Two vertical buildings of heights $30$ m and $38$ m are $15$ m apart. A cable is pilled taut and attached from the top of one building to the top of the other building. Find the length of the cable.

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Solution

$d i f f e r e n c e o f h e i g h t o f b u i l d i n g = 38 - 30 = 8 m g a p b e t w e e n b u i l d i n g = 15 m ∴ l e n g t h o f c a b l e = \sqrt{8^{2} + 15^{2}} = 17 m$

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Two vertical buildings of heights 30 m and 38 m are 15 m apart. A cable is pilled taut and attached from the top of one building to the top of the other building. Find the length of the cable.

differenceofheightofbuilding=38−30=8mgapbetweenbuilding=15m∴lengthofcable=√82+152=17m

Two vertical buildings of heights $30$ m and $38$ m are $15$ m apart. A cable is pilled taut and attached from the top of one building to the top of the other building. Find the length of the cable.

$d i f f e r e n c e o f h e i g h t o f b u i l d i n g = 38 - 30 = 8 m g a p b e t w e e n b u i l d i n g = 15 m ∴ l e n g t h o f c a b l e = \sqrt{8^{2} + 15^{2}} = 17 m$