Given: isosceles trapezoid efgh prove: δfhe ≅ δgeh it is given that trapezoid efgh is an isosceles trapezoid. we know that fe ≅ gh by the definition of . the base angle theorem of isosceles trapezoids verifies that angle is congruent to angle . we also see that eh ≅ eh by the property. therefore, by , we see that δfhe ≅ δgeh.

Given: An Isosceles trapezoid EFGH in which EF =GH

To prove: ΔFHE ≅ ΔGEH

Proof: In Isosceles trapezoid EFGH, Considering two triangles ΔFHE and ΔGEH

  1. FE ≅ G H   → [ Given]

  2. ∠H = ∠E

→ Draw GM⊥HE and FN ⊥EH, and In Δ GMH and ΔFNE,

                          GH=FE [Given]

          ∠M+∠N=180° so GM║FN and GF║EH, So GFMN is a rectangle.]

∴ GM =FN [opposite sides of rectangle]

∠GMH = ∠FNE [ Each being 90°]

Δ GMH ≅ ΔFNE  [ Right hand side congruency]

→∠H =∠E [CPCT]

→ Side EH is common i.e  EH ≅ EH .

→ΔFHE ≅ ΔGEH. [SAS]

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}