Holt Algebra 1 Homework and Practice Workbook 1st Edition Chapter 2
Page 15 Problem 1 Answer
Given ΔABC∼ΔDEF
To Find : x in the given diagram
In order to find the solution, we need to find EF by applying similar to rule.
ΔABC∼ΔDEF∼ is known as similar sign, where we now know that,
AB/DE=BC/EF=AC/DF But since, the value of AB and DE
has not been provided, then the equation will change into,
BC/EF=AC/DF
Now replace the equation with the values given in the diagram:
5/x=9/27
Use cross multiplication, where N1 is multiplied by D2 and D1 is multiplied by N2. The new equation will be
5×27=9×x
5×27=9×x
Now, to find the x solve the equation
=135=9x
Divide 135 with 9
=135/9
=x
=x=15 cm
The value of x=15 cm
Page 15 Problem 2 Answer
FGHJK∼MNPQR
To Find: x
In order to find the solution, we need to find PQ by applying similar to rule.
Since FGHJK∼MNPQR∼ is known as similar sign, where we now know that
FG/MN=GH/NP
=HJ/PQ
=JK/QR
But since, HJ, JK, PQ and QR values has been provided, then the equation will change into,
HJ/PQ=JK/QR
We need to find the value of PQ.
Replace the equation with the values given in the diagram:
2/x=5/8
Use cross multiplication, where is N1 multiplied by D2 by and D1 is multiplied by N2 . The new equation will be
=2×8=5×x
=2×8=5×x
Now, to find the x solve the equation.
=16=5x
Divide 5 with 16
=16/5
=x
=x=3.2 cm
The value of x=3.2cm
Page 15 Problem 3 Answer
Given the height of the first pole and both the shadow.
To find: height of the second pole.
In order to find the height of the second pole:
we will substitute the height with the xconvert the equation into fraction and cross multiply it to find the solution of x
Since we know the both shadow lengths, let’s convert it into fraction, as,4ft/20ft
Let’s make the height of the second utility pole be x, and convert it into fraction,5.5ft/x
To find x, combine the equation together
4ft/20ft=5.5ft/x
In order to find x, Cross multiply the given equation
=4×x=20×5.5
=4x=110
Divide 110 with 4
x=110/4ft
x=27.5ft
The height of the utility pole is 27.5 ft.
Page 15 Problem 4 Answer
Given: Radius of cylinder=3cm
Length of cylinder=10cm
and every dimension of cylinder is multiplied by 3 to form a new cylinder.
To find the ratio of the volumes related to the ratio of corresponding dimension
We find the new dimension and using that we find new volume and proceed.
Original radius=3cm;Length=10cm
New Radius =3× 3 cm =9cm
new length = 10× 3 = 30cm
we know Volume of cylinder = 2πrh2
Therefore, ratio of volume= 2π(3)2(10)2π(9)2(30)=1/27=1/33
Hence the ratio of volume is 3 times the ratio of dimension
Hence the ratio of volume is 32 times the ratio of dimension
Page 15 Problem 5 Answer
Given: Area of both the rectangle.
To find: Scale factor of the rectangle.
In order to find the solution,
Compare both the rectangle.Making the scale factor xFinding the x, we will find the scale factor
Since, the two rectangles will be similar figures, we can determine the scale factor that was used by comparing the squares of the ratio of the sides with the ratio of their areas.
Let b be the original breadth
Let the scale factor be x
b2(b×x)2=48/12
=(b/bx)2
=4/1
=(1/x)2
=4/1 [Find the square root of both sides]
=1/x
=2/1 [Invert to find x]
=x=1/2 The scale factor of the rectangle is 1/2.