Find the equation of the line that goes through (1, 2) and (3, 4).
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- <u>The </u><u>right </u><u>angled </u><u>below </u><u>is </u><u>formed </u><u>by </u><u>3</u><u> </u><u>squares </u><u>A</u><u>, </u><u> </u><u>B </u><u>and </u><u>C</u>
- <u>The </u><u>area </u><u>of </u><u>square </u><u>B</u><u> </u><u>has </u><u>an </u><u>area </u><u>of </u><u>1</u><u>4</u><u>4</u><u> </u><u>inches </u><u>²</u>
- <u>The </u><u>area </u><u>of </u><u>square </u><u>C </u><u>has </u><u>an </u><u>of </u><u>1</u><u>6</u><u>9</u><u> </u><u>inches </u><u>²</u>
- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>area </u><u>of </u><u>square </u><u>A</u><u>? </u>
The right angled triangle is formed by 3 squares
<u>We </u><u>have</u><u>, </u>
- Area of square B is 144 inches²
- Area of square C is 169 inches²
<u>We </u><u>know </u><u>that</u><u>, </u>
Let the side of square B be x
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
Thus, The dimension of square B is 12 inches
<h3><u>Now, </u></h3>Area of square C = 169 inches
Let the side of square C be y
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
Thus, The dimension of square C is 13 inches.
<h3><u>Now, </u></h3>It is mentioned in the question that, the right angled triangle is formed by 3 squares
The dimensions of square be is x and y
Let the dimensions of square A be z
<h3><u>Therefore</u><u>, </u><u>By </u><u>using </u><u>Pythagoras </u><u>theorem</u><u>, </u></h3>- <u>The </u><u>sum </u><u>of </u><u>squares </u><u>of </u><u>base </u><u>and </u><u>perpendicular </u><u>height </u><u>equal </u><u>to </u><u>the </u><u>square </u><u>of </u><u>hypotenuse </u>
<u>That </u><u>is</u><u>, </u>
<u>Here</u><u>, </u>
- Base = x = 12 inches
- Perpendicular = z
- Hypotenuse = y = 13 inches
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
Thus, The dimensions of square A is 5 inches
<h3><u>Therefore</u><u>,</u></h3>Area of square
Hence, The area of square A is 25 inches.