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What is the scale factor of the perimeters of triangle ABC and triangle WXY?
APPLYING THE SCALE FACTOR OF SIMILAR TRIANGLES TO THE PERIMETER The scale factor of two similar triangles (or any geometric shape, for that matter) is the ratio between two corresponding sides. In today's lesson, we will show that this same scale factor also applies to the ratio of the two triangles' perimeter. This is fairly easy to show, so today's lesson will be short. PROBLEM Two triangles, ΔABC and ΔADE are similar, ΔABC∼ ΔADE. The scale factor, AB/AD is 6/5. Find the ratio of the perimeters of the two triangles. Similar triangles in geometry STRATEGY We will use the definition of the scale factor to define one set of sides in terms of the other set of sides, Then, apply the definition of the perimeter. and write out the perimeter of both triangles using one set of sides. SOLUTION (1) ΔABC∼ ΔADE //Given (2) AB/AD = 6/5 //Given (3) BC/DE = 6/5 //(1), (2), scale factor is the same for all sides in similar triangles. (4) AC/AE = 6/5 //(1), (2), scale factor is the same for all sides in similar triangles. (5) AB = 6/5*AD // rearrange (2) (6) BC = 6/5*DE // rearrange (3) (7) AC = 6/5*AE // rearrange (4) (8) PABC=AB+BC+AC //definition of perimeter (9) PADE=AD+DE+AE //definition of perimeter (10)PABC=6/5AD+6/5DE+ 6/5*AE //(8), (5), (6) , (7), Transitive property of equality (11)PABC=6/5*(AD+DE+AE) //(10), Distributive property of multiplication (12) PABC=6/5*PADE //(11), (9), Transitive property of equality (13) PABC/PADE=6/5 And so we have easily shown that the scale factor of similar triangles is the same for the perimeters.
How do you get the product of binomial and trinomial?
You multiply each element of the binomial into each element of the trinomial and then combine like terms. For example, (ax + b)*(cx2 + dx + e) = acx3 + adx2 + aex + bcx2 + bdx + be = acx3 + (ad + bc)x2 + (ae + bd)x + be
When is a polynomial in one variable in standard form?
The standard form of a polynomial in x isa(n)*x^n + a(n-1)*x^(n-1) + ... + a(1)*x + a(0) = 0where the a(i) are constants, for i = 0, 1, 2, ..., n.
Find the acute angle between the diagonals of rectangle whose side are 5cm and 7cm methods?
Name the rectangle ABCD,letting the larger side be AB(7cm) and the smaller side be AC(5cm). Draw the two diagonals across the rectangle. Name the point of intersection of the two diagonals (which resembles an X) F. Name the midpoint between the smaller side AC, to be E. AFC is the acute angle you are looking for, angle AFE is half of that. So AFE is a right angled triangle and AE is equal to half of 5cm, which is 2.5; FE is equal to half of 7cm, which is 3.5. Now, taking triangle AFE, Tan AFE=2.5/3.5 AFE=35.537 degrees Since the angle AFE is half of the acute angle AFC that you are looking for, multiply AFE into two So, 35.537*2 =71.075 degrees =71.1 degrees (to 1 decimal place) AFC= 71.1 (ANS)
