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23 Mar 2005, 15:26
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In a IIgm (parallelogram) the lengths of adjacent sides are 12 and 14 respectively. If the length of one diagonal is 16 find that of the other.
(a) 21
(b) 20.6
(c) 20.4
(d) 20.1
(e) 19.5
(a) 21
(b) 20.6
(c) 20.4
(d) 20.1
(e) 19.5
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23 Mar 2005, 16:28
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anirban16
(B) 20.6
The figure becomes too complex, but I'd post the solution now, and the figure later (from home)
This holds true if the smaller diagonal is 16.
Assume distance between the 14-14 parallel sides is z, and dropping the perpendicular from the 12-14 vertex onto the 14 side results in a triangle with a base x. (Of course its height would be z, and hypotenuse would be 12.
Therefore, 144 = x^2 + y^2
Also, in the right triangle formed by the smaller diagonal,
16^2 = z^2 + (14-x)^2
or 256 = z^2 + 196 + x^2 - 28x
Solving them, we get x = 3.
Putting x back in, we get z = sqrt(134).
Applying these values to the right triangle formed by the larger diagonal,
diagonal^2 = z^2 + (14+x)^2
= 134 + 17^2 = 134 + 289 = 423
=> longer diagonal = sqrt (423) ~ 20.6
24 Mar 2005, 06:53
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p^2+q^2=2(a^2+b^2) => where p is diag2 and q is diag1, a and b are the adjacent sides. => insert the values => 256+q^2=2(144+196) => q^2=424=~20,6
