Rectangle ABCD: Express (X-Y)² Using Side Lengths and Triangle Area S

Question

Given the rectangle ABCD

AB=Y AD=X

The triangular area DEC equals S:

Express the square of the difference of the sides of the rectangle

using X, Y and S:

YYYXXXAAABBBCCCDDDEEE

Video Solution

Solution Steps

00:00 Express the square of the rectangle side differences
00:11 Substitute in the side expressions according to the given data and proceed to open parentheses
00:20 Triangle area expression according to the given data
00:27 Triangle height equals rectangle side X
00:30 Apply the formula for calculating the area of the triangle
00:33 (height x base) divided by 2
00:40 Determine the expression for X
00:46 Determine the expression for Y
00:54 Determine the expression for the product of the sides
01:02 Substitute these expressions into our shortened multiplication formula
01:17 Pay attention to squaring both the numerator and the denominator
01:29 Remove the common factor from the parentheses
01:44 This is the solution

Step-by-Step Solution

Since we are given the length and width, we will substitute them according to the formula:

(ADAB)2=(xy)2 (AD-AB)^2=(x-y)^2

x22xy+y2 x^2-2xy+y^2

The height is equal to side AD, meaning both are equal to X

Let's calculate the area of triangle DEC:

S=xy2 S=\frac{xy}{2}

x=2Sy x=\frac{2S}{y}

y=2Sx y=\frac{2S}{x}

xy=2S xy=2S

Let's substitute the given data into the formula above:

(2Sy)22×2S+(25x)2 (\frac{2S}{y})^2-2\times2S+(\frac{25}{x})^2

4S2y24S+4S2x2 \frac{4S^2}{y^2}-4S+\frac{4S^2}{x^2}

4S=(S2y+S2x1) 4S=(\frac{S^2}{y}+\frac{S^2}{x}-1)

Answer

(xy)2=4s[sy2+sx21] (x-y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}-1\rbrack