Answer:
Yes, Mina is correct
Step-by-step explanation:
Let the triangles be A and B
Given
Triangle A:

Triangle B:

Required
Is Mina's claim correct?
First, we calculate the third angle in both triangles.
For A:


For B:


For triangle A, the angles are: 34, 57 and 89
For triangle B, the angles are: 34, 57 and 89
<em>Since both triangles have the same angles, then by the postulate of AAA (Angle-Angle-Angle), the triangles are similar.</em>
It would be at least 70 yph I am probably sure
The answer is B because u always start of like that
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.
Let the length of one of the diagonals be 2x and the other be 2y, then
cos (76/2) = x/10
x = 10 cos 38 = 7.88 cm
sin (76/2) = y/10
y = 10 sin 38 = 6.16 cm