Kenji uses the diagram to determine the quotient of StartFraction 9 Over 10 EndFraction and Three-fifths. A fraction bar labeled
3 answers:
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<h3>Answer with explanation:</h3>
It is given that:
Circle 1 has center (−4, −7) and a radius of 12 cm.
Circle 2 has center (3, 4) and a radius of 15 cm.
Two circles are said to be similar if by some translation and dilation it could be placed over the other to form the same circle.
The circles are similar because the transformation rule ( x,y ) → (x+7,y+11) can be applied to Circle 1 and then dilate it using a scale factor of 5/4
( Since, as the center of circle 1 is (-4,-7)
so,
(-4+7,-7+11) → (3,4)
( Since, the radius of circle 1 is 12 and that of circle 2 is 15 cm.
so, let the scale factor be k .
that means :
)
The artistic crop isn't helpful; it cuts off some vertex names.
The circumcenter H is the meet of the perpendicular bisectors of the sides, helpfully drawn. We have right triangle ELH, right angle L, so
EH² = HL² + EL²
EL = √(EH² - HL²) = √(5.06²-2.74²) ≈ 4.2539393507665339
Since HL is a perpendicular bisector of EF, we get congruent segments FL=EL.
Answer: 4.25
