ADDITIONAL SOLVED PROBLEMS
UNIT-02 PLANE GEOMETRY
Prob.1 The diagram shows two concentric circles. If the area of
the annulus (area bound by the two circles) equals 15.0 cm2, what
is the value of R?
1.0 cm
R
Solution:
15 = πR2- π(R-1)2 = πR2 - π(R2+1-2R) = π(-1+2R)
R = (15+π)/2π = 2.9 cm.
Prob.2. Triangle ABC is a right-angle triangle with AC = 30.0 cm and AB = 40.0 cm. AD
is perpendicular to CB. Determine the area of triangle ACD. (Hint: locate two similar
triangles in the diagram and use the condition of similarity of triangles to determine what
you need to calculate the area.)
Solution:
CB = [(30.0)2 +(40.0)2]1/2 = 50.0 cm
ΔACD and ΔABC are similar since they are right-angle triangles and α is common to
both.
C
Therefore, AC/AB = CD/AD = ¾
α
or CD = ¾ AD
D
30cm
For ΔACD,
(AD)2 + (CD)2 = (AD)2 + (3/4* AD)2
2
2
= 25/16*(AD) =(30.0) = 900 cm
2
A
Therefore AD =(900*16/25)1/2 = 24.0 cm and CD = 18.0 cm.
Thus the area of ΔACD = 1/2 {CD*AD} = ½[18.0 *24.0] = 216 cm2
40cm
B
Prob.3. A circular piece of cardboard is cut into two pieces along
the line AB with AB = R, the radius of the circle. Determine the
ratio of the area of the smaller piece to the area of the larger
piece of the cardboard.
A
R
C
O
Solution:
CO = [R2-(R/2)2]1/2 = [3/4]1/2R
B
Area of ΔABO = CO*R/2 = [3/16]1/2R2
Area of the smaller piece, A1 = πR2(60/360) - [3/16]1/2R2
= (1/6) πR2 - [3/16]1/2R2
(Note that the ΔABO is an equilateral triangle. Therefore, each of its internal angles
equals 600)
Area of the larger piece, A2 = πR2 -A1 = (5/6)πR2 + [3/16]1/2R2
A1/A2 = [(1/6)π - [3/16]1/2]/ [(5/6) π + (3/16)1/2] = 3x10-2
Prob. 4. Express the area of the figure ABCDOA outline in bold in terms of the radius R
of the circle. BD and AC are two diameters of the circle intersecting at right angle.
Solution:
Note that the outlined figure ABCDOA consists of three
right-angle triangles each of altitude R and base R.
Therefore, the area enclosed by the figure outlined in bold
= 1.5R2
Prob. 5. In the figure, ac = 7m, cd = 3m, cb = 5m. What is
the length of the hypotenuse ab?
Solution:
(ab)2 = (ac +cd)2 +(db)2
= (ac)2 +(cd)2 + 2ac*cd +(db)2
= (ac)2 +(cb)2 + 2ac*cd = 49 + 25 + 42 = 116
Therefore ab = 10.8 m
Prob.6 Consider a piece of cardboard in the shape of a trapezoid ABCD. All dimensions
are given in cm.
B
C
[a] Calculate the area of the trapezoid and
express it units of m2.
10
Area of trapezoid ABCD = ½[22+10]*10 =
160 cm2 = 1.6*10-2m2
[b] What is the area (in m2) of the largest
triangle that can be cut from the trapezoid?
D
A
10
5
15
22
Area = ½[22*10] = 110 cm2 = 1.1*10-2 m2
Prob.7 The triangle ABC is cut from a
rectangular piece of cardboard. All
dimensions are given in cm units.
Express the area of triangle ABC as a
fraction of the area of the rectangular
cardboard.
Area DABC/Area of rectangle
= [15*12]/2*25*12 = 0.3
25
B
12
A
C
10