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Gem Carnelian
#1 Posted : Monday, October 17, 2011 3:48:15 AM
Rank: Newbie

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Joined: 10/17/2011
Posts: 1

Please help me solve this problems.

1. From a 100cm by 100cm and 60cm block of marble, a pyramid with a square base of 100cm and a height of 60cm was carved. What is the volume of the pyramid? What is the volume of marble removed?

2. If three vertices of a parallelogram are A (2,0), B (4,4), C (6,0), find the fourth vertex D if D is in the 4th quadrant.

3. Two vertices of isosceles triangle ABC are A (0,2) and B (2,2)and AB is the base. Find the vertex C if it is on the x-axis.

Thank you for the help extended.
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lisa
#2 Posted : Tuesday, November 08, 2011 4:44:22 PM
Rank: Administration

Groups: Administration , Administrators, Member, Registered Users

Joined: 3/16/2011
Posts: 33

HI!

I will try to help you as well as I can

1. The volume of a pyramide you will find using the following formula found on this page http://www.mathplanet.co...sms,-cylinders-and-cones

\\V=\frac{1}{3}\cdot B\cdot h

where h is the height (in our case 60 cm)

and B is the base area (in our case the area of a square base of 100 cm)

We begin by getting the value of B

The area of a square is the base times the height i.e.

\\B=b\cdot h=100\cdot100=1000cm^{2}

Now we have a value of h and a value of h and can insert them into the formula of the volume of a pyramid

\\V=\frac{1}{3}\cdot B\cdot h=\frac{1}{3}\cdot10000\cdot60=\frac{600000}{3}=200000cm^{3}

 

2. Plot the 2 coordinates in a coordinate plane. In a parallelogram two opposite sides are parallel which means that the slope of the line L1 that intersects vertex B and C is the same as the slope of the line L2 that intersects veritces A and D. This means that if we find the slope of the line L1 we can use that to find the coordinates for D

We call the slope m which gives

\\m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-4}{6-4}=\frac{-4}{2}=-2

The slope-intersect form of a linear function as L1 and L2 is

y=mx+b

This means that the function of L2 is

L2:y=-2x+b

To get the value of b we can insert the coordinates for one point on the line (in this case (2, 0) and solve for b

\\0=-2\cdot2+b\\0=-4+b\\0{\color{Red} +4}=-4+b{\color{Red} +4}\\4=b

Which gives us the linear function of L2

L2:y=-2x+4

Then we can do the same but for lines L3 (intesects A and B) and L4 (intersects C and D)

m_{L3}=\frac{y_{B}-y_{A}}{x_{B}-x_{A}}=\frac{4-0}{4-2}=\frac{4}{2}=2\\\\L4:y=2x+b\\0=2\cdot6+b\\0=12+b\\0{\color{Red} -12}=12+b{\color{Red} -12}=-12=b\\\\L4:y=2x-12

The coordinates of D you will find where L2 and L4 intersect which gives us the following system of equations

\left\{\begin{matrix} y=-2x+4\\ y=2x-12 \end{matrix}\right.

Solve for x

\\2x-12=-2x+4\\2x-12{\color{Red} +2x}=-2x+4{\color{Red} +2x}\\4x-12=4\\4x-12{\color{Red} +12}=4{\color{Red} +12}\\4x=16\\\\\frac{4x}{4}=\frac{16}{4}\\\\x=4

To find the corresponding y-value we insert the value of x in either L2 or L4. We use L4

\\y=2\cdot4-12\\y=8-12\\y=-4

i.e. the coordinates for vertex D is (4, .4)

 

3.Begin by plotting the points in a coordinate plane.

In an isosceles triangle the two sides that are not tha base are congruent i.e. of the same length.

This means that the distance between A and C is the same as the distance between B and C and we can use this fact together with the distane formula to solve this problem.

The distance formula

d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}

We know that the vertex C is on the x-axis which means it has the y-coordinate 0 i.e. the coordinates (x, 0). We also know tha values of A and B which gives us the following equations

\\d_{A\rightarrow C}=\sqrt{(x_{C}-x_{A})^{2}+(y_{C}-y_{A})^{2}}=\sqrt{(x-0)^{2}+(0-2)^{2}}=\\\\=\sqrt{x^{2}+2^{2}}=\sqrt{x^{2}+4}\\\\\\d_{B\rightarrow C}=\sqrt{(x_{B}-x_{C})^{2}+(y_{B}-y_{C})^{2}}=\sqrt{(2-x)^{2}+(2-0)^{2}}=\\\\=\sqrt{4-4x+x^{2}+4}=\sqrt{x^{2}-4x+8}

Since we know that

d_{A\rightarrow C}=d_{B\rightarrow C}

we can form the following system of equations

\left\{\begin{matrix} d_{A\rightarrow C}=\sqrt{x^{2}+4}\\\\ d_{B\rightarrow C}=\sqrt{x^{2}-4x+8} \end{matrix}\right.

and solve it for x

\\\sqrt{x^{2}+4}=\sqrt{x^{2}-4x+8}\\\\\left (\sqrt{x^{2}+4} \right )^{2}=\left (\sqrt{x^{2}-4x+8} \right )^{2}\\\\x^{2}+4=x^{2}-4x+8\\\\ {\color{Red} \not}{x^{2}}+4={\color{Red} \not}{x^{2}}-4x+8\\\\4=-4x+8\\\\4{\color{Red} -8}=-4x+8{\color{Red} -8}\\\\-4=-4x\\\\\frac{-4}{-4}=\frac{-4x}{-4}\\\\1=x

This means that the coordinates for vertex C is (1, 0)
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