1) First try to find a perfect square number that divides evenly into the number under the square root. Perfect square numbers include 4, 9, 16, 25, 36, 49, etc. because they are squares of the numbers 2, 3, 4, 5, 6, 7, etc., respectively.
2) Separate the root expression into the two numbers you found as factors, with each number now under its own square root symbol. For example, sqrt(12) = sqrt(4) times sqrt(3).
3) Keep simplifying. The square root of a perfect square becomes just a number, without the radical square root symbol. sqrt(4)*sqrt(3) becomes 2*sqrt(3).
4) Check to see if anymore simplifying can be done, either by combining constants or finding another perfect square number that divides evenly into your factors. If no more simplifying can be done, you have your final "simplified radical form" (SRF) answer, 2*sqrt(3).
If there is a square root anywhere in the DENOMINATOR, you must "rationalize" the denominator, in other words, somehow get rid of the root in the bottom, in order for the expression to be considered simplest. For example, if you have 4/sqrt(2) you must use the trick of simplifying by multiplying the expression by sqrt(2)/sqrt(2). This will clear the square root in the bottom. If you multiply straight across, you would get 4*sqrt(2)/2, which simplifies again to 2*sqrt(2).
<span>If you have a more complicated fraction with a square root in the denominator, like (4+sqrt(3))/(5-sqrt(3)), you will need to use the "conjugate" to simplify. This means multiplying top and bottom by the expression in the denominator, but with the middle sign flipped. Here you would multiply by (5+sqrt(3))/(5+sqrt(3)), multiply straight across on top and bottom using FOIL-ing and distribution. Some terms will cancel in the bottom, leaving you with a simpler denominator that has no square roots.</span>
The relationship occurs because having two congruent Pentagons and generating a segment within them (or outside), congruence is extrapolated to the triangles generated within them. Thus, if there is congruence among the pentagons, it will exist between the formed triangles. In other words since the two pentagons are congruent, the corresponding angle pair is congruent. also the two corresponding side pairs are also congruent.
In the attached image for example, the ABCDE and KLMNO pentagons are congruent, so all of their internal division lines are also congruent (AC and KM)
yes, triangle DEF is similar to triangle DBC, BC corresponds to EF, and angle DCB corresponds to angle F.
Step-by-step explanation:
Part A: Angle D is congruent to angle D by the reflexive property. Since line BC is parallel to line EF then angle DCB = angle DFE by corresponding angles. Hence triangle DEF is similar to triangle DBS by the AA Similarity Postulate.
Part B: BC corresponds to EF because they are in the same order and the triangles listed DEF and DBC
Part C: Angle DCB corresponds to angle F since they are corresponding angles with the two given parallel lines BC and EF.