Line RS intersects triangle BCD at two points and is parallel to segment DC. Triangle B C D is cut by line R S. Line R S goes th
rough sides D B and C B. Lines D C and R S are parallel. width= Which statements are correct? Select three options. △BCD is similar to △BSR. StartFraction B R Over R D EndFraction = StartFraction B S Over S C EndFraction If the ratio of BR to BD is Two-thirds, then it is possible that BS = 6 and BC = 3. (BR)(SC) = (RD)(BS) StartFraction B R Over R S EndFraction = StartFraction B S Over S C EndFraction
1 answer:
Answer:
The correct statements are;
1) ΔBCD is similar to ΔBSR
2) BR/RD = BS/SC
3) (BR)(SC) = (RD)(BS)
Step-by-step explanation:
1) Since RS is parallel to DC, we conclude that;
∠BDC = ∠BRS (Angles formed on the same side of the transversal)
Furthermore;
∠BCD = ∠BSR (Angles formed on the same side of the transversal)
∠CBD = ∠CBD (Reflexive property)
Thus;
ΔBCD ~ ΔBSR by the Angle-Angle-Angle (AAA) similarity criterion.
2) Given that ΔBCD ~ ΔBSR, we obtain;
BC/BS = BD/BR → (BS + SC)/BS = (BR + RD)/BR = 1 + SC/BS = RD/BR + 1
1 + SC/BS = 1 + RD/BR thus, SC/BS = 1 + BR/RD - 1
SC/BS = RD/BR
By inverting both sides we find;
BR/RD = BS/SC
3) From BR/RD = BS/SC, we apply cross multiplication;
BR/RD = BS/SC leads to;
BR × SC = RD × BS → (BR)(SC) = (RD)(BS).
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Answer:
The three errors are:
1) Incorrectly swapping the variables x and y.
2) Failure to use the ± symbol.
3) The domain is wrong; it should be x ≤ 0.
Step-by-step explanation:
Review the following steps.
Step 1:
The initial step is incorrect because the variables x and y were switched improperly; it should be:
Step 2:
Step 3:
Step 4: , for x ≥ 0
Step 4 contains an error as the ± sign is required. Additionally, the function's domain is inaccurately stated.
The radicand must be nonnegative, implying that x must satisfy x ≤ 0.
Response:
The probability that a student has a pet, given they do not have any siblings is:
Option: D ( 60%)
Step-by-step breakdown:
Let A represent the situation where a student lacks a sibling.
Let B signify the occurrence that a student has a pet.
Consequently, A∩B refers to the event in which a student is without siblings but possesses a pet.
Let P denote the chance of an event happening.
We need to determine:
P(B|A)
From our knowledge:
From the data provided:
P(A)=0.25
and P(A∩B)=0.15
Thus,
which expressed as a percentage is:
Therefore, the probability is:
60%