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3 years ago

a rectangle is 4 centimeters longer than it is wide. the area is 117 square centimeters. find the dimensions of the rectangle

Mathematics

1 answer:

vekshin13 years ago

7 0

Answer:

width: 9 cm
length: 13 cm

Step-by-step explanation:

Integer factors of 117 include ...

... 117 = 1×117 = 3×39 ≈ 9×13

The last factor pair is two factors that differ by 4. We can take these to be the dimensions of the rectangle.

_____

If you want to write an equation for width (w), it might be ...

... w(w+4) = 117

... w² +4w -117 = 0

The factorization problem for this quadratic is the problem of finding two factors of 117 that differ by 4. That is what we have done, above.

If you want to solve this by completing the square, you can to this:

... w² +4w = 117

... w² +4w +4 = 121 . . . . . add 4 = (4/2)² to complete the square

... (w+2)² = 121

... w + 2 = ±√121 = ±11 . . . . take the square root

... w = -2 ± 11 . . . . . we're only interested in the positive solution

... w = 9, then w+4 = 13 and the dimensions are 9 cm by 13 cm.

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3 years ago

Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)

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The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

$A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ABC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|$

On solving we get:

$Area\,of\,triangle\,=25.$

In Δ ADC:

$A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)$

Now, to get the area of triangle ADC:

$Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

$Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|$

On solving it by same process as above we get:

$Area\,of\,triangle\,=114$

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

$25+114\\=25+114\\=139\ unit^2$

Therefore, area of quadrilateral ABCD is 139 unit^2.

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