Showing posts with label Geometry. Show all posts
Showing posts with label Geometry. Show all posts
Friday, June 5, 2015
Sunday, May 31, 2015
Friday, May 29, 2015
Thursday, May 28, 2015
find the unknown length in the right triangle one side is 9m and the other side 15m.
If we assume that 15m is the longest side then a side of 12m will give us a right triangle (why?)
:)
Friday, January 2, 2015
Tuesday, May 27, 2014
Wednesday, May 21, 2014
Which of the points A(10,4),B(7,6),C(-3,5) is nearest to P(3,-2)?
If we plot all four points carefully we can see that CP=AP
so when compared to BP they are both longer.
Therefore
is shortest
so when compared to BP they are both longer.
Therefore
Monday, April 21, 2014
The sides of an equilatersl triangle are 3x-1, 5y, and x+13. Find the measures of each side.
equilateral => 3x-1=x+13 => x=7
and x+7 turns into 7+13=20
and so all sides are 20 units
:)
and so all sides are 20 units
:)
In Triangle ABC, Angle B = 15 degrees less than twice angle A and Angle C is more than angle A. Find angle C.
B=2A-15
C>A
and A+B+C=180
which becomes...
A+(2A-15)+C=180
C=180+15-3A
C=165-3A
so

Which sets our boundary to test for valid triangles. After trial and error (there's probably a better way) we can deduce the following.
The least value for C under these conditions is C=42 and the greatest is a little more than 142
so...these are pretty good approximations

and A+B+C=180
which becomes...
A+(2A-15)+C=180
C=180+15-3A
C=165-3A
so
Which sets our boundary to test for valid triangles. After trial and error (there's probably a better way) we can deduce the following.
The least value for C under these conditions is C=42 and the greatest is a little more than 142
so...these are pretty good approximations
Friday, April 4, 2014
the perimeter of a rectangle is 102 inches , and the length of the diagonal is 39 inches. find the dimensions of the rectangle.
1)
P=2(l+w)=102
l+w=51
2)
l+w=51
2)
l^2+w^2=39^2
using substitution...
(l-51)^2+l^2=1521
l={15, 36}
so the dimensions are 15in x 36in
using substitution...
(l-51)^2+l^2=1521
l={15, 36}
so the dimensions are 15in x 36in
Perimeter of a rectangle if its length is (4c+7) units and its width is (c-3) units
P=
2(L+W)=
2(4c+7+c-3)=
2(5c+4)=
10c+8
where c > 3
2(L+W)=
2(4c+7+c-3)=
2(5c+4)=
10c+8
where c > 3
Thursday, April 3, 2014
The length of a rectangle is 5 inches more than double its width. If the perimeter is 70 inches, find the dimensions of the rectangle by using a system of equations.
length of a rectangle is 5 inches more than double its width means:
l=2w+5
the perimeter is 70 inches means:
2l+2w=70
or l+w=35
subbing in for l and simplifying...
(2w+5)+w=35
3w=30
so w=10
then that means
l+10=35
and so l=25
the dimensions must be: 25 inches x 10 inches
the perimeter is 70 inches means:
2l+2w=70
or l+w=35
subbing in for l and simplifying...
(2w+5)+w=35
3w=30
so w=10
then that means
l+10=35
and so l=25
the dimensions must be: 25 inches x 10 inches
The length and width of a rectangle are consecutive even integers. The perimeter is 44 meters. Find the length and width.
2(l+w)=P
2(l+l+2)=44
2l+2=22
2l=20
l=10
length:10, width:12
2(l+l+2)=44
2l+2=22
2l=20
l=10
length:10, width:12
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