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In a regular pentagon ABCDE and regular hexagon MNOPQR, 2AB = 3MN and CD + QR = 10. Which is 4BC + 5PQ? Answer 44 42 38 33
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well write the equations in terms of the same sides since both polygons are regular so they can now be written and solved as simultaneous equations 2AB - 3MN = 0 AB + MN = 10 so multiply the 2nd equation by 3 and add them 2AB - 3MN = 0 3AB + 3MN = 30 --------------- 5AB = 30 so bow you can solve for AB then go and find MN... and then realise 4BC + 5PQ = 4AB + 5PQ hope it helps
oops should read 4AB + 5MN
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