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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Step by Step Explanation:
|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.^@ |
|At ^@B^@ construct ^@\angle CBX = 90^\circ.^@|
|From ^@B^@ cut off ^@BD = 23 \space cm^@.|
|Join ^@CD^@ and draw perpendicular bisector of ^@CD^@ intersecting ^@BD^@ at ^@A^@.|
|Join ^@AC^@ to get the required triangle ^@ABC^@.|