"I thought the study techniques in Get The Best Grades With the Least Amount of Effort would be just as useless as all the books my Dad had already bought me. But this time I thought I would try just 3 of the tips for 1 week to see what happened. I couldn't believe how much better I felt about studying. I only have one piece of advice for other students – just do it!" ~ Wilson, United States
How many more squares in the figure above must be shaded so that the ratio of the shaded squares to the unshaded squares is the same as the ratio of the length of the shorter side of the rectangle to the length of the longer side?
The ratio of short : long sides is 3:6 which can also be expressed as 1:2
However, this doesnt mean half, it means 1 for every 2 (ie, there are 3 total)!!!
The entire rectangle contains 3*6 = 18 squares. To get the same ratio as above of shaded:unshaded, you need to divide that into three even portions (obviously groups of 6), where one portion is shaded, the other two are not….
if you have 6 shaded and 12 unshaded then you have 6:12 which reduces to 1:2, same as above.
If there are already 5 shaded, then you need one more to make six – the answer is D
DC || AB since trapezoid
< CAB = < ACD opposite angles of the line intersecting parallel lines are equal
< ACD =
AD = DC – opposite sides of equal angles are equal in a triangle.
For #8 – I am rusty on precise wording of geometry theorems, but will give you the general idea here:
Basically, to prove this parallelogram is a rhombus, you have to prove that all the sides are equal.
Given that
Now you can say that the two triangles are congruent by angle side angle, since AC = AC.
In △ABF
∠A + ∠B + ∠F = 180
(x-10) + x + 120 = 180
2x + 110 = 180
2x = 70
x = 35
Answer: (C) 35
——————–
#8
I. a + b = 180
Let angle just above b = x
Corresponding angles are congruent. So a = x
b = 180 – x = 180 – a
So statement I is true
II. a = f
If we were to increase c without changing a, then f would decrease
So a cannot be equal to f in all cases
So statement II is false
III. c + b = e + d
AB || CD & transversal BD : alternate interior angles are congruent, so c = e
AB || CD & transversal EF : alternate interior angles are congruent, so b = d
Therefore c + b = e + d
So statement III is true
Answer: (D) I and III only
——————–
#10
I. x = y
BD || EH, transversal AG: alternate exterior angles are congruent: x = y
So statement I is true
II. AC = FG
AC = AF – CF
FG = CG – CF
But we know that AF = CG, so AC = FG
Statement II is true
III. AB = GH
AC = GF (from II)
BC = HF (given)
Included angles x = y (from I)
So triangles ABC and GHF are congruent, and AB = GH
Statement III is true
Answer: (E) I, II, and III
——————–
#14
In a way, this question is slightly misleading, in that it provides information that is totally unnecessary to solve the problem. It doesn’t matter that △ABC is isosceles or that AB ≅ AC. All that matters is that CD is perpendicular to AB and x = 65
In △ACD:
∠CAD = x = 65 (given)
∠ADC = 90 (since CD⊥AB)
y = ∠ACD = 180 – 65 – 90
y = 25
——————–
#20
2y = 3x + 8
y = 3/2 x + 8
This line has slope 3/2
Line parallel to this will also have slope 3/2
Equation of line passing through (4,16) with slope 3/2:
y – 16 = 3/2 (x – 4)
y = 3/2 x – 6 + 16
y = 3/2 x + 10
Since this is equation of line in slope-intercept form: y = mx + b, where b = y-intercept, then
y-intercept = 10
—–
Here’s an alternate method:
First line has slope 3/2
Parallel line also has slope 3/2 and passes through (4,16)
It intercepts y-axis at point (0,y)
So slope from (4,16) to (0,y) = 3/2
(y-16) / -4 = 3/2
2y – 32 = -12
2y = 20
y = 10
Amsco’s geometry! Help plz?
Do somebody have the link the all the answers or know them?
i need problems.
Pg 13 #2,4
Pg 92 #13,14,16
PG 133 #16
Pg 172, #11,12
Pg 203 #15
Pg 208 #15
Pg 340 #15
Pg 341 #21
pg 352 #29,33
Pg 372 #18,19
Pg 376 325
Pg 378 #14
Sat geometry problem help?
problem number 3
How many more squares in the figure above must be shaded so that the ratio of the shaded squares to the unshaded squares is the same as the ratio of the length of the shorter side of the rectangle to the length of the longer side?
(A) 4 (B) 3 (C) 2 (D) 1 (E) 0
figure link
http://www.jmap.org/JMAP/SupportFiles/Amsco_Textbooks/Geometry/SAT_Prep/Chapter_12_SAT.pdf
http://www.jmap.org/JMAP/SupportFiles/Amsco_Textbooks/
Geometry/SAT_Prep/
Chapter_12_SAT.pdf
The ratio of short : long sides is 3:6 which can also be expressed as 1:2
However, this doesnt mean half, it means 1 for every 2 (ie, there are 3 total)!!!
The entire rectangle contains 3*6 = 18 squares. To get the same ratio as above of shaded:unshaded, you need to divide that into three even portions (obviously groups of 6), where one portion is shaded, the other two are not….
if you have 6 shaded and 12 unshaded then you have 6:12 which reduces to 1:2, same as above.
If there are already 5 shaded, then you need one more to make six – the answer is D
For #7 we have:
DC || AB since trapezoid
< CAB = < ACD opposite angles of the line intersecting parallel lines are equal
< ACD =
For #8 – I am rusty on precise wording of geometry theorems, but will give you the general idea here:
Basically, to prove this parallelogram is a rhombus, you have to prove that all the sides are equal.
Given that
Now you can say that the two triangles are congruent by angle side angle, since AC = AC.
And, since
7.
AD and BC are parallel, and AC is a transversal, so angle DCA = angle CAB
Also, angle CAB = angle DAC (as AC bisects angle DAB)
So, in triangle DAC, angle DAC = angle DCA, making it isosceles. The sides opposite these angles will be equal.
So, AD = DC
QED
8.
A parallelogram is also a trapezoid. Using the above, you can prove that AD = DC.
Now, in a parallelogram, opposite sides are equal, so AD = BC, AB = DC.
So, all the four sides are equal, making it a rhombus.
Geometry Help Please?????
okay so i am having trouble with these 2 questions on a take home test i have recieved, due tomorrow.
they are #7 and #8 on part 2 of this test: http://www.jmap.org/JMAP/SupportFiles/Amsco_Textbooks/Geometry/Chapter_Tests/Chapter_10_Test.pdf
does anyone think they could help me out with these proofs as i am utterly lost on how to prove what they are asking.
thanks in advance.
Geometry Help???????
okay so i am having trouble with these 2 questions on a take home test i have recieved, due tomorrow.
they are #7 and #8 on part 2 of this test: http://www.jmap.org/JMAP/SupportFiles/Amsco_Textbooks/Geometry/Chapter_Tests/Chapter_10_Test.pdf
does anyone think they could help me out with these proofs as i am utterly lost on how to prove what they are asking.
thanks in advance.
Sat math problems geometry help with explanations please?
http://www.jmap.org/JMAP/SupportFiles/Amsco_Textbooks/Geometry/SAT_Prep/Chapter_9_SAT.pdf
figures are above link
problem 6
In the figure above, what is the value of x?
(A)60 (B) 55 (C) 35 (D) 30 (E) 25
problem 8
In the figure above, AB parallel to CD and EF parallel to GH Which of the following must be true?
I. a +b = 180
II. a = f
III. c + b = e + d
(A) I only (B) II only (C) I and II only (D) I and III only (E) I, II, and III
Problem 10
In the figure above BD parallel to EH and BC = HF. If AF = CG, which of the following must be true?
I. x = y
II. AC = FG
III. AB = GH
(A) I only (B) II only
(C) III only (D) II and III only
(E) I, II, and III
problem 14
In isosceles triangle ABC above, AB congruent to AC and CD is perpendicular to
AB. If x = 65, what is the value of y?
Problem 20
What is the y-intercept of the line parallel to 2y = 3x + 8 and passing through (4, 16)?
6
∠CFE = 180 – 90 – 30 = 60
∠AFB = 180 – ∠CFE = 180 – 60 = 120
In △ABF
∠A + ∠B + ∠F = 180
(x-10) + x + 120 = 180
2x + 110 = 180
2x = 70
x = 35
Answer: (C) 35
——————–
#8
I. a + b = 180
Let angle just above b = x
Corresponding angles are congruent. So a = x
b = 180 – x = 180 – a
So statement I is true
II. a = f
If we were to increase c without changing a, then f would decrease
So a cannot be equal to f in all cases
So statement II is false
III. c + b = e + d
AB || CD & transversal BD : alternate interior angles are congruent, so c = e
AB || CD & transversal EF : alternate interior angles are congruent, so b = d
Therefore c + b = e + d
So statement III is true
Answer: (D) I and III only
——————–
#10
I. x = y
BD || EH, transversal AG: alternate exterior angles are congruent: x = y
So statement I is true
II. AC = FG
AC = AF – CF
FG = CG – CF
But we know that AF = CG, so AC = FG
Statement II is true
III. AB = GH
AC = GF (from II)
BC = HF (given)
Included angles x = y (from I)
So triangles ABC and GHF are congruent, and AB = GH
Statement III is true
Answer: (E) I, II, and III
——————–
#14
In a way, this question is slightly misleading, in that it provides information that is totally unnecessary to solve the problem. It doesn’t matter that △ABC is isosceles or that AB ≅ AC. All that matters is that CD is perpendicular to AB and x = 65
In △ACD:
∠CAD = x = 65 (given)
∠ADC = 90 (since CD⊥AB)
y = ∠ACD = 180 – 65 – 90
y = 25
——————–
#20
2y = 3x + 8
y = 3/2 x + 8
This line has slope 3/2
Line parallel to this will also have slope 3/2
Equation of line passing through (4,16) with slope 3/2:
y – 16 = 3/2 (x – 4)
y = 3/2 x – 6 + 16
y = 3/2 x + 10
Since this is equation of line in slope-intercept form: y = mx + b, where b = y-intercept, then
y-intercept = 10
—–
Here’s an alternate method:
First line has slope 3/2
Parallel line also has slope 3/2 and passes through (4,16)
It intercepts y-axis at point (0,y)
So slope from (4,16) to (0,y) = 3/2
(y-16) / -4 = 3/2
2y – 32 = -12
2y = 20
y = 10
What text book?
Amsco’s geometry! Help plz?
Do somebody have the link the all the answers or know them?
i need problems.
Pg 13 #2,4
Pg 92 #13,14,16
PG 133 #16
Pg 172, #11,12
Pg 203 #15
Pg 208 #15
Pg 340 #15
Pg 341 #21
pg 352 #29,33
Pg 372 #18,19
Pg 376 325
Pg 378 #14