h_1 + h_2).
Short answer:
The top shaded triangle is 1/2 the top rectangle, the bottom shaded triangle is 1/2 the bottom rectangle, and together they take up half the big rectangle.
So double 10 cm^2 to get 20 cm^2.
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
$endgroup$
The area of the rectangle is 20 cm^2.
Detailed answer:
Since the line segment AX is parallel to the base, the height of the triangles remain constant. So we can move point A to point B without changing the areas.
As for the area of the rectangle let's look at the top rectangle which has height h_1 and the bottom rectangle which has height h_2, and together these rectangle form the whole. They both have the same base b, and the rectangles dimensions are b and h_1 + h_2.
The top rectangle has area A_1 = 1/2 bh_1.
The bottom rectangle has area A_2 = 1/2 bh_2.
You already said that A_1 + A_2 = 10.
So A_1 + A_2 = 1/2 b(h_1 + h_2)
The area of our desired rectangle is b(h_1 + h_2).
So we have 10 = 1/2 b(h_1 + h_2)
Which means 20 = b(h_1 + h_2).
Short answer:
The top shaded triangle is 1/2 the top rectangle, the bottom shaded triangle is 1/2 the bottom rectangle, and together they take up half the big rectangle.
So double 10 cm^2 to get 20 cm^2.
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
answered Jan 20 at 6:18
gunsnfloydgunsnfloyd
31
31
add a comment |
add a comment |
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$begingroup$
When you ask a question, it is better to include what you have tried so far. Perhaps this is the reason why people give you downvotes and are not willing to answer your question. Anyway, I could show you how to solve questions like this. Whether point A or B is outside the rectangle or not doesn't affect the area of the yellow triangles. You could think of the two yellow triangles as right triangles. Then the area of the rectangle is basically twice the sum, which is 20 $cm^2$.
$endgroup$
– Larry
Jan 20 at 6:16