Examples with solutions for Area of a Rectangle: Using short multiplication formulas

Exercise #1

Given the rectangle ABCD

AB=X the ratio between AB and BC is equal tox2 \sqrt{\frac{x}{2}}

We mark the length of the diagonal A A with m m

Check the correct argument:

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Video Solution

Step-by-Step Solution

Let's find side BC

Based on what we're given:

ABBC=xBC=x2 \frac{AB}{BC}=\frac{x}{BC}=\sqrt{\frac{x}{2}}

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

2x=xBC \sqrt{2}x=\sqrt{x}BC

Let's divide by square root x:

2×xx=BC \frac{\sqrt{2}\times x}{\sqrt{x}}=BC

2×x×xx=BC \frac{\sqrt{2}\times\sqrt{x}\times\sqrt{x}}{\sqrt{x}}=BC

Let's reduce the numerator and denominator by square root x:

2x=BC \sqrt{2}\sqrt{x}=BC

We'll use the Pythagorean theorem to calculate the area of triangle ABC:

AB2+BC2=AC2 AB^2+BC^2=AC^2

Let's substitute what we're given:

x2+(2x)2=m2 x^2+(\sqrt{2}\sqrt{x})^2=m^2

x2+2x=m2 x^2+2x=m^2

Answer

x2+2x=m2 x^2+2x=m^2

Exercise #2

Given the rectangle ABCD

AB=X

The ratio between AB and BC is x2 \sqrt{\frac{x}{2}}

We mark the length of the diagonal A the rectangle in m

Check the correct argument:

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Video Solution

Step-by-Step Solution

Given that:

ABBC=x2 \frac{AB}{BC}=\sqrt{\frac{x}{2}}

Given that AB equals X

We will substitute accordingly in the formula:

xBC=x2 \frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}

x2=BCx x\sqrt{2}=BC\sqrt{x}

x2x=BC \frac{x\sqrt{2}}{\sqrt{x}}=BC

x×x×2x=BC \frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC

x×2=BC \sqrt{x}\times\sqrt{2}=BC

Now let's focus on triangle ABC and use the Pythagorean theorem:

AB2+BC2=AC2 AB^2+BC^2=AC^2

Let's substitute the known values:

x2+(x×2)2=m2 x^2+(\sqrt{x}\times\sqrt{2})^2=m^2

x2+x×2=m2 x^2+x\times2=m^2

We'll add 1 to both sides:

x2+2x+1=m2+1 x^2+2x+1=m^2+1

(x+1)2=m2+1 (x+1)^2=m^2+1

Answer

m2+1=(x+1)2 m^2+1=(x+1)^2

Exercise #3

Given the rectangle ABCD

AB=Y AD=X

Triangular area DEC equal to S

Expresses the square of the difference of the sides of the rectangle

band means of X, Y and S

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Video 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

Exercise #4

Look at the rectangle in the figure.

x>0

The area of the rectangle is:

x213 x^2-13 .

Calculate x.

x-4x-4x-4x+1x+1x+1x²-13

Video Solution

Answer

x=3 x=3

Exercise #5

Shown below is the rectangle ABCD.

AB = y

AD = x

Express the square of the sum of the sides of the rectangle using the area of the triangle DEC.

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Video Solution

Answer

(x+y)2=4s[sy2+sx2+1] (x+y)^2=4s\lbrack\frac{s}{y^2}+\frac{s}{x^2}+1\rbrack