Triangle similarity criteria

 AAA(which is same as AAS or ASA)
SAS
SSS
RHL

Note that these are same as triangle congruence criteria but for congruence we ensure that sides are equal and for similarity we ensure that corresponding sides have the same ratio.

Comments

Post a Comment

Popular posts from this blog

IOQM 2024 Paper solutions (Done 1-21, 29)

Image
Q1 (number-theory). The smallest positive integer that does not divide 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 is: Solution : This product has all the integers from 1 to 9. So smallest integer not dividing it would be the next prime after 9. Which is 11. Answer: 11 Q2 (combinatorics): The number of four-digit odd numbers having digits 1, 2, 3, 4, each occurring exactly once, is: Solution: Last digit has to be 1 or 3: 2C1 Choosing remaining 3 digits: 3C3 Arranging those 3 digits: 3! Answer: 2C1*3C3*3! = 12 Q3 (number-theory): The number obtained by taking the last two digits of \(5^{2024}\) in the same order is: Solution: To get last 2 digits from number, divide by 100 and take the remainder. So we have to compute mod 100. 25 mod 100 = 25 25.5 mod 100 = 125 mod 100 = 25 5^3.5 mod 100 = 25.5 mod 100 = 25 5^4.5 mod 100 = 25.5 mod 100 = 25 ..... So: 5^2024 mod 100 = 25. Q4 (geometry): Let ABCD be a quadrilateral with \(\angle ADC = 70^\circ\), \(\angle ACD = 70^\circ\), \(\angle ACB = 10^\circ\...
Read more

Combinatorics DPP - RACE 6 - Q16 pending discussion

Image
Q1 . A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is: Solution: One wrong way to solve this: For the mandatory questions: 7C3*8C4 Now 3 more questions need to be answered from 4 of A and 4 of B. Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B 4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1 = 4 + 24 + 24 + 4 = 56 Finally: 7C3*8C4*56 = 35*70*56 = 137200 Why is it wrong? Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B). But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting. Correct Solution : Simply create the cases for A,B: (3,7),(4,6),(5,5),(6,4) 7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926. Q2 . A kindergarten teacher has 25 kids in her...
Read more

Algebra DPP 1.3 Quadratics

 Q1 . For what values of b do the equations: 1988x² + bx + 8891 = 0 and 8891x² + bx + 1988 = 0 have a common root? Solution: We are given two quadratic equations: 1988 x 2 + b x + 8891 = 0 1988x^2 + bx + 8891 = 0 8891 x 2 + b x + 1988 = 0 8891x^2 + bx + 1988 = 0 We are to find the values of b b for which they have a common root . Step 1: Let α \alpha be the common root. Then α \alpha satisfies both equations. So, From equation (1): 1988 α 2 + b α + 8891 = 0 (i) 1988\alpha^2 + b\alpha + 8891 = 0 \quad \text{(i)} From equation (2): 8891 α 2 + b α + 1988 = 0 (ii) 8891\alpha^2 + b\alpha + 1988 = 0 \quad \text{(ii)} Step 2: Subtract equation (i) from equation (ii): ( 8891 α 2 − 1988 α 2 ) + ( b α − b α ) + ( 1988 − 8891 ) = 0 (8891\alpha^2 - 1988\alpha^2) + (b\alpha - b\alpha) + (1988 - 8891) = 0 ( 8891 − 1988 ) α 2 − ( 8891 − 1988 ) = 0 (8891 - 1988)\alpha^2 - (8891 - 1988) = 0 6903 α 2 − 6903 = 0 6903\alpha^2 - 6903 = 0 α 2 = 1 ⇒ α = ± 1 \alpha^2 = 1 \Rightar...
Read more