Mathematics Question and Solution

By Ogheneovo Daniel Ephivbotor
10th June, 2024

Let the equal side lengths be a.


14² = a²+16²-2*16acosb

196 = a²+256-32acosb --- (1).


21² = 2a²+2a²cosb

cosb = (441-2a²)/2a² --- (2).


Substituting (2) in (1).


196 = a²+256-32a(441-2a²)/2a²

32(441-2a²)/2a = a²+256-196

32(441-2a²)/2a = a²+60

14112-64a² = 2a³+120a

2a³+64a²+120a-14112 = 0


It implies;


a = 12 units.

a is the required length.

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