A triangle ^@ABC^@ with ^@\angle ABC = 90^\circ^@, has the length of the side ^@AB = 128 \space cm^@ and of ^@BC = 96 \space cm^@. What is the length of the perpendicular line from side ^@AC^@ to point ^@B^@?


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Answer:

^@ 76.8 \space cm^@

Step by Step Explanation:

  1. In the right angled triangle ^@ABC, AB = 128 \space cm, BC = 96 \space cm^@
    ^@ \begin {align} Now, \space AC & = \sqrt{ (AB)^2 + (BC)^2 } && \ldots \text { (As per Pythagoras Theorem) } \\ & = \sqrt{(128)^2 + (96)^2} \\ & = \sqrt{(16384 + 9216)} \\ & = \sqrt{(25600)} \\ & = 160 \space cm \end {align}^@
  2. ^@ \begin {align} \text { The area of the right angled triangle ABC, when base is } BC & = \dfrac { AB \times BC} {2} \\ & = \dfrac { 128 \times 96 } {2} \\ & = 6144 \space cm^2 && \ldots(1) \end {align}^@
  3. ^@ \begin {align} \text { The area of the right angled triangle ABC, when base is AC } & = \dfrac { AC \times BD } {2} \\ & = \dfrac { 160 \times BD } {2} && \ldots(2) \end {align}^@
  4. Since, the area in the equations ^@(1)^@ and ^@(2)^@ should be the same, equating the area in equation ^@(2)^@, we get:
    ^@ \begin {align} & \implies \dfrac { 160 \times BD } {2} = 6144 \\ & \implies 160 \times BD = 12288 \\ & \implies BD = \dfrac { 12288 } { 160 } \\ & \implies BD = 76.8 \space cm \end {align}^@
  5. Therefore, the length of the perpendicular line from side ^@AC^@ to point ^@B^@ is ^@ 76.8 \space cm^@.

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