Word Problems?

Question

I am trying to use the Pythagorean theorem, but I cannot get an answer. The word problem is: the length of a rectangle is 1cm longer than its width. If the diagonal of the rectangle is 4cm, what are the dimensions (width and length) of the rectangle?

Answer 1

The width is x and the length is x + 1. Using the Pythagoerean Theorem, x^2 + (x + 1)^2 = 4^2, so x^2 + x^2 + 2x + 1 = 16. Simplify that to 2x^2 + 2x - 15 = 0, and you have a simple quadratic expression to solve.

Answer 2

I'll first state the equations that you might need to apply.

a^2 = b^2 + c^2, where a is the length of the slant side or the hypotheneuse.

To start this question, we must first give the width a variable and lets call it "x". Thus, we can form the equations,

Width = x
Length = x + 1
Diagonal/Hypotheneuse = 4

These 3 equations then apply using the Pythagoras Theorem, where "a" will be the length of the diagonal, "b" and "c" the width and length. Thus, substituting the values into the Pythagoras equation.

4^2 = x^2 + (x + 1)^2
16 = x^2 + x^2 + 2x + 1
16 = 2x^2 +2x + 1
2x^2 +2x - 15 = 0

You cannot solve this by the traditional cross-out method. To operate this equation you need to apply the General Solution for Quadratic Equation formula, which is given below. In this case, a = 2, b = 2, and c = -15.

x = [- b + (b^2 - 4ac)^(1/2)]/(2a)
and
x = [- b - (b^2 - 4ac)^(1/2)]/(2a)

Substituting the values,

x = [- 2 + ((2)^2 - 4(2)(-15))^(1/2)]/(2(2))
= 2.28
and
x = [- 2 - ((2)^2 - 4(2)(-15))^(1/2)]/(2(2))
= -3.28, rejected because of the negative sign.

From these solutions, the only possible answer would be x = 2.28cm. Thus the dimensions of the length and width are 3.28cm and 2.28cm respectively.

Hope this helps.

Answer 3

Let L be the length (one leg of the right triangle) and W be the width (the other leg).

L can be expressed in terms of the width:

L = W + 1

The hyponenuse = 4

W^2 + (W + 1)^2 = 4^2
W^2 + W^2 + W + W + 1 = 16
2W^2 + 2W - 15 = 0

I think the quadratic formula must be used here. I get W = 2.284 (approximately).

Answer 4

4² = x² + (x+1)²
4² = 2x²+ 2x + 1
0 = 2x² + 2x - 15

use the quadratic formula to solve for x

Answer 5

2.5 cm

Answer 6

w^2+(w+1)^2=4^2
2w^2+2w=15
2w^2+2w-15=0
w=[-2+/-rt(4+120)]/4
=(1+rt31)/2

Answer 7

l=w+1
diagonal=4
d^2=l^2+w^2
4^2=(w+1)^2+w^2
16=w^2+2w+1+w^2=2w^2+2w+1
2w^2+2w-15=0
w=(-2=sqrt(4+4*2*15))/4
w=(-2+sqrt(124))/4=2.284
l=w=1=3.284

check
2.284^2+3.284^2=16.001
sqrt(16.001)=4