I'll take care of #2 first.
the common formula is
A = P(1 + r)^n , where r is the rate per interest period expressed as a decimal (e.g. 3% --> .03 )
and n is the number of interest periods
I don't know where your a comes from.
so
1690 = 1000(1 + r)^2
(1+r)^2 = 1.69
take √ of both sides
1+r = √1.69 = 1.3
r = .3 or 30%
( wow, let me know what bank that would be)
Can you please show you solution and final answer here. Thanks!
1.Find the altitude,perimeter and area of an isosceles trapezoid whose sides have lengths 10 cm, 20 cm, 10 cm and 30 cm.
2. The formula A=P(a+r)^2 is used to find the amount of money A in an account after P pesos have been invested in the accounting paying r % annual interest rate for 2 years. Find the interest ate r if P1000 invested grows to P1690 in two years.
5 answers
From my previous reply to this problem
http://www.jiskha.com/display.cgi?id=1418471605
(it was called #3 )
I had given you
"I drew a figure ABCD , with AD || BC
AD = 10
AB = 20
BC = 30, and
CD = 10
From A draw a line AP, so that AP is || to DC and P is on BC
So AP=10 and APCD Is a rhombus (all sides = 10)
Now look at triangle ABP , it is isosceles with sides 20,20, and 10
We need an angle, I found angle APB
cos APB =5/20
angle APB = appr 75.52°
So angle APC = 104.48°
Take it from there. "
I don't know what part you did not understand, was it the finding of angle APB ?
let's do it another way:
In triangle ABP , let angle ABP = Ø
by the cosine law:
10^2 = 20^2 + 20^2 - 2(20)(20)cosØ
100 = 800 - 800cosØ
800cosØ = 700
cosØ = 7/8
Ø = 28.955°
and it is easy to see that angle APB = (180-28.955)/2 or 75.522° like before.
and every angle can now be found
So back in the parallogram APCD , angle C = 75.522°
we can find the altitude (h) :
sin 75.522 = h/10
h = 9.682
and the area of the parallelogram = base x heigh
= 10(9.682) = 96.82 cm^2
In the isosceles triangle, using angle B and the contained sides
area = (1/2)(20)(20)sin 28.955
= 96.519
total area = 106.2 cm^2
btw, I question your labeling of the original figure as an isosceles trapezoid
To me that means that the two non-parallel sides would be equal, for yours they are not.
http://www.jiskha.com/display.cgi?id=1418471605
(it was called #3 )
I had given you
"I drew a figure ABCD , with AD || BC
AD = 10
AB = 20
BC = 30, and
CD = 10
From A draw a line AP, so that AP is || to DC and P is on BC
So AP=10 and APCD Is a rhombus (all sides = 10)
Now look at triangle ABP , it is isosceles with sides 20,20, and 10
We need an angle, I found angle APB
cos APB =5/20
angle APB = appr 75.52°
So angle APC = 104.48°
Take it from there. "
I don't know what part you did not understand, was it the finding of angle APB ?
let's do it another way:
In triangle ABP , let angle ABP = Ø
by the cosine law:
10^2 = 20^2 + 20^2 - 2(20)(20)cosØ
100 = 800 - 800cosØ
800cosØ = 700
cosØ = 7/8
Ø = 28.955°
and it is easy to see that angle APB = (180-28.955)/2 or 75.522° like before.
and every angle can now be found
So back in the parallogram APCD , angle C = 75.522°
we can find the altitude (h) :
sin 75.522 = h/10
h = 9.682
and the area of the parallelogram = base x heigh
= 10(9.682) = 96.82 cm^2
In the isosceles triangle, using angle B and the contained sides
area = (1/2)(20)(20)sin 28.955
= 96.519
total area = 106.2 cm^2
btw, I question your labeling of the original figure as an isosceles trapezoid
To me that means that the two non-parallel sides would be equal, for yours they are not.
For the trapezoid, all we really need is the altitude. You can surely then find the area and perimeter.
The long base is 10 cm longer than the shorter one.
So, draw altitudes at the end of the short base and you will see two right triangles with base 5 and hypotenuse 10.
So, the trapezoid's altitude is the same as the triangles', which is 5√3.
The long base is 10 cm longer than the shorter one.
So, draw altitudes at the end of the short base and you will see two right triangles with base 5 and hypotenuse 10.
So, the trapezoid's altitude is the same as the triangles', which is 5√3.
The fact that the sides were not listed sequentially made me interpret the question in an entirely different way, so go with Steve' solution.
Anyway, my interpretation led to a nice problem, lol
Anyway, my interpretation led to a nice problem, lol
So the perimeter is 70 cm and the area is 125√3 sq.cm?
3 answers