A parallelogram has consecutive sides with lengths 9 and 7 and diagonals of integral length How...

A parallelogram has consecutive sides with lengths 9 and 7 and diagonals of integral length How long are these diagonals?

Expert Solution

By the properties of the parallelogram, d₁² + d₂² = 2a² + 2b², where d1 and d2 are the diagnols of the parallelogram and a and b are the lengths of the sides of the parallelogram.

Here a = 9 and b = 7

Therefore d₁² + d₂² = 2*9² + 2*7² = (2 * 81) + (2 * 49) = 162 + 98 = 260

Since the diagnols are integers therefore they have to be perfect squares whose sum is = 260

d1²	d2²	d1²+d2²	d1	d2
1	259	260	1	16.093477
4	256	260	2	16
9	251	260	3	15.84298
16	244	260	4	15.620499
36	224	260	6	14.96663
49	211	260	7	14.525839
64	196	260	8	14
81	179	260	9	13.379088
100	160	260	10	12.649111
121	139	260	11	11.789826
144	116	260	12	10.77033
169	91	260	13	9.539392
196	64	260	14	8
225	35	260	15	5.9160798
296	-36	260	17.204651	#NUM!

We see only two cases cropping up, where both diagnols are integers.

Case 1: d1 = 2 and d2 = 16. This can be discarded as the lengths of the diagnol 1 is too small to form a parallelogram.

Case 2: d1 = 8 and d2 = 14. This is the correct and only possible option.

milcah answered 2 years ago

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