^@ \triangle ABC ^@ is a triangle such that ^@ \angle C = 90^\circ. ^@ Suppose ^@ AC = 12 \space cm, AB = 13 \space cm ^@ and the perpendicular distance from ^@ C ^@ to ^@ AB ^@ is ^@ x \space cm. ^@ Find the value of ^@ x. ^@


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Answer:

^@ 4.62 ^@

Step by Step Explanation:
  1. We need to find ^@ x, ^@ the perpendicular distance from ^@ C ^@ to ^@ AB, ^@ if ^@ \angle C = 90^\circ, AC = 12 \space cm , ^@ and ^@ AB = 13 \space cm. ^@ Let ^@ D ^@ be the point where the perpendicular from ^@ C ^@ meets ^@ AB. ^@
    Let ^@ D ^@ be the point where the perpendicular from ^@ C ^@ meets ^@ AB. ^@
    A C B D 12 cm 13 cm
  2. Using Pythagoras' theorem in triangle ^@ ABC ^@
    ^@ BC = \sqrt{ 13^2 - 12^2 } = 5 ^@
  3. Since,
    ^@ \begin{align} & \angle ACB = \angle ADC = 90^\circ \\ & \angle BAC = \angle CAD && [\text{ Common angle }] \\ & \angle CBA = \angle DCA && [\text{ Angle sum property }] \\ & \triangle ACB \text{ and } \triangle ADC \text{ are similar by } AAA \text{ criterion } \\ & \implies \dfrac{ BC }{ AB } = \dfrac{ CD }{ AC } \\ & \implies \dfrac{ 5 }{ 13 } = \frac{ x }{ 12 } \\ & \implies x = 4.62 \end{align} ^@
  4. Hence, the value of ^@ x ^@ is ^@ 4.62. ^@

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