Which transformation will map an isosceles trapezoid onto itself? rotation by 180° about its center, reflection across a diagona
l, reflection across a line joining the midpoints of the nonparallel sides,rotation by 360° about its center
2 answers:
Answer:
The correct option is 4.
Step-by-step explanation:
The non parallel sides of an isosceles trapezoid are congruent.
The image of an isosceles trapezoid is same as the preimage of isosceles trapezoid if
1. Reflection across a line joining the midpoints of parallel sides.
2. Rotation by 360° about its center.
3. Rotation by 360° about origin.
If we rotate the trapezoid by 180° about its center, then the parallel sides will interchanged.
If we reflect the trapezoid across a diagonal, then the resultant figure will be a parallelogram.
If we reflect across a line joining the midpoints of the nonparallel sides, then the parallel sides will interchanged.
After rotation by 360° about the center, we always get an onto figure.
Therefore option 4 is correct.
Answer: The answer is (d) rotation by 360° about its centre.
Step-by-step explanation: As shown in the attached figure, AB and CD are parallel sides of an isosceles trapezoid ABCD with centre 'O' and AD and BC are equal and non-parallel sides.
(i) If ABCD is rotated by 180° about its centre, then the new figure will not coincide with the original one. So, this option is not correct.
(ii) If ABCD is reflected across a diagonal, then also the new figure will be reverse of the original one and so this option is also incorrect.
(iii) If ABCD is reflected across a line joining the mid-points of the non-parallel sides, then also the two figures will be opposite of each other. This option will also not work.
(iv) If we rotate ABCD through 360° about centre 'O', the both the figures will coincide with each other.
Thus, the correct option is (d) rotation by 360° about its centre.
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On the other side, if b is negative and gets closer to zero, then 1/b will be negative and those negative values will decrease without bound. So 1/b approaches negative infinity if we approach 0 on the left (or negative) side.
The graph of y = 1/x shows this. See the diagram below. Note the vertical asymptote at x = 0. The portion to the right of it has the curve go upward to positive infinity as x approaches 0. The curve to the left goes down to negative infinity as x approaches 0.
