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spayn [35]
2 years ago
13

Describe how the area of a rectangle will change if its width remains constant but its length is halved

Mathematics
1 answer:
IRISSAK [1]2 years ago
8 0

Answer:

Step-by-step explanation:

Area of a rectangle is given as length times breadth. Mathematically,

A = l * b

If the length is halved and the breadth is constant, then we have

A = l/2 * b

A = lb/2

From the above, the area of a rectangle is lb, while the area of the hypothetical rectangle will be lb/2

This means that, if the length is halved, and the width is constant, then the area of the rectangle reduces by half.

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2 years ago
Brady made a scale drawing of a rectangular swimming pool on a coordinate grid. The points (-20, 25), (30, 25), (30, -10) and (-
djverab [1.8K]

Answer:

Length = 50 units

width = 35 units

Step-by-step explanation:

Let A, B, C and D be the corner of the pools.

Given:

The points of the corners are.

A(x_{1}, y_{1}})=(-20, 25)

B(x_{2}, y_{2}})=(30, 25)

C(x_{3}, y_{3}})=(30, -10)

D(x_{4}, y_{4}})=(-20, -10)

We need to find the dimension of the pools.

Solution:

Using distance formula of the two points.

d(A,B)=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}----------(1)

For point AB

Substitute points A(30, 25) and B(30, 25) in above equation.

AB=\sqrt{(30-(-20))^{2}+(25-25)^{2}}

AB=\sqrt{(30+20)^{2}}

AB=\sqrt{(50)^{2}

AB = 50 units

Similarly for point BC

Substitute points B(-20, 25) and C(30, -10) in equation 1.

d(B,C)=\sqrt{(x_{3}-x_{2})^{2}+(y_{3}-y_{2})^{2}}

BC=\sqrt{(30-30)^{2}+((-10)-25)^{2}}

BC=\sqrt{(-35)^{2}}

BC = 35 units

Similarly for point DC

Substitute points D(-20, -10) and C(30, -10) in equation 1.

d(D,C)=\sqrt{(x_{3}-x_{4})^{2}+(y_{3}-y_{4})^{2}}

DC=\sqrt{(30-(-20))^{2}+(-10-(-10))^{2}}

DC=\sqrt{(30+20)^{2}}

DC=\sqrt{(50)^{2}}

DC = 50 units

Similarly for segment AD

Substitute points A(-20, 25) and D(-20, -10) in equation 1.

d(A,D)=\sqrt{(x_{4}-x_{1})^{2}+(y_{4}-y_{1})^{2}}

AD=\sqrt{(-20-(-20))^{2}+(-10-25)^{2}}

AD=\sqrt{(-20+20)^{2}+(-35)^{2}}

AD=\sqrt{(-35)^{2}}

AD = 35 units

Therefore, the dimension of the rectangular swimming pool are.

Length = 50 units

width = 35 units

7 0
3 years ago
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