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Makapuu

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Makapu‘u Beach and Waimānalo Bay beyond seen from the highway overlook at Makapu‘u
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Makapu‘u Beach and Waimānalo Bay beyond seen from the highway overlook at Makapu‘u

Makapu‘u is the extreme eastern end of the Island of O‘ahu in the Hawaiian Islands, comprising the remnant of a ridge that rises 647 feet (197 meters) above the sea. The cliff at Makapu‘u Point forms the eastern tip and is the site of a prominent lighthouse. The place name of this area, meaning "bulging eye" in Hawaiian, is thought to derive from the name of an image said to have been located in a cave here called Keanaokeakuapōloli. The entire area is quite scenic and a panoramic view is presented at the lookout on Kalanianaole Highway (State Rte. 72) where the roadway surmounts the cliff just before turning south towards leeward O'ahu and Honolulu.

Features of special interest in this area are:

The Makapu‘u area is reached approximately 2 kilometres (1.2 mi) east of Waimānalo Beach on Kalanianaole Highway (State Rte. 72) or from the Honolulu side (south shore; Hawaii Kai) travelling east along the same highway beyond Sandy Beach. The Wayside Park (road for hiking in to the lighthouse) is about midway up the draw on the right-hand side coming from the Honolulu side.

See also

 


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