29.  In the adjoining figure, ABCD is a rectangle.  Its diagonal AC =  15 cm and $\angle$ACD = $\alpha$.  If cot $\alpha =\frac{3}{2}$ , find the perimeter and area of the rectangle.

30.  Using the measurements given in figure alongside,
(a) find the values of :
(1) sin $\varphi$ (ii) tan $\theta$.
(b) write an expression for AD in terms of $\theta$.
Hint: (b) CD =  5. Draw DE perpendicular to AB, BE = 5, EA = 9.

 

Solution to 29

cot(x) = base/perpendicular
cot(ACD) =  base(CD)/Perpendicular(AD) = 3/2
CD/AD = 3/2
CD = (3/2)x  # assume AD to be x

AC² = CD²+ AD² 
15^2 = ( (3/2)x)^2 + x^2 =
225 = (9/4)x^2 + x^2
225 = (13/4)x^2
x^2  = 225*(4/13)
x = 30/sqrt(13)

AD = 30/sqrt(13)
CD = (3/2) * AD
         = (3/2) * 30/sqrt(13) = 45/ sqrt(13)

Perimeter is 2( AD + CD) = 2(30/sqrt(13) + 45/sqrt(13) ) = 150/sqrt(13)

Area is AD*CD = 30/sqrt(13) * 45/sqrt(13) = 1350/13 sq cm

30
DC²= DB² - BC²
         = 13² - 12² =  169 - 144 = 25
DC = 5 cm

If we draw DE perpendicular to AB such that E is on AB
BCDE is a rectangle
BC =DE = 12 cm
CD = BE = 5 cm
so AE = AB - BE = 14 - 5 = 9 cm

in triangle ADE ( right angled at D)
AE = 9 cm DE = 12 cm
AD^2 = AE^2 + DE^2 = 9^2 + 12^2 = 81 + 144 = 225
AD = 15 cm
sin(EAD) = DE/DA = 12/15 = 4/5  #sin = perpendicular/hypotenuse
tan(EAD) = DE/EA = 12/9 = 4/3 # tan = perpendicular/base
sin(EAD) = DE/DA
AD = DE/sin(EAD) = DE*csec(EAD) because 1/sin(EAD) = csec(EAD)

Cot @=3/2 means DC/AD=3/2
Let AD be 2k
DC be 3k where ki is positive real no
In ADC
Using pythagoras theorem
(3k)^2+(2k)^2=15^2
13k^2=225
K=15/?13
AD=BC=2*15/?3=30/?3
AB=DC=45/?3
Perimeter=2(45/?3+30/?3)
=2*75/?3=150?3/3=50?3cm
Area=1350/3=450 cm2