Question

the length of a rectangle is 3m more than twice the width, and the area is 180m². find the dimensions of the rectangle​

Answer 1

The breadth (width) of a rectangle, w, is 3m less than its length, l. If the area, A, is 180m square, what is the dimension of the rectangle?

Given:

A = 180 m^2

w = l - 3 m

Find:

l

We have one unknown in this scenario, representing length, l. So, we only need to write one equation for a numerical solution.

Since we know how the area, A, of a rectangle is related to its length, l, and breadth (width), w, we can write the equation as follows:

A = lw

Substituting our given values into the equation:

180 = l x (l - 3)

180 = l^2 - 3l

0 = l^2 - 3l - 180

This is a simple quadratic equation. We can solve it by brute force using the Quadratic Formula which states:

The roots of an equation in the for Ax^2 + Bx + C = 0 are

(-B + or minus SQRT[B^2 - 4AC])/2A

But, if we are on a timed test, there is an easier way to get to this answer rapidly.

We know the length and breadth (width) are probably integers, and that they lie three spaces apart on the number line, l = w + 3. We also know the area, A = 180, so 180 = lw. Since the SQRT[A] would equal l and w if they were the same, that’s a good first approximation.

SQRT[180] = 13.4

So, w will be less than 13.4, and l will be more.

Let’s try w = 13, and l = 13 + 3 = 16. 13 x 16 = 208 > 180 No

Let’s try w = 12, and l = 12 + 3 = 15. 12 x 15 = 180 = 180 Yes

A good test taker will be able to do this math in their head, without doing any of the calculations we’ve shown here. Let’s run through the mental steps.

180 = lw

w = l + 3

SQRT[180] = 13.4 = lw so l and w straddle 13.4

the last two digits of l and w must have a product that is a multiple of ten since 180 is a multiple of ten. 2 and 5 are three apart and a multiple of 10, so try 12 x 15 = 180 Check.The breadth (width) of a rectangle, w, is 3m less than its length, l. If the area, A, is 180m square, what is the dimension of the rectangle?

Given:

A = 180 m^2

w = l - 3 m

Find:

l

We have one unknown in this scenario, representing length, l. So, we only need to write one equation for a numerical solution.

Since we know how the area, A, of a rectangle is related to its length, l, and breadth (width), w, we can write the equation as follows:

A = lw

Substituting our given values into the equation:

180 = l x (l - 3)

180 = l^2 - 3l

0 = l^2 - 3l - 180

This is a simple quadratic equation. We can solve it by brute force using the Quadratic Formula which states:

The roots of an equation in the for Ax^2 + Bx + C = 0 are

(-B + or minus SQRT[B^2 - 4AC])/2A

But, if we are on a timed test, there is an easier way to get to this answer rapidly.

We know the length and breadth (width) are probably integers, and that they lie three spaces apart on the number line, l = w + 3. We also know the area, A = 180, so 180 = lw. Since the SQRT[A] would equal l and w if they were the same, that’s a good first approximation.

SQRT[180] = 13.4

So, w will be less than 13.4, and l will be more.

Let’s try w = 13, and l = 13 + 3 = 16. 13 x 16 = 208 > 180 No

Let’s try w = 12, and l = 12 + 3 = 15. 12 x 15 = 180 = 180 Yes

A good test taker will be able to do this math in their head, without doing any of the calculations we’ve shown here. Let’s run through the mental steps.

180 = lw

w = l + 3

SQRT[180] = 13.4 = lw so l and w straddle 13.4

the last two digits of l and w must have a product that is a multiple of ten since 180 is a multiple of ten. 2 and 5 are three apart and a multiple of 10, so try 12 x 15 = 180 Check.