Remember that this week there will be presentations! For the blog posts, I want you to post every single problem of the practice test of your assigned chapter of the AGMath link. If you have 2 chapters, then you post the solutions for both tests. Otherwise, you have to do your chapter plus a random one that you won't need to do a presentation on. Here are the assigned chapters and groups (which I slightly changed) again:
- Aishwarya, Sanjana -> Chapter 1 and Chapter 14 (don't do practice test and matrices for this chapter)
- Christine, Akshara, Julia -> Chapter 2 (this chapter is perhaps the most important b/c it applies to all problems not just Geometry)
- Vivian, Alice -> Chapter 4 and Chapter 5
- Jeffrey, Jacky, Mihir -> Chapter 6
- Maya, Julianna -> Chapter 8
- Adiyan, Vignesh, Minhoo -> Chapter 9
- Anindit, Christopher -> Chapter 10
- Melinda and Lucy -> Chapter 11
- Allan, Ming, Stephanie -> Chapter 12
Random Chapter:
- Aishwarya, Sanjana, -> Chapter 9
- Christine, Akshara, Julia -> Chapter 12
- Vivian, Alice -> none
- Jeffrey, Jacky, Mihir -> Chapter 2
- Maya, Julianna -> Chapter 11
- Adiyan, Vignesh, Minhoo -> Chapter 4
- Anindit, Christopher -> Chapter 8
- Melinda and Lucy -> Chapter 5
- Allan, Ming -> Chapter 6
Christopher-
ReplyDeleteChapter 10: Volume Part 1
For questions 1-9, find the volume.
#1: A triangular prism with right triangle base with legs 5 and 12 in. and 7 in. high.
Solution: The volume for a prism is bh, or base * height. Since the base is a right triangle, we know that if 12 in. is the base of the triangle, 5 in. is the height, being perpendicular to the base. So, using the formula for the area of a triangle 1/2 bh, we can determine that the area of the base is 30 sq. in. and therefore the volume of the prism is 30 * 7 = 210 cubic in.
Answer: 210 cubic in.
#2: A trapezoidal prism, where the trapezoidal base has bases 5 and 17 in. with height 8 in. and the height of the prism is 8 in.
Solution: Using the formula for the area of a trapezoid (b1 + b2)/2 * h, where b1 and b2 are the two bases and h is the height, we find that the area of the trapezoid is (17 + 5)/2 * 8 = 88 sq. in. So, the volume of the prism is 88 * 8 = 704 cubic in.
Answer: 704 cubic in.
#3: A rectangular prism measuring 10*6*10 m, with a hole with diameter 4 m drilled so that its depth is 10 m.
Solution: The volume of a rectangular prism is the product of its side lengths, so the volume is 10*6*10 = 600 cubic meters. The hole has a diameter of 4 m, so it has a radius of 2 m. Using the formula for the area of a circle, which is pi*r^2, the area of the base of the imaginary cylinder is 4 pi sq. meters. Since the height of the cylinder is 10 m, the volume is 10 * 4 pi or 40 pi cubic meters. So, the volume of the box is 600 - 40 pi cubic meters, or about 475.1 cubic meters.
Answer: 475.1 cubic meters
#4: A cone with radius 18 m and height 24 m.
Solution: Using the formula for the volume of a cone, which is 1/3*pi*r^2*h, we find that the volume is 1/3*pi*18^2*24, which is 2592 pi, or about 8178.4 cubic meters.
Answer: 8178.4 cubic meters
#5: An oblique triangular pyramid with right triangle base with legs 10 and 24 in., and a height of 8 in.
Solution: The area of the base is (10*24)/2 or 120 sq. in, so the volume is 8*120 or 960 cubic in.
Answer: 960 cubic in.
#6: A square pyramid where every side length is 8 cm.
Solution: In order to find the height, we must find the slant height of each triangular face. Using the pythagorean theorem, we find that the height of each equilateral triangle face is sqrt(8^2 - 4^2), or 4*sqrt(3) cm. After we find the slant height, we can use the pythagorean theorem again to find the height using the slant height and 4 cm (half the side length). So, the height is sqrt((4*sqrt(3))^2 - 4^2), or 4*sqrt(2) cm. Afterward, we can calculate the volume as 1/3*8^2*(4*sqrt(2)), or (256*sqrt(2))/3. This rounds to 120.7 cubic cm.
Answer: 120.7 cubic cm.
Christopher-
ReplyDeleteChapter 10: Volume Part 2
#7: A sphere with radius 3 m.
Solution: Using the formula for the volume of a sphere 4/3*pi*r^3, we find that the volume is 4/3*pi*3^3, which 36 pi, which is about 113.1 cubic meters.
Answer: 113.1 cubic meters
#8: A semi-sphere with diameter 12 in.
Solution: This semi-sphere has a radius of 6 in. So, using the formula again, 1/2(4/3*pi*6^3) = 144 pi. This rounds to 452.4 cubic in.
Answer: 452.4 cubic in.
#9: A semi-sphere on top of a cylinder of height 6 in, both with a radius of 6 in.
Solution: The volume of the semi-sphere is 4/3*pi*6^3 or 144 pi. The volume of the cylinder is pi*6^2*6 = 216 pi. So, the volume of the figure is 360 pi or about 1130.9 cubic in.
Answer: 1130.9 cubic in.
For questions 10-12, find the surface area.
#10. Same problem as #7.
Solution: The formula for the surface area of a sphere is 4*pi*r^2, so the surface area is 4*pi*3^2, which is 36 pi, which rounds to 113.1 sq. meters.
Answer: 113.1 sq. meters
#11: Same problem as #8.
Solution: Using the formula, the surface area of the spherical portion of the semi-sphere is 1/2(4*pi*6^2) or 72 pi. The base, which is a circle, has an area of pi*6^2 or 36 pi. So, the entire surface area is 108 pi, or about 339.3 sq. in.
Answer: 339.3 sq. in.
#12: Same problem as #9.
Solution: The surface area of the semi-sphere is 1/2(4*pi*6^2) or 72 pi. The surface area of the base of the cylinder is pi*6^2, or 36 pi. The non-base portion of the cylinder has a surface area that can be expressed as a rectangle with the length of the circumference of the base and the width of the height of the cylinder. So, the surface area of that portion is 2*pi*6 (the formula for the circumference of a circle where 6 is the radius) * 6 (the height), which is 72 pi. This adds up to 180 pi, or about 565.5 sq. in.
Answer: 565.5 sq. in.
Christopher-
ReplyDeleteChapter 10: Volume Part 3
#13: A steel cube with 7 in. edges is dropped into a cylinder of water, causing the level to rise 3 in. What is the radius of the cylinder?
Solution: According to displacement, the increase in height multiplied by the base of the cylinder must equal the volume of the cube. The volume of the cube is 343 cubic in., and if the level rose by 3 in., then the area of the base must be 114 1/3 sq. in. Using the formula for the area of a circle, which is pi*r^2, we find that r, the radius, must equal about 6.03 in.
Answer: 6.03 in.
#14: A wooden dowel (cylindrical rod of wood) is 15 cm long with a 1.5 cm radius. The dowel weighs 77.7 grams. If the dowel floats upright, how many centimeters will above the surface of the water?
Solution: The density of water is 1 g/cm^3. According to the rules of physics, the percentage of an object that is underwater is determined by the density of the object in g/cm^3. The volume of the cylinder is 15*pi*1.5^2, or 33.75 pi. So, the density is 77.7/33.75 pi, which is about 77.7/106.02875, which rounds to 0.73282. This means that 73.282% of the cylinder, which is 15 cm long, is underwater. This is 10.9923 cm underwater, which means that 4.0077, or about 4.01, cm is above the surface of the water.
Answer: 4.01 cm
#15: A cubic foot of water is about 7.5 gallons. A hexagonal fish tank with base side length 18 in. holds 65 gallons of water. How many inches deep is the water in the tank?
Solution: 65 gallons is 8 2/3 cubic ft. To convert this to cubic in., you have to multiply it by 12^3 or 1728, to get 14976 cubic in. The base of the fish tank is the sum of the areas of 6 equilateral triangles with side length 18 in. Because an equilateral triangle can be bisected into two 30-60-90 triangles, we can use this triangle's properties to realize that the height of the triangle is 9*sqrt(3). Therefore, the area of the base is 6*((9*sqrt(3)*18)/2) or 486*sqrt(3). Since the volume of the tank (14976) is the product of the base (486*sqrt(3)) and the height, we can estimate the height to 17.79 in.
Answer: 17.79 in.
#16: A sphere with a density of 2.4g/cm^3 weighs 90 grams. What is its surface area?
Solution: Since it weighs 90 grams, its volume is 90/2.4 or 225/6 cubic cm. Using the formula for the volume of a sphere, 4/3*pi*r^3, we find that the radius is about 2.07648. Therefore, using the formula for the surface area of a sphere, 4*pi*r^2, we find that the surface area is about 54.17 sq. cm.
Answer: 54.17 sq. cm
#17: A dodecahedron has 12 faces, each of which is a regular pentagon. How many vertices are there on a dodecahedron?
Solution: You can find the number of vertices in a 3-D figure by making a net, finding each isolated face, and assigning n vertices to each, where n is the number of edges in a face. Afterward, you omit extra vertices. If you make a net of a dodecahedron, you can find six visibly isolated (not touching any other face) faces. So, there is 30 vertices to be counted altogether. However, if you create the dodecahedron, the five outer "isolated" faces actually meet each other at an entire edge each, or two vertices. So, for each of the five faces, you must omit two vertices each. That leaves 20 non-redundant vertices.
Answer: 20 vertices
Not Random Questions: Chapter 2
ReplyDelete~Christine T^T
~Problems 11-18 (i'm not finishing the rest)
~The explanations may be confusing to follow
11) How many toothpicks will be in the nth figure?
Answer: n(n+1)
If we consider 2 toothpicks that look like the top part of a triangle as one, then hopefully you can see a pattern. The 1st shape has one, 2nd shape has 1 on the top row and 2 on the second row, the 3rd shape has 1 on the 1st row, 2 on the 2nd row, and 3 on the third row, and so on. The numbers add from 1 to n. n is also the nth figure
The formula for that is…
n(n+1)/2
But then, one "part of the triangle" consists of 2 toothpicks…
n(n+1)/2 x 2
=n(n+1)
12) How many toothpicks will be in the 20th figure?
Answer:420 toothpicks
Just take the formula for Problem 11, n(n+1), then substitute 20 for n because n represents the nth figure.
20(20+1)
=20 x 21
=420
13) How many diamonds (small only) will be in the nth figure?
Answer: {n(n+1)}-3/3
If you examine the figures closely, you can see that the 2 toothpicks on the base of the figure (one on the left, one on the right), play no part in making a diamond. If you look at figure 3, there are 12 toothpicks in total. If we subtract 2 from 12 (because the 2 toothpicks are useless) then we are left with 10 toothpicks. You can easily see that in figure 3 there are 3 diamonds. Two of the diamonds each take 3 toothpicks to build but the one diamond takes 4 toothpicks, 1 more than the other two. To make it fair (so that no diamond gets more toothpicks than the others), we can subtract 1 from the remaining 10 toothpicks which leaves us with 9 toothpicks. Now each diamond takes 3 toothpicks, so 9/3=3 diamonds.
So first, we subtracted 2 from the total amount of toothpicks. The formula for amount of toothpicks is, n(n+1)
{n(n+1)}-2
Second, we subtracted 1 more toothpick…
{n(n+1)}-2-1
= {n(n+1)}-3
Then, we divided it by 3…
= {n(n+1)}-3/3
14) In the nth figure, how many white dots will be on the bottom row?
Answer: n+1
For this question, I just looked at the bottom row for each figure. The 1st had 2 white dots, 2nd had 3, 3rd had 4, 4th had 5, and so on.
That must mean that it is n+1
15) How many white dots will there be in the nth figure?
Answer: n^2+3n+2/2 or (n+1)(n+2)/2
I thought that this could be solved by closely looking at the figures. You can see that on the 1st figure there are 1 white dot and then 2 white dots. 2nd figure has 1 white dot, 2 white dots, then 3 white dots. You can come to a conclusion that a figure has 1 more "white dot row" than it's nth place. The formula for adding 1 to n is…
n(n+1)/2
But now, n turns into n+1.
(n+1)(n+1+1)/2
=(n+1)(n+2)/2
=n^2+3n+2/2
16) How many white dots will be there in the 25th figure?
Answer: 351
This one is easy. Substitute 25 for n in the formula above.
25^2+3(25)+2/2
=625+75+2/2
=700+2/2
=702/2
=351
17) How many black AND white dots will there be in the nth figure?
Answer: n^2+2n+1
The formula for white dots is n^2+3n+2/2. The formula for black dots is n(n+1)/2. If you add the formulas…
n^2+3n+2/2+n(n+1)/2
=n^2+3n+2+n^2+n/2
=2n^2+4n+2/2
=2(n^2+2n+1)/2
=n^2+2n+1
18) How many black AND white dots will there be in the 50th figure?
Answer: 2601
Substitute 50 for n in the formula in Problem 17.
50^2+2(50)+1
=2500+100+1
=2601
Christopher-
ReplyDeleteChapter 8: Area Part 1
For questions 1-7, find the area.
#1: A trapezoid with bases 54 and 30 ft and height 18 ft.
Solution: The area of a trapezoid is ((b1+b2)/2)*h, so the area is ((30+54)/2)*18 = 756 sq. ft.
Answer: 756 sq. ft
#2: A square with 4 in. sides and a semi-circle on top with the same diameter as a side of the square.
Solution: The area of the square is 4^2 or 16 sq. in. The area of the semi-circle can be used by determining half of the formula for a circle, which is pi*r^2. The radius of the circle is 2 in. So, 1/2(pi*2^2) = 2 pi. The area of the figure is 16 + 2 pi sq. in.
Answer: 16 + 2 pi sq. in.
#3: A heptagon with sidelength 80 ft and apothem 83 ft.
Solution: Since a heptagon can be divided into 7 triangles with base 80 ft and height 83 ft, we can determine the area as 7(80*83/2) or 23240 sq. ft.
Answer: 23240 sq. ft
#4: A hexagon with side length 10 cm, but missing a hexagon in the middle that has half the apothem.
Solution: The area of the larger hexagon without the other hexagon is the sum of the areas of 6 equilateral triangles with side length 10 cm. Since an equilateral triangle can be divided into two 30-60-90 triangles, we find that the apothem is 5*sqrt(3). So, the area is 6(10*5*sqrt(3)/2) or 150*sqrt(3). Since the apothem of the inner hexagon is half the apothem of the outer hexagon and the regular hexagons are similar, the area of the inner hexagon is 1/4 the area of the outer one. So, 150*sqrt(3) - 1/4(150*sqrt(3)) = (225*sqrt(3))/2 sq. cm.
Answer: (225*sqrt(3))/2 sq. cm
#5: A 2:1 rectangle with two circles with radius 6 in. inscribed inside, with a square inscribed in each of those. Find the area of the the figure without the circles but with the squares.
Solution: Since the radius of the circles is 6 in., the dimensions of the rectangle are 24*12 in. This is 288 sq. in. The area of the circles is 2(pi*6^2) or 72 pi. The diagonal of the squares is equal to the diameter of the circles, which is 12 in., so we can use the pythagorean theorem or the properties of a 45-45-90 triangle (which is what a diagonal of a square cuts a square into) to find that the side length of the squares is 6*sqrt(2), and the area of both squares is 2(6*sqrt(2))^2, or 144. So, the area is 288-72 pi+144 = 432-72 pi sq. in.
Answer: 432-72 pi sq. in.
#6: Triangle ABC with area 180 sq. cm, and X is on AB so that AX = 4BX. Find the area of triangle CXB.
Solution: Since BX is a fourth or AX, it is a fifth of AB. If AB is the base, then AB*height = 180. So, 1/5*AB*height = 36 sq. cm.
Answer: 36 sq. cm
#7: A circle inscribed with a square with side length 12. A circle sharing its diameter with a side of the square is also present. Find the area of the large square, not including where the two circles overlap.
Solution: The area of the square is 12^2 or 144. The diagonal of the square is 12*sqrt(2), which is also the diameter of the large square, so the area of the large square is pi*(6*sqrt(2))^2, which is 72 pi. The diameter of the small circle is the same as the side length of the square, which is 12, so the area of the small circle is pi*6^2 or 36 pi. The area of the region is the sum of 3/4 of the area of the large circle without the square, plus the square and without half of the small circle. This is 3/4(72 pi - 144) + 144 - 1/2(36 pi), or 36 + 36 pi sq. units.
Answer: 36 + 36 pi sq. units
Christopher-
ReplyDeleteChapter 8: Area Part 2
For questions 8-10, find the geometric probability.
#8: Getting in a quarter-circle inscribed in a square.
Solution: If the side of the square is s, then the geometric probability of landing in the quarter-circle is 1/4(pi*s^2)/s^2, which is pi/4, which is about 0.8.
Answer: 0.8
#9: A triangle WVX where Y is on VX and Z is on YW. VY=XY and YZ=WZ. Find the geometric probability of landing on VZY.
Solution: Because VY=XY and triangles WYX and WYV share the same height, WYV is half of triangle WVX. Because YZ=WZ and triangles VZW and VZY share the same height, VZY is half of WYV, or a quarter of WVX. The geometric probability is 0.25.
Answer: 0.25
#10: A regular octagon inscribed in a square. Find the geometric probability of landing in the square, but not in the octagon.
Solution: If the side length of the octagon is s, then we can find the side length of the square by calculating the sum s + 2x, where x is the measure of the legs of the 45-45-90 triangles that we are finding the geometric probability for. Since the hypotenuse of those triangles is s, the legs are s*sqrt(2)/2, and the side length of the square is s + s*sqrt(2). We can also find the apothem of the octagon, because it is half of the side length of the square. So, the geometric probability is (s + s*sqrt(2))^2 - 8(1/2*s*(s+s*sqrt(2)/2)) / (s + s*sqrt(2))^2, which simplifies to 1/2*sqrt(2)+2, or about 0.2.
Answer: 0.2
#11: A triangular pyramid with equilateral triangle base with side length 12 ft and slant height of 25 ft. Find the surface area of the base.
Solution: Knowing that equilateral triangles can be divided into two 30-60-90 triangles, we find that the height of the triangle is 6*sqrt(3), and so the area of the triangle is 1/2(12*6*sqrt(3)), or 36*sqrt(3), which rounds to 62.4 sq. ft.
Answer: 62.4 sq. ft
#12: Find the total surface area of the triangular pyramid in question #11.
Solution: The areas of the non-base triangles are 1/2(25*12), or 150, so the base (about 62.4) + 3(150) is about 512.4 sq. ft.
Answer: 512.4 sq. ft
#13: A cone with radius 10 ft and slant height 26 ft, but cut on the top so the top base has a radius of 5 ft and the missing cone has a slant height of 13 ft. Find the surface area of the top and bottom.
Solution: The area of the bottom is pi*10^2 or 100 pi. The area of the top is pi*5^2 or 25 pi. The total surface area is 125 pi, or about 392.2 sq. ft.
Answer: 392.2 sq. ft
#14: Find the total surface area for the decapitated cone in question #13.
Solution: The area of the non-base portion of a cone can be expressed as the area of a triangle with a base of the circumference of the circle and a height of the slant height of the cone. So, the surface area of the non-decapitated cone would be 1/2((2*pi*10)*26), or 260 pi. However, the original cone is missing a smaller cone of radius 5 and slant height 13, so the surface area lost is 1/2((2*pi*5)*13), or 65 pi. So, the non-base surface area is 260-65 or 195 pi, and the entire surface area is about 392.2(calculated earlier) + 195 pi, or about 1005.3 sq. ft.
Answer: 1005.3 sq. ft
sorry for posting late, did not know there was a blog post. Christopher has already done most of the answers but here are my methods:
ReplyDeleteVOLUME, Part 1:
1) The question asked to find the volume of a triangular prism. The prism has a height of 7 and a triangular face of 5 by 12 by 13 inches. Knowing that it is a Pythagorean triplet, we can say that 12 is the height and 5 is the width. therefore, the area of the base of the prism is .5(12)(5) which equals 30. Multiply that by 7 (the height) to get an answer of 210.
Answer: 210 cubic inches
2)This is a trapezoidal prism. They give you a lot of information that you don't need for this problem. We all know that the volume formula for all prisms is bh. the base in this case is a trapezoid, and the area of a trapezoid is .5(h)(b1 + b2). Substituting for this case, we get the area of the trapezoid to be:
.5(8)(17+5)
4(22)
=88
the height of this prism is given as 8. Therefor, the volume is 88*8
= 704.
Answer: 704 cubic inches
3) For this problem, we are given 2 shapes: a rectangular prism and a cylinder. The volume of the shape we want to find is nothing but the volume of the prism - the volume of the cylinder:
volume of the rectangle prism: hlw
= 10*10*6
= 600
volume of cylinder: bh
= pi(2)^2 * 10
= 40(pi)
=125.663....
volume of shape:
600-125.663...
which is about 474.3 cubic meters
Answer: 474.3 cubic meters
4)The area of a cone is 1/3bh, so we could just substitute in our given numbers:
1/3 (24)* (18^2)(pi)
this is about 2592 pi, or to the nearest 10th, 8143 cubic meters.
Answer: 8143 cubic meters.
5) To find the volume of any pyramid, the formula is 1/3(b)(h)
substituting, in this case we get:
1/3 *(.5)(10)(24) * 8
= 1/3 * 120 * 8
= 40 * 8
= 320.
Answer: 320 cubic inches
6) This is a bit tricky because they don't give us everything, so we must be careful not to make mistakes.
We know that the formula for the volume of a prism is 1/3 bh.
So, the base is easy as it is a square. simply do 8*8 to get 64.
To find the height, we first have to find the "slant height" of the pyramid. Since the triangles are equilateral, we know if we draw an altitude it forms 2 30-60-90 triangles. using the properties of a 30-60-90 triangle, we can find that the slant height is in fact 4(sqrt(3)). Now, we can use the Pythagorean theorem to find the height. if we call x the height, we can say that:
(slant height) ^2 = x^2 + 4^2
48 = x^2 + 16
x^2 = 32
x = 4(sqrt(2))
so, the height is 4 (sqrt(2))
substituting back into our original equation, we find that the volume of the prism is 64*4(sqrt(2)/3, or 256(sqrt(2)/3
This rounds to 120.8 cubic centimeters
Adiyan Kaul
ReplyDeletePage 19 Geometry
1. Use the Pythagorean Theorem
to solve for x.
This was the equation I set up.
(x+1)^2=1^2+(x+(2-2x))^2
(x+1)^2=1^2+(x+2-2x)^2
(x+1)^2=1^2+(-x+2)^2
1+2x=1+4-4x
2x+4x=5-1
6x=4
x=2/3
2. 29^2=(6+x)^2+y^2
25^2=x^2+y^2
y^2=29^2-(6+x)^2
29^2-(6-x)^2=25^2-x^2
29^2-[36+x^2-12x]=25^2-x^2
841-36+12x=625
805-625=-12x
180=12x
x=15
y^2=25^2-15^2=400 y=20
3. y^2=(x+2)^2+2^2
4^2=y^2+x^2
16=(x+2)^2+4+x^2
16=x^2+4+4x+4+x^2
16=x^2+8+4x+x^2
8=x^2+4x+x^2
8=2x^2+4x
2x^2+4x-8=0
using quadratic equation x=1.2
Chapter one: Aishwarya Laddha
ReplyDelete( use figure given in packet)
1. angles FGA are ____ angles
answer: supplementary
2. Ray CG is same as ray____
answer: AC
3. Name a pair of complementary angels
answer: < VWY and TWV
4. Triangle AGE is classified as __ ____
Answer: obtuse scalene
5: using correct notation, label a pair of congruent segments
answer: AB ≅BE
6. How many obtuse angles are there on the figure to the left?
solution: <AGE
answer: 1 obtuse angle
7. Angles TWV and ______ are vertical angles
answer: <XWZ
8. quadrilateral STWX is a ______.
answer: trapezoid
9. If the m<TWX is 133˚, what is the m<VWY?
answer: 43˚
10. Ray TW is the same as ray ____.
answer: Ray XS
(use figure given in packet)
11. name a point of tangency
answer: point D
12. what is the measure of arc XY if <XWZ and <XWY are congruent angles?
answer: 3/4πr
13. Name two semicircles.
answer: arc EB and arc BE
14.How many arcs are on circle W?
solution: arcs: YZ, YX, ZX, ZY, XZ, XY
answer: 6 arcs
15. name three chords on circle A
answer: chords EB ED DC
16. answer 18 posts
17. regular ocatgon ABCDEFGH, are segments AB and EF parallel, perpendicular, or neither?
answer: parallel
18. How many sides on a triangular prism?
answer: 4
19. How many edges on a triangular prism
answer: 15 edges
20. (see image provided in packet)
what figure is formed by the net given?
answer: triangular pyramid
Adiyan Kaul
ReplyDeletePage 21 Geometry
1. CA^2=CX^2+XA^2
CA^2=5^2+4^2=41
CC^2=CX^2+CX^2
5^2+6^2=61
CB^2=CX^2+XB^2=61
AB^2=5^2+4^2=41
AB+AC+CC+CB=2(sq rt 41+sq rt 61)
Adiyan Kaul
ReplyDeletePage 21 Geometry
2.<DCB=30 degrees
<ACB=60 degrees
Since angle DAC=30 degrees and angle ACD=30 degrees therefore DA=DC=8 centimeters
sin 60 degrees=CB/8 therefore CB=6.4
8^2=CB^2+BD^2
64-40.96=23.04 therefore BD=4.8
Area BAC= 1/2*12.8*6.4=40.96
Random Questions:Chapter 12 (Trigonometry)
ReplyDelete~Christine T^T
~Problems 1- 3 (and I'm not finishing the rest. Akshira and Julia's job)
1)What is the smallest angle measure in a 4-5-6 triangle?
Answer: 48
The angles of a triangle adds up to 180…
4x+5x+6x=180
15x=180
x=12
The smallest measure would be 4x.
4(12)
=48
2)The hypotenuse in a right triangle measures 14cm. One of the
two acute angles measures 34o. What is the perimeter of the
triangle?
Answer:21+√147
If the base is b and the height is h, then you can find b by using cosine.
cos62=b/14
Using your scientific calculator…
cos62=0.46947156…
b=0.46947156… x 14
b=6.57260188…
About 7 if you round it to the nearest whole number. Using the pythogoreum method you can find out h..
14^2=7^2+h^2
196=49+h^2
h^2=147
h=√147
√147+14+7
=21+√147
3)Trains have a very difficult time climbing steep inclines. If the
maximum angle that a train can climb is 3o, how many miles
long must the tracks be to climb 7,000 feet up over a mountain
pass? Round to the nearest mile.
Answer: 9 miles
You can find the hypothenuse by doing sine.
7000/x=sin3
x=49603.17ft.
1mile=5280 ft.
49603.17/5280
=9.39454≈9 miles
Adiyan Kaul
ReplyDeletePage 22 Geometry
1. 55^2=44^2+x^2
3025=1936+x^2
1089=x^2
x=33
2. 10^2=8^2=x^2
164=x^2
x=2 sq rt 42
3. 10^2=5^2+x^2
75=x^2 .
x= 5 sq rt 3
4. 12^2=2x^2
144=2x^2
x= 6 sq rt 2
5. Since in a 30 60 90 triangle the sides are in the ratio of 1:2:sq rt 3 therefore 8^2=4^2+x^2
x=4 sq rt 3
Adiyan Kaul
ReplyDeletePage 22 Geometry
6. 1:2: sq rt 3
x^2=x/2^2+21^2
x^2=x^2/4+21^2
4x^2-x^2/4=21^2
3x^2=1764
x=14 sq rt 3
This comment has been removed by the author.
ReplyDeleteHi Ashish,
ReplyDeleteDue to ignorance / three tests tomorrow, Allan, Ming, and I will probably be unable to do the blogpost.
Sorry.
~Stephanie
Adiyan Kaul
ReplyDeletePage 22 Geometry
7. AB=sq rt((-7+3)^2+(6-2)^2)
sq rt 16+16
4 sq rt 2
8. AC=sq rt((5-4)^2+(2+4)^2)= sq rt 37
BC= sq rt ((9-4)^2+(-1+4)^2))= sq rt 34
AB=5 Since the sides are not equal it is a scalene triangle.
Chapter 8
ReplyDeletePractice Test
Problem 1)
I drew a line from the corner of the slanted line to the other side. Then I found the area of the rectangle and the area of the right triangle.
Answer: 756
Problem 2)
The area of the square is 4*4=16. The area of the semicircle is 2pi.
Answer: 16 + 2pi
Problem 3)
I divided the hexagon into six triangles. The area of each one is (80*83)/2, which is 3320. Then, I multiplied this by 6.
Answer: 23240
Problem 4)
I divided the hexagons into six triangles, and found the area of each by using the equilateral triangle formula. I subtracted the two areas.
Answer: 113.5 * sqrt 3
Problem 5)
I found the area of each square and subtracted it from the circle. Then I subtracted this value from the area of the entire rectangle.
Answer: 432 - 72pi
Problem 6)
I divided the triangle into two right triangles by drawing an altitude from C.
Answer: 36
Problem 7)
I found the area of the square and the area of the circle, and used this to find the area of the shaded region.
Answer: 288 pi
Problem 8)
I assumed that the side of the square was 4 (to make the calculation easy). Then, the area of the quarter circle would 4pi, and the area of the entire square would be 16. 4pi / 16 is equal to pi/4.
Answer: pi/4
Julianna~~~~~
ReplyDeleteUnrandom questions: Chapter 8
#1- This figure can be split into an 18 ft x 30 ft rectangle and a right triangle with base 18 ft, height 24 ft, and hypotenuse 30 ft. The area of the rectangle is 540 square feet, and the area of the triangle is (1/2)(18 ft)(24 ft)=216 square feet. The combined area is 540+216=756 square feet.
ANSWER- 756 square feet
#2- The square has an area of 4^2=16 square inches. The semicircle has diameter 4 inches, so it has radius 2 inches. The area of the semicircle is (1/2)(2^2)(pi)=2 pi. The total area is the sum of these two.
ANSWER- (16 + 2 pi) square inches
#3- The heptagon is regular, so all the sides have length 80 feet. You can split this into 7 isosceles triangles by connecting the center to each of the 7 vertices. Each isosceles triangle has base 80 feet and height 83 feet. The area of each is 40*83=3320 square feet. The total area is 3320*7=23240 square feet.
ANSWER- 23240 square feet
#4- Each hexagon can be split into 6 equilateral triangles. For the larger hexagon, each of the triangles has side length 10 cm. The area of each is 25 root 3. The total area of the larger hexagon is 150 root 3. The side length of each of the equilateral triangles that can be formed in the smaller hexagon is 5, because it is half of that of the larger one. The area of each triangle in the smaller hexagon is 25 root 3 over 4. The total area is 75 root 3 over 2. The area of the shaded region is the area of the larger hexagon minus the area of the smaller one. 150 root 3 minus 75 root 3 over 2 is 225 root 3 over 2.
ANSWER- (225 root 3)/2
#5- The large rectangle has side lengths 24 inches x 12 inches which is an area of 288 square inches. The area of each of the circles is 36 pi, and the total area of both circles is 72 pi. The area of the shaded region inside the rectangle but outside of the circles is 288-72 pi. Each of the squares inscribed has diagonal 12 cm. Therefore the area of each is 72. The total area is (288-72 pi)+177=465-72 pi.
ANSWER- (465-72 pi) square centimeters
#6- The area of triangle CXB is one-fifth the area of the total triangle. The total area is 180 square centimeters. One-fifth of that is 36. The area is 36 square centimeters.
ANSWER- 36 square centimeters
OMG WHAT HAPPENED ALL MY POSTS JUST GOT DELETED
ReplyDelete-Jacky-
ReplyDeleteCircle Properties Practice Test
Refer to your diagram
1) <WCM = 50°
<WCY = 180-50 = 130°
Arc WY = 130°
<WZY = 130/2 = 65°
Ans: 65°
2) <ZYA = 90/2 = 45°
<LZY = 180-(90+45) = 45°
Arc LY = 45*2 = 90°
Ans: 90°
3) Arc ANH = 50*2 = 100°
Arc x = 180-100 = 80°
Ans: 80°
4) <x = (52+(180-52)/2)/2
<x = (52+64)/2 = 116/2 = 58°
Ans: 58°
5) Top left arc = 45*2 = 90°
Arc x = 180-90 = 90°
Arc x = (π6^2)/4 = 36π/4 = 9π
Ans: 9π
6) Small arc = 55° = 11π
Diameter = 11π*(360/55) = 11π*6.5454... = 72π
Radius = sqrt(72π/π) = sqrt(72)
Ans: sqrt(72)
7) 30miles per hour is 2640feet per minute
2640ft = 500rotation
1 rotation = 5.28
πr^2 = 5.28
r^2 = 1.7
r = 1.3
1.3*12 = 16
Ans: 16
8) I don’t know
9) (2π(11))/2+(2π(5))/2+6(2)
11π+5π+12
16π+12
Ans: 16π+12
10)2a^2 = 20^2
a^2 = 200
a = sqrt(200)
a = 10*sqrt(2)
(10*sqrt(2))/2
5*sqrt(2)
2(5*sqrt(2))^2 = b^2
2(50) = b^2
100 = b^2
b = 10
π(10/2)^2
π(5)^2
25π
Ans: 25π
11)x^2+(2x)^2 = 2*sqrt(5)
x = 2
Diameter of circle = 2
2(4)-π1^2
8-π
Ans: 8-π