Construct a right triangle whose base is ^@17 \space cm^@ and sum of its hypotenuse and another side is ^@23 \space cm^@.


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Answer:

C B X D A

Step by Step Explanation:
  1. Let ^@ABC^@ be the required triangle such that ^@BC = 17 \space cm, \angle B = 90 ^ \circ^@ and ^@AC + AB = 23 \space cm^@. Draw a ^@BC = 17 \space cm.^@ C B
  2. At ^@B^@ construct ^@\angle CBX = 90^\circ.^@ C B X
  3. From ^@B^@ cut off ^@BD = 23 \space cm^@. C B X D
  4. Join ^@CD^@ and draw perpendicular bisector of ^@CD^@ intersecting ^@BD^@ at ^@A^@. C B X D A
  5. Join ^@AC^@ to get the required triangle ^@ABC^@. C B X D A

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