Please help me solve this problem ASAP
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[tex]\bold{\huge{\blue{\underline{ Solution }}}}[/tex]
The right angled triangle is formed by 3 squares
We have,
We know that,
[tex]\bold{ Area \: of \: square = Side × Side }[/tex]
Let the side of square B be x
Subsitute the required values,
[tex]\sf{ 144 = x × x }[/tex]
[tex]\sf{ 144 = x² }[/tex]
[tex]\sf{ x = √144}[/tex]
[tex]\bold{\red{ x = 12\: inches }}[/tex]
Thus, The dimension of square B is 12 inches
Area of square C = 169 inches
Let the side of square C be y
Subsitute the required values,
[tex]\sf{ 169 = y × y }[/tex]
[tex]\sf{ 169 = y² }[/tex]
[tex]\sf{ y = √169}[/tex]
[tex]\bold{\green{ y = 13\: inches }}[/tex]
Thus, The dimension of square C is 13 inches.
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
That is,
[tex]\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}[/tex]
Here,
Subsitute the required values,
[tex]\sf{ (z)² + (x)² = (y)² }[/tex]
[tex]\sf{ (z)² + (12)² = (169)² }[/tex]
[tex]\sf{ (z)² + 144 = 169}[/tex]
[tex]\sf{ (z)² = 169 - 144 }[/tex]
[tex]\sf{ (z)² = 25}[/tex]
[tex]\bold{\blue{ z = 5 }}[/tex]
Thus, The dimensions of square A is 5 inches
Area of square
[tex]\sf{ = Side × Side }[/tex]
[tex]\sf{ = 5 × 5 }[/tex]
[tex]\bold{\orange{ = 25\: inches }}[/tex]
Hence, The area of square A is 25 inches.