Question

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in right triangle ( abc ), ( ac=sqrt{5}mathrm{cm} ) and ( overline{bc} ) is 1 cm longer than ( overline{ab} ). ( bc=) cm.

Explanation:

Step1: Let \(AB = x\) cm

Since \(BC\) is 1 cm longer than \(AB\), then \(BC=(x + 1)\)cm.

Step2: Apply the Pythagorean theorem

In right - triangle \(ABC\) with \(\angle A=90^{\circ}\), by the Pythagorean theorem \(AB^{2}+AC^{2}=BC^{2}\).
We know that \(AC = \sqrt{5}\)cm, \(AB=x\)cm and \(BC=(x + 1)\)cm.
Substitute these values into the Pythagorean theorem: \(x^{2}+(\sqrt{5})^{2}=(x + 1)^{2}\).

Step3: Expand the equation

Expand \((x + 1)^{2}\) using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\), where \(a=x\) and \(b = 1\).
\(x^{2}+5=x^{2}+2x+1\).

Step4: Solve for \(x\)

Subtract \(x^{2}\) from both sides of the equation:
\(x^{2}-x^{2}+5=x^{2}-x^{2}+2x + 1\).
We get \(5=2x+1\).
Subtract 1 from both sides: \(2x=5 - 1=4\).
Divide both sides by 2: \(x = 2\).

Step5: Find the length of \(BC\)

Since \(BC=x + 1\) and \(x = 2\), then \(BC=2+1=3\)cm.

Answer:

\(3\)