50_Math_Problems_With_Solution_Geometry_1_Maths_Solutions,_me_Aju

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Math Problems
With Solution
 
 
Geometry
1
50
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M a t h s s o l u t i o n s 0 2
Contents
Questions................................................03
 Solutions................................................20
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M a t h s s o l u t i o n s 0 3
AB = 16 cm, AC = 14 cm & BC = 10 cm
PQR is incircle of triangle
In figure
Find the area of blue region
QUESTION 002
QUESTION 001
ABCDEF is a regular hexagon
AB = 10 cm 
AQE is a sector ( center at F )
PD is tangent of sector
In figure
Find the length of red line
Three semicircles
PA = 10 cm & PB = 20 cm
In figure
Find the radius of circle
QUESTION 003
A P B
A B
C
D
F
E
P
Q
A B
C
Q
R
P
Questions
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M a t h s s o l u t i o n s 0 4
QUESTION 004
QUESTION 005
ABCD is a square
AB = 80 cm
Two quarter circles
Circle is passing through P,Q & R
In figure
Find the length of PQ
ABCDEF is a regular hexagon
PA = PB
DQ⊥PC & PC⊥CR
In figure
Find PQ:QR:RC
Two squares
AB = 20 cm & BE = 40 cm
In figure
Find the blue shaded area
QUESTION 006
BA
D C
P Q
R
A B
C
CD
D
P
Q
R
A
C
B
D
G F
E
O
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M a t h s s o l u t i o n s 0 5
QUESTION 007
QUESTION 008
In figure
A regular octagon & 4 square
sides = 4 cm
Find the area of blue shaded square
ABCD is a square 
AB = 8 cm
O - center of semicircle
PB & CQ are tangent of semicircle
In figure
Find the area of blue quadrilateral
A regular pentagon
Altitude of pentagon is (a+b)
In figure
Find a:b
QUESTION 009
O
D C
BA
P
Q
a
b
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M a t h s s o l u t i o n s 0 6
AB, BC, CD & DA are tangents of smaller
circle
PK = 2 cm & QC = 3 cm
In figure
Find the area of blue quadrilateral
QUESTION 010
QUESTION 011
AB = 16 cm, AC = 14 cm & BC = 10 cm
O - center of incircle
In figure
Find x+y+z
AXB is a sector ( O - center )
OA = OB = 6 cm
∠QPR = 60°
Circle is passing through O, A & B
In figure
Find the area of blue region
QUESTION 012
A C
B
K
D
P Q
A B
C
x z
y
O
A B
O
X
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M a t h s s o l u t i o n s 0 7
QUESTION 014
QUESTION 015
A semicircle
A square is inscribed inside the semicircle
In figure
Find the area of blue shaded region
QUESTION 013
ABCDE is a regular pentagon
AB = 4 cm
APD is sector and BQC is semicircle
PQ is the tangent of both curves
In figure
Find PQ
AOB is a quarter circle
PQ ∥ RS ∥ BO
OA = 10 cm
AP = PR = RB
In figure
Find the area of blue region
4 cm
A B
CE
D
P
Q
A
B
O
P
Q S
R
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M a t h s s o l u t i o n s 0 8
QUESTION 016
∠ACH = 30° & ∠BCH = 45°
∠CAD = ∠HAD
Circumcircle of ΔABC = 1 cm
In figure
Find the area of ΔBDH
QUESTION 018
A rectangle and a quarter circle
In figure
Find the radius of quarter circle
QUESTION 017
Three semicircles
In figure
Find the radius of large semicircle
4 
cm
2 cm
4 cm
6 cm
A B
C
D
H
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QUESTION 019
M a t h s s o l u t i o n s
ABCD is a Square
AB = 6 cm
HEF is a semicircle
GDH & CFG are quarter circle
In figure 
Find the blue shaded area
0 9
QUESTION 020
ABCD is a parallelogram
AB = 4 cm & AD = 2 cm
DQ & PB arc of circle with center A
In figure
Find the area of blue shaded region
ABCD is a parallelogram
AB = 6 cm & AD = 4 cm
∠BAD =60°
 OA = OC
In figure
Find the blue area 
QUESTION 021
A B
CD P
Q
60°
A
D P
Q B
C
O
D C
BA
H
E
F
G
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M a t h s s o l u t i o n s
ADB is a semicircle
AC = 8 cm & BC = 2 cm
CD ⊥ AB
PCQ is an arc ( radius = CD )
In figure
Find the length of PQ
1 0
QUESTION 022
QUESTION 023
Area of ΔABC is 6 cm²
AB = 4 cm
AB ⊥ AC & AD ⊥ BC
In figure
Find the area of ΔABD
A regular hexagon & a square
Area of square = 16 cm²
In figure
Find the area of blue triangle
QUESTION 024
A B
C
D
P
A BC
D
Q
P
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M a t h s s o l u t i o n s
AMQ & APB are semicircles
AB ⊥ PQ
BM is tangent of semicircle AMQ
BQ = 32 cm & BM = 40 cm
In figure
Find area of blue region
1 1
QUESTION 025
QUESTION 026
A quarter circle & a semicircle
OA = 4 cm
In figure
Find the radius of blue circle
AB = 16 cm, AC = 14 cm & BC = 10 cm
PQR is incircle
In figure
Find the area of blue region
QUESTION 027
A B
P
Q
M
O A
B
A B
C
P
Q
R
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M a t h s s o l u t i o n s
ABCD is a square
AM = MD = DN = ND
BP & BQ are tangent of circle
In figure
Find AX : XY : YC
1 2
QUESTION 028
QUESTION 029
Three Squares
In figure
Find Blue area : Red area
QUESTION 030
A B
CD
P
Q
M
N
X
Y
AB = 16 cm, AC = 14 cm & BC = 10 cm
In figure
Find the area of blue region
A B
C
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M a t h s s o l u t i o n s 1 3
QUESTION 031
QUESTION 032
∠CAD = 15°
AB = BD & AC = BC
Radius of circle = 6 cm
In figure
Find the area of ΔABC
QUESTION 033
AB = 112 cm, AC = 98 cm & BC = 70 cm
AB ⊥ PC, AC ⊥ BR & BC ⊥ AQ
In figure
Find the radius of circle
A B
C
P
Q
R
D
C
A B
R - center of semicircle
AB = 10 cm, AC = 8 cm & BC = 6 cm
AC & BC are tangents
In figure
Find the area of blue triangle A B
C
P
Q
R
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M a t h s s o l u t i o n s 1 4
QUESTION 034
QUESTION 035
A square & two semicircles
In figure
Find a : b : c
Two incircles
AB = 5 cm, AC = 4 cm & 3 cm
AB ⊥ CD
In figure
Red Area : Blue Area = ?
QUESTION 036
AB = 16 cm, AC = 14 cm & BC = 10 cm
AQ = CQ
∠BAP = ∠CAP
In figure
Find the length of XY
A B
C
PQ
X
Y
A B
CD
X
Y
P
Q
c
b
a
C
BA D
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QUESTION 038
QUESTION 039
M a t h s s o l u t i o n s 1 5
QUESTION 037
AB ⊥ CD
AO = 4 cm, BO = 3 cm & CO = 2 cm
In figure
Find the radius of circle
ABCD & AEFG are square
∠EAB = 120°
AB = 4 cm
In figure
Find the area of blue region
Two Squares
In figure
Find PQ : XY
A B
C
D
O
A B
CD
E
F
G
P
Q
X
Y
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QUESTION 042
M a t h s s o l u t i o n s 1 6
QUESTION 040
QUESTION 041
AOB is a quarter circle
Two semicircles
In figure
Find Red Area : Blue Area
ABCDE regular pentagon
AB = 10 cm
BEFG is a square
BEH equilateral triangle
In figure
Find the area of blue triangle
ABCDE is a regular pentagon
AB = 2 cm & CR = 1 cm
PD & PB are tangent of circle
In figure
Find ∠BPD A
B
C
D
E
Q
R
P
A
F
E
D
C
B
G
H
O A
B
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M a t h s s o l u t i o n s 1 7
QUESTION 043
QUESTION 044
AXB & CXD are semicircles
In figure
Find ∠APC
AB = 10 cm, AC = 8 cm & BC = 6 cm
AB ⊥ LP, AC ⊥ NR & BC ⊥ MQ
AN = CN, CM = BM & BL = AL
In figure
Find LP + NR + MQ
QUESTION 045
A
B
C
D
P
X
A
C
B
R
Q
P
L
MN
ABCD is a rectangle
PQ = 3 cm & QR = 2 cm
In figure
FInd radius of 3rd circle
D
A
C
B
Q
P
R
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QUESTION 046
QUESTION 048
M a t h s s o l u t i o n s 1 8
QUESTION 047
A circular sector
∠AOB = 60°
OB = 6 cm
In figure
Find the area of blue area
PQRS & ABCD are squares
In figure
Find AB : PQ
A & B are center of circles
AP = 3 cm & BP = 6 cm
AY & BX are tangents
In figure
Find the area of ΔABQ
A BP
X
Y
Q
O
P Q
RS
A B
CD
B
A
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M a t h s s o l u t i o n s 1 9
QUESTION 049
QUESTION 050
A, B, C, D & E are corner of star
In figure
∠A + ∠B + ∠C + ∠D + ∠E = ?
Radius of circle PAC = 4 cm
Radius of circle PQR = 2 cm
BA &BC are tangent of both circles
In figure
Then find the area of ΔABC
B
A
C
P
Q
R
A
D
C
B
E
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M a t h s s o l u t i o n s 2 0
SOLUTION 001
Solutions
A B
C
A
C
B
P
R
Q
Blue area = Area of APOQ - Area of red sector O
Let AQ = x, then
AP = x ( AQ & AP are tangents of circle start from same point )
CP = CR = 14-x
QB = BR = 16-x
We also have
BR + CR = 10
(16-x) + (14-x) = 10
x = 10 cm 16 - xx
x
14 
cm
14 - x
16 - x
Let s = ½ perimeter of ΔABC, r = Radius of incircle,
Δ = Area of ΔABC & a,b,c are sides of triangle
We know, Δ = sr
s = ½(a+b+c) = ½(10+14+16) = 20 cm
Δ² = s(s-a)(s-b)(s-c) {Heron's formula}
Δ² = 20(20-10)(20-14)(20-16)
Δ² = 4800
Δ = 40√3 cm²
We get, 40√3 = 20r
r = 2√3 cm
16 cm
10 cm
14 
- x
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M a t h s s o l u t i o n s 2 1
We know AB = 10 cm ∠DEF = ∠FAB = 120°
Let DQ = x, PQ = y & PA = z
then PD = x+y & PB = 10-z
∠FEX = ∠FAY = 180° - 120° = 60°
so  FE = EX = FA = AY = 10 cm 
From figure
DE×DX = DQ² {Secant-Tangent Theorem}
DE×(DE+EX) = x²
10(10+10) = x² = 200 ⇒ x = 10√2 cm
PA×PY = PQ² {Secant-Tangent Theorem}
z(z+10) = y² ⇒ y² = z²+10z......................eq(1)
SOLUTION 002
10 cmA Q
P
O
B
C
R
2√
3 cm
Area of APOQ = 10×2√3 = 20√3 cm² = 34.641 cm²
From figure
tan(∠OAQ) = (2√3)/10 ⇒ ∠OAQ = 19.106°
so ∠AOQ = 90° - 23.413° = 70.894°
Then
∠POQ = 2×∠AOQ = 2×70.894° = 141.788°
A Q
O
P
2√
3 cm
B
R
C
10 cm
Area of red sector = (141.788/360)×π(2√3)² = 14.848 cm²
Area of APOQ = 34.641 cm²
Blue area = Area of APOQ - Area of the red sector
Blue area = 27.713 - 13.945 = 13.678 cm²
Blue area = 19.792 cm²
A BP
D
C
E
F
Q
Y
X 10 cm
x
y
z
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M a t h s s o l u t i o n s 2 2
From ΔBCD
∠C = 120° & CB = CD = 10 cm
BD² = BC² + BD² - 2×BC×BD cos 120 {cosine rule}
BD² = 10² + 10² - 2×10×10×(-½) = 300
From ΔPBD
PD² = PB² + BD²
(x+y)² = (10-z)² + 300
200+20y√2+y² = 100-20z+z² + 300
20y√2 + 20z + y² = 200 + z²
20y√2 + 20z + z² + 10z = 200 + z²
2y√2 = 20 - 3z...........................................................eq(a)
Squre both sides then,
8y² = 400 - 120z + z²..................................................eq(2)
eq(2)×8 then,
8y² = 8z² + 80z...........................................................eq(3)
From eq(2) & eq(3)
8z²+80z = 400 - 120z + 9z²
z² - 200z + 400 = 0....................................................eq(3)
Solving eq(4) then we get, z = 100 + 40√6 & 100 - 40√6
also z is smaller than side of polygon so, z = 100 - 40√6
From Eq(a) 
 2y√2 = 20 - 3z
y = (20 - 3(100 - 40√6))/2√2
y = 60√3 - 70√2 
We know, PD = x+y
So PD = (10√2) + (60√3 - 70√2)
PD = 60√3 - 60√2 cm
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M a t h s s o l u t i o n s 2 3
Let r is radius of circle & PX = QX = x
Then PQ = 2x
From ΔAOR
AO² = AR² + OR²
(40-r)² = 20² + r² {pythagorean theorem}
1600-80r+r² = 400 + r²
r = 15 cm
SOLUTION 004
From figure 
r - Radius of circle
OX = 5+r, OY = 15-r, OZ = 10+r, XY = 10 & XZ = 15
From ΔXOY 
OY² = XO² + XY² - 2×XO×XY×cos(∠OXP)
(15-r)² = (5+r)² + 10² - 20(5+r)cos(∠OXP)
225-30r+r² = 25+10r+r² + 100 - 20(5+r)cos(∠OXP)
cos(∠OXP) = (2r-5)/(5+r)......................eq(1)
From ΔXOZ
OZ² = XO² + XZ² - 2×XO×XZ×cos(∠OXP)
(10+r)² = (5+r)² + 15² - 30(5+r)cos(∠OXP)
100+20r+r² = 25+10r+r² + 225 - 30(5+r)cos(∠OXP)
cos(∠OXP) = (15-r)/(3(5+r))......................eq(2)
From eq(1) & eq(2)
(2r-5)/(5+r) = (15-r)/(3(5+r))
2r-5 = (15-r)/3
6r-15 = 15-r
7r = 30
r = 30/7 cm
SOLUTION 003
A P B
O
X Y Z555
10
5
r
r
r
A B
P Q
R
O
r
r
40
-r
x X
20 cm 20 cm
CD
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SOLUTION 005
M a t h s s o l u t i o n s 2 4
ΔAOR & ΔPOX are similar
so, 
20/x = (40-r)/r
20r = 40x-xr
20×15 = 40x-15x = 25x
x = 12
so PQ = 2x = 2×12
PQ = 24 cm
A
C
B
DE
F
P
Q
R
x
2x
x
x√7
2x
2x√
3
x√
7
2x
√
3
Let sides = x, PQ = a, QR = b & RD = c, Then PA=PB=x
From ΔFAP
PF² = AF²+AP²-2×AF×AP×cos(∠FAP)
PF² = (2x)²+x²-2×2x×x×cos(120)
PF² = 4x²+x²-4x²×(-½) = 7x² ⇒ PF = x√3
PF = PC = x√7 {Due to symmetry}
From ΔEDF
FD² = (2x)²+(2x)²-2×2x×2x×cos(120)
FD² = 8x²-8x²×(-½) = 12x² ⇒ FD = 2x√3
FD = BD = 2x√3
From ΔPQF & ΔDQF
FQ² = (x√7)²-a² {Pythagorean theorem}
FQ² = (2x√3)²-(b+c)² {Pythagorean theorem}
(x√7)²-a² = (2x√3)²-(b+c)²
7x²-a² = 12x²-(b+c)²
(b+c)²-a² = 5x²
(b+c+a)(b+c-a) = 5x²....................................eq(1)
c
b
a
2x
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M a t h s s o l u t i o n s 2 5
From ΔPCR & ΔDCR
RC² = (2x)²-c² {pythagorean theorem}
FQ² = (x√7)²-(a+b)² {pythagorean theorem}
(2x)²-c² = (x√7)²-(a+b)²
4x²-c² = 7x²-(a+b)²
(a+b)²-c² = 3x²
(a+b+c)(a+b-c) = 3x²....................................eq(2)
From ΔBDP
(a+b+c)² = (2x√3)²+x² {pythagorean theorem}
(a+b+c)² = 12x²+x² = 13x²
a+b+c = x√13.............................................eq(3)
Put eq(3) in eq(1)
(b+c-a)x√13 = 5x²
b+c-a = 5x/√13..........................................eq(4)
Put eq(3) in eq(2)
(a+b-c)x√13 = 3x²
a+b-c = 3x/√13............................................eq(5)
eq(4) + eq(5)
(b+c-a) + (a+b-c) = 8x/√13
2b = 8x/√13 ⇒ b = 4x/√13
eq(4) - eq(5)
(b+c-a) - (a+b-c) = 2x/√13
2c - 2a = 2x/√13
c-a = x/√13................................................eq(6)
Put b = 4x/√13 in eq(3)
a + 4x/√13 + c = x√13
a+c = x√13 - 4x/√13 = 9x/√13................eq(7)
 
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Put b = 4x/√13 in eq(4)
4x/√13 + c-a = 5x/√13
c-a = x/√13...............................................eq(8)
eq(7) + eq(8)
(a+c) + (c-a) = 8x/√13 ⇒ a = 4x/√13
Put a = 4x/√13 in eq(8)
c-4x/√13 = x/√13 ⇒ c = 5x/√13
a : b : c = 4x/√13 : 4x/√13 : 5x/√13
a : b : c = 4 : 4 : 5
M a t h s s o l u t i o n s 2 6
Blue area = Area of AOGCDA
Area of AOGCDA = Total area - Yellow area
SOLUTION 006
A EB
CD
FG
From fig(2)
ΔAOE & ΔGOF similar
AE/GF = PO/QO
60/40 = (40-x)/x
60x = 1600-40x
100x = 1600 ⇒ x = 16 cm
Blue area = Total area - (Area of ΔAEF + Area of ΔOFG)
Total area = 20²+40² = 400+1600
Total area = 2000 cm²
Area of ΔAEF = ½×AE×EF
Area of ΔAEF = ½×60×40 = 1200 cm²
Area of ΔOFG = ½×GF×x = ½×40×16
Area of ΔOFG = 320 cm²
Blue Area = 2000-1200-320 = 480 cm²
Blue Area = 480 cm²
A B E
CD
FG
20 cm 40 cm
x
40
-x
40 cm
O
P
Q
Fig(1)
Fig(2)
O
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M a t h s s o l u t i o n s 2 7
Area of Kite = longer side × smaller side
Area of Kite = 4x cm²
From figure 
Let PY = y & PZ = x Then
We know PB = AB = 8 cm {Tangents from B}
From ΔBYC
BY² = CY² + BC²
(8+y)² = (8-y)² + 8²
64+16y+y² = 64-16y+y² + 64
32y = 64 ⇒ y = 2 cm & BY = 10 cm
ΔBZX is similar to ΔCZY
so we get BZ = YZ 
8-x = 2+x ⇒ x = 3 cm
Area of Kite = 4×3 = 12 cm²
Area of Kite = 12 cm²
SOLUTION 007
From ΔABC
∠CAB = 135 -90 = 45°
∠ABC = 135 -90 = 45°
We know AB = BD = FH = 4 cm,
BC = AC = AB sin 45 = 2√2 cm
CD = BD-BC = 4-2√2 cm  ⇒ FE = 4-2√2 cm
GH = 4-2√2 cm
Side of square = EG = FH - FE - GH = 4 - (4-2√2) - (4-2√2)
EG = 4√2 - 4
Area of square = (4√2 - 4)² = 48-32√2
Area of square= 48-32√2 cm²
A
B DC
E
F
G
H
4 cm
2√2 cm
SOLUTION 008
A B
D C
X
Q
P
O Z
Y
4 
cm
4 
cm
4 c
m
4 cm
x
8 
cm
2 cm
y 8-y
8-x
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Area of Kite = ½×AC×BD
AC = AK+CK
We know PK = 2 cm & QC = 3 cm
PC = CK-PK = 6-2 = 4 cm
From ΔCPN 
sin(∠PCN) = 2/4 ⇒ ∠PCN = 30°
NC² = 4²-2² = 12 ⇒ NC = 2√3 cm
ΔCPN similar to ΔCKD
CD/NC = KC/PC
CD/2√3 = 6/4 ⇒ CD = 3√3 cm
ND = √3 cm
ND = LD = √3 cm
Let side of pentagon = 2x then
AD² = AE²+DE²-2×AE×DE×cos 108
AD² = 4x²+4x²-8x²×¼(1-√5) ⇒ AD² = (√5 +1)² x²
AD = EC = 3.236 x
DF² = AD²-AF²
(a+b)² = (√5 +1)² x² - x² = (5+2√5)x²
a+b ≈ 3.078 x
∠DEC = (180-108)/2 = 36°
a = 2x sin 36° ≈ 1.175 x
b = 3.078 x - 1.175x = 1.903 x
a/b = 1.175 x/1.903 x = 1175/1903
a:b ≈ 1175:1903
M a t h s s o l u t i o n s 2 8
SOLUTION 009
A B
C
D
E O
a
b
F
2x
x
2x
SOLUTION 010
A C
B
D
PK QM
N
2 
cm
3 cm1 cm
2√3 cm
√3 cm
L
√
3 
cm
1.
5√
3 
cm
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s
From ΔCMD
DM = CD×sin 30
DM = 3√3 × (½) = 1.5√3 cm ⇒ BD = 3√3 cm {Due to symmetry}
CM = CD×cos 30 = 3√3×cos 30 = 4.5 cm
KM = KC - MC = 6 - 4.5 = 1.5 cm
Let AL = p & AK = q from figure 
AL² = AK×AC
p² = q(q+4)
p²-q² = 4q............................................................eq(1)
From ΔADM
AD² = AM²+DM²
(√3+p)² = (1.5+q)²+(1.5√3)²
3 + 2p√3 + p² = (2.25+3q+q²) + 6.75
p²-q² = 3q - 2p√3 + 6........................................eq(2)
From eq(1) & eq(2)
4q = 3q - 2p√3 + 6
6-q = 2p√3 ⇒ 12p² = 36-12q+q²....................eq(3)
eq(1)×12 ⇒ 12p²-12q² = 48q..........................eq(4)
From eq(3) & eq(4)
36-12q+q² - 12q² = 48q
11q²+60q-36 = 0 solving this equation we get, q = -6 & 6/11
q not become negative so, q = 6/11
AC = AK+KC = q+KC = 6/11 + 6 = 72/11 cm
Area of Kite = ½×AC×BD = ½×(72/11)×3√3 = 17.005 cm²
Area of Kite = 17.005 cm²
2 9
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 3 0
SOLUTION 011
Let AQ = p, then
AP = p {AQ & AP are tangents of circle start from same point}
CP = CR = 14-p
QB = BR = 16-p
We also have
BR + CR = 10
(16-p) + (14-p) = 10
p = 10 cm
Let s = ½ perimeter of ΔABC, r = Radius of incircle,
Δ = Area of ΔABC & a,b,c are sides of triangle
We know, Δ = sr
s = ½(a+b+c) = ½(10+14+16) = 20 cm
Δ² = s(s-a)(s-b)(s-c) {Heron's formula}
Δ² = 20(20-10)(20-14)(20-16)
Δ² = 4800
Δ = 40√3 cm²
We get, 40√3 = 20r
r = 2√3 cm
From ΔAQO
x² = p²+r² = 10²+(2√3)² = 100+12 = 112
x=4√7 cm
From ΔBQO
z²=(16-p)²+r² = 6²+(2√3)² = 36+12 = 48
z=4√3 cm
From ΔCPO
y²=(14-p)²+r² = 4²+(2√3)² = 16+12 = 28
y = 2√7 cm
x+y+z = 6√7 + 4√3 cm
A
C
B
O
x z
y R
Q
P
p 16-p
p
14-
p
16-p
14-p
r
r
r
airesshahin@gmail.com 04 Jun 2021
SOLUTION 013
M a t h s s o l u t i o n s
Blue Area = Area of circle - Yellow Area -
 Green Area - White Area
3 1
SOLUTION 012
Radius of circle = 6/(2 sin 60) = 2√3 cm
Area of circle = π(2√3)² = 12π cm²
From figure
∠AOB = ∠AOC = 120°
Green Area = (⅓)π(2√3)² = 4π cm²
Yellow Area = (⅓)π(2√3)² = 4π cm²
White Area = Area of sector - Area of ΔAOC
 - Area of ΔAOB
Area of ΔAOC = ½(2√3)²sin 120 = 3√3 cm²
Area of ΔAOB = ½(2√3)²sin 120 = 3√3 cm²
Area of sector BXC = ⅙π(6)² = 6π cm²
White Area = 6π-3√3-3√3 = 6π-6√3 cm²
Blue Area = 12π-4π-4π-(6π-6√3) = 6√3-2π cm²
Blue Area = 6√3-2π cm²
A
C B
O
6 cm
6 
cm
6 cm
2√3 cm 2√
3 c
m
2√
3 cm
x
A
BC
O
2×Blue Area = Yellow Area - Red Area
Yellow Area = Area of circle - Green Area
airesshahin@gmail.com 04 Jun 2021
SOLUTION 014
From figure
PN = QN = 2 cm {QPNM is a Rectangle}
PQ = MN {QPNM is a Rectangle}
EQ = 4 cm 
From ΔCDE
CE² = ED²+CD²-2×ED×CD×cos 108
CE² = 4²+4²-2×4×4×¼(1-√5) ⇒ CE² = 24+8√5
From ΔCNE
NE² = CE²-NC² {Pythagorean theorem}
NE² = 24+8√5 - 2² = 20+8√5
From figure
AP×PB = PS²
4×(x+4) = x²
4x+16 = x²
x²-4x-16 = 0 solving this equation, then we get
x = 2±2√5 length not become negative, so
x = 2+2√5
2r = 4+x+4 = 4+2+2√5+4 = 10+2√5 ⇒ r = 5+√5 cm
Area of circle = π(5+√5)² = 30+10√5 ≈ 164.496 cm²
Red Area = ½×π(2+2√5)² = 12π+4π√5 ≈ 65.798 cm²
sin(∠POS) = x/r = (2+2√5)/(5+√5) = ⅖√5
cos(∠POS) = (½x)/r = (1+√5)/(5+√5) = ⅕√5
∠MOS = 2×∠POS {Due to symmetry}
sin(∠MOS) = ⅖√5 × ⅕√5 = ⅘
∠MOS ≈ 126.869° ⇒ 126.869/360 ≈ 0.352
Green area = Area of sector SAM - area of ΔSOM
Green area = 0.352×π(5+√5)² - ½(2+2√5)² ≈ 57.970 - 20.944 = 37.011 cm²
Yellow area = 164.496 - 37.011 = 127.485 cm²
2×Blue Area = 127.485 - 65.798 = 61.687 cm²
Blue Area = 30.8435 cm²
M a t h s s o l u t i o n s 3 2
A P BQ 4 cm
x
4 cm
RS
O½x
r
NM
BA
D
C
P
E
Q
M
N
4 cm
2 
cm
2 
cm2 cm
2 cm
4 cm4 c
m
4 cm
airesshahin@gmail.com 04 Jun 2021
SOLUTION 015
M a t h s s o l u t i o n s
From ΔMNE
MN² = NE²-ME² {Pythagorean theorem}
MN² = 20+8√5 - 6² = 8√5 - 16
MN = PQ = 1.374 cm
3 3
From figure
∠AOP = ∠POR = ∠ROB = 30°
Blue area = Area of ΔOQP + Area of sector POR
 - Area of ΔOSR
From  ΔOQP
OP = 10 cm
OQ = OP×cos 30 = 12cos 30 = 6√3 cm
Area of ΔOQP = ½×12×6√3×sin 30
 = 18√3 cm²
Area of sector POR = 30×π×12²/360
 = 12π cm²
From  ΔOSR
OS = OP×cos 60 = 12cos 60 = 6 cm
Area of  ΔOSR = ½×12×6×sin 60
 = 18√3 cm²
Blue area = 18√3 + 12π - 18√3 = 12π cm²
Blue area = 12π cm²
B
Q SA O
P
R
10 cm
airesshahin@gmail.com 04 Jun 2021
SOLUTION 016
M a t h s s o l u t i o n s
Let r is radius of circle
From ΔOPQ
(r-4)²+(r-2)² = r²
r²-8r+16 + r²-4r+4 = r²
r²-12r+20 = 0 or (r-10)(r+2) = 0
so, r = 10 & r = -2
Length not become negative so r = 10 cm
Radius of circle = 10 cm
3 4
SOLUTION 017
A
PO B
QR
4 
cm
r-
4 
cm
r c
m
r-
4 
cm
2 cmr-2 cm
From Figure
AP = 2 cm, QR = 3 cm & RS = 3 cm
From ΔPOQ
PQ² = OP²-OQ² {Pythagorean theorem}
PQ² = PQ² = 5²-3²  ⇒ PQ = 4 cm
From Figure
AR = AP+PQ+QR = 2+4+3 = 9 cm
AR×RB = RS² {Chord theorem}
9×RB = 3² = 9 ⇒ RB = 1 cm
Diameter of circle = AR+RB = 9+1 = 10 cm
Radius of large semicircle = 5 cm
A BQ RP
O S
2 cm2 cm
3 cm3 cm
3 
cm
3 
cm
2 c
m
3 c
m
airesshahin@gmail.com 04 Jun 2021
SOLUTION 018
Blue area = Yellow area - Red area -
 Green area - Area of orange circle
SOLUTION 019
M a t h s s o l u t i o n s 3 5
A B
C
D
H
Let consider BH = x
From figure
∠ACH = 30° & ∠BCH = 45°
BH = x & CH = x {Because ∠BCH = 45°}
AH = CH×tan 30 {Because ∠ACH = 30°}
AH = x×tan 30 = ⅓x√3
∠CAH = 90°-30°= 60°
∠DAH = ½∠CAH = 30°
DH = AH×tan 30° = ⅓x√3×⅓√3 = ⅓x
Diameter of circumcircle = AB/sin (∠ACB)
AB = Diameter of circumcircle × sin (∠ACB)
⅓x√3 + x = 2×sin (75) = ½√2 (1+√3)
x = ½√6 cm
Area of ΔBDH = ½×DH×BH = ½×⅓x×x = ⅙x² = ⅙(½√6)² = ¼ cm²
Area of ΔBDH = ¼ cm²
x
airesshahin@gmail.com 04 Jun 2021
Blue area = Area of sector PAB
 + Area of ΔADP
 - Area of sector DAQ
From ΔADY
DY = 2×sin 60 = √3
DY = PX = √3
From ΔAPX
sin(∠PAX) = PX/AP = √3/4
∠PAX = 25.66°
∠DAP = 60-25.66 = 34.34°
Let r is the radius of orange circle
From figure
PG = EG-(EO+PO) = 6-(r+r) = 6 - 2r
From ΔOGC 
OC² = OG²+CG² {pythagorean theorem}
(3+r)² = (6-r)²+3²
9+6r+r² = 36-12r+r²+9
18r = 36 ⇒ r = 2
Yellow area = 3×6 + ½π×3² = 18+4.5π cm²
Red area = ¼π×3² = 2.25π cm²
Green area = ¼π×3² = 2.25π cm²
Area of orange circle = π×2² = 4π cm²
Blue area = 18+4.5π - 2.25π -2.25π - 4π = 18 - 4π cm²
Blue area = 18-4π cm²
SOLUTION 020
M a t h s s o l u t i o n s 3 6
A B
CD
E
H F
G
O
P
rr
3 cm3 cm
3 
cm
A
C
B
D P
Q X
4 cm2 
cm
Y
√
3 
cm
√
3 
cm
airesshahin@gmail.com 04 Jun 2021
SOLUTION 021
M a t h s s o l u t i o n s 3 7
Area of sector PAB = (25.66/360)π×4² = 3.58 cm²
Area of ΔADP = ½×2×4×sin 34.34 = 2.25 cm²
Area of sector DAQ = ⅙×π×2² = 2.09 cm²
Blue area = 3.58+2.25-2.09 = 3.74 cm²
Blue area = 3.74 cm²
From figure
AQ = AD×cos 60 = 4×½ = 2 cm
DQ = AD×sin 60 = 4×½√3 = 2√3 cm
QB = AB-AQ = 6-2 = 4 cm
QB = DP = 4 cm
BD² = DQ²+QB² = (2√3)² + 4² = 28
BD = 2√7 cm ⇒ OD = √7 cm 
tan(∠PQD) = DP/QD = 4/(2√3)
∠PQD = 49.106°
∠POD = 2×∠PQD = 98.214°
Blue area = 2×Area of segment PXD
Area of Segment PXD = Area of sector POD - Area of ΔPOD
Area of sector POD = (98.214/360)π×(√7)² = 6.000 cm²
Area of ΔPOD = ½×√7×√7×sin 98.214 = 3.464 cm²
Area of segment PXD = 6.000-3.464 = 2.536 cm²
Blue area = 2×2.536 = 5.072 cm²
Blue area = 5.072 cm²
BA
CD
Q
P
O
60°
4 
cm
Y
X
airesshahin@gmail.com 04 Jun 2021
SOLUTION 023
M a t h s s o l u t i o n s
From figure 
AC+BC = 8+2 = 10 cm
OA = OB = OD = OQ =10/2 = 5 cm
OC = AC-OA = 8-5 = 3 cm
CD² = AC×BC
CD² = 8×2 = 16 ⇒ CD = 4 cm
DQ = 4 cm {Radius of circle}
QR = PR {O & R centers and 
 PQ is a chord }
From ΔODQ
Area² = s(s-a)(s-b)(s-c) {Heron's formula}
Here a = 5 cm, b = 5 cm & c = 4 cm 
2s = a+b+c = 5+5+4 = 14 ⇒ s = 7 cm
Area² = 7(7-5)(7-5)(7-4) = 7×2×2×3 = 84 cm² ⇒ Area = 2√21
We can also find area of triangle in another way that is
Area = ½×OD×QR = ½×5×QR
½×5×QR = 2√21
QR = ⅘√21 ⇒ PQ = 2×⅘√21 = 1.6√21 cm
PQ = 1.6√21 cm
3 8
SOLUTION 022
From ΔABC
Area = ½×AB×AC
6 = ½×4×AC ⇒ AC = 3 cm
BC² = AB²+AC² = 4²+3²  ⇒ BC = 5 cm
From figure
PA=PD {BC is diameter and AD⊥BC}
A C B
P
D
Q
O5 cm 3 cm 2 cm
R
5 
cm
5 cm
4 cm
A B
C
D
P
4 cm
3 
cm
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s
We can also find area of ΔABC in another way that is
Area of ΔABC = 6 = ½×BC×PA = ½×5×PA ⇒ PA = 12/5 = 2.4 cm
AD = 2(12/5) = 24/5 = 4.8 cm
From ΔABP
BP² = AB²-PA² = 4²- (2.4)² = 10.24 ⇒ BP = 3.2 cm
Area of ΔABD = ½×AD×BP = ½×4.8×3.2 = 7.68 cm²
Area of ΔABD = 7.68 cm²
3 9
SOLUTION 024
From figure
Area of square = 16 cm²
Side of square = AB = AQ = 4 cm
Area of blue triangle = Area of ΔABQ 
 + Area of ΔAFQ
 + Area of ΔABF
Area of ΔABQ = ½×AB×AQ = ½×4×4 = 8 cm²
 
Area of ΔAFQ = ½×AF×AQ×sin 30 
Area of ΔAFQ = ½×AF×AQ×sin 30 = ½×4×4×½
Area of ΔAFQ = 4 cm²
Area of ΔABF = ½×AF×AB×sin 120 
Area of ΔABF = ½×4×4×sin 120 = 4√3 cm²
Area of blue triangle = 8+4 - 4√3 = 12-4√3 cm²
Area of blue triangle = 12-4√3 cm²
A
C
B
E D
F
Q P
4 cm
4 
cm
4 cm
airesshahin@gmail.com 04 Jun 2021
From figure
Blue area = Area of sector ARP - Area of ΔPQR - Area of small semicircle
BQ×BA = BM²
32×BA = 40² ⇒ BA = 50 cm
Radius of large semicircle = 50/2 = 25 cm
or RA = RB = RP = 25 cm
AQ = BA -BQ = 50-32 = 18 cm
BQ×AQ = PQ²
32×18 = PQ² ⇒ PQ = 24 cm
Radius of small semicircle = 18/2 = 9 cm
QR = RA-AQ = 25-18 = 7 cm 
tan(∠PRQ) = PQ/QR = 24/7  ⇒ ∠PRQ = 73.740°
Area of sector ARP = (73.740/360)π×25² = 402.189 cm²
Area of ΔPQR = ½×24×7 = 84 cm²
Area of small semicircle = ½π×9² = 127.235 cm²
Blue area = 402.189-84-127.235 = 190.954 cm²
Blue area = 190.954 cm²
SOLUTION 026
M a t h s s o l u t i o n s 4 0
SOLUTION 025
A B
P
Q
M
R
25 cm
18 cm 25 cm7 cm
40 cm
O A
B
Let radius of circle = R
From figure OP = YX
From ΔPOX
OP² = OX²-PX² = (4-R)²-R² = 16-8R........eq(1)
From ΔXYZ
YX² = XZ²-YZ² = (2+R)²-(2-R)²  = 8R.......eq(2)
From eq(1) & eq(2)
16-8R = 8R ⇒ R = 1 cm
R
R
R
2-R
2 cm
2 cm
R
Z
Y X
P
Q
4-R
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 1
Blue Area = Yellow Area - Red Area 
SOLUTION 027
Let radius of Blue circle is x
From ΔUVZ
(2+x)²-(2-x)² = UV² ⇒ UV = 2√(2x)
From ΔXWV
(1+x)²-(1-x)² = VW² ⇒ VW = 2√x
From eq(1)
OP² = 16-8R = 16-8 = 8 ⇒ OP = 2√2 cm
OP = UW = UV+VW = 2√(2x) + 2√x = 2√2
√(2x) + √x = √2
√x × (√2 + 1) = √2
√x = 2-√2 ⇒ x = 6-4√2 cm
Radius of circle = 6-4√2 cm
A
B
O
X
Z
P
U
2+x
x x
1-x
x
2-
x
WV
1+x
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 2
A
C
B
R
Q
P
O
Let AP = x, then
AR = x ( AP & AR are tangents of circle start from same point )
CR = CQ = 14-x
BP = BQ = 16-x
We also have
BQ + CQ = 10
(16-x) + (14-x) = 10
x = 10 cm
Let s = ½ perimeter of ΔABC, r = Radius of incircle,
Δ = Area of ΔABC & a,b,c are sides of triangle
We know, Δ = sr
s = ½(a+b+c) = ½(10+14+16) = 20 cm
Δ² = s(s-a)(s-b)(s-c) {Heron's formula}
Δ² = 20(20-10)(20-14)(20-16)
Δ² = 4800
Δ = 40√3 cm²
We get, 40√3 = 20r
r = 2√3 cm
A
C
B
Q
R
Px
x
16-x
14-
x
16-x
14-x
2√
3 c
m
2√3 cmS
6 cm
6 cm
4 cm
4 c
m
10 
cm
10 cm
From ΔBOQ
∠BQO = 90°
From ΔABC
AB² = AC²+BC²-2×AC×BC× cos C {Cosine rule}
16² = 14²+10²-2×14×10× cos C
256 = 296 - 280 cos C  ⇒ cos C = ⅐
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 3
SOLUTION 028
C
From ΔACQ
AQ² = AC²+CQ²-2×AC×CQ× cos C {Cosine rule}
AQ² = 14²+4²-2×14×4×⅐ = 196 ⇒ AQ = 14 cm 
That is AQ = AC = 14 cm ⇒ ∠C = ∠AQC
cos C = ⅐ ⇒ ∠C = ∠AQC = 81.787°
∠OQS = 180-∠OQB-∠AQC = 180-90-81.787
∠OQS = 8.213°
From ΔOSQ
∠OSQ = ∠OQS = 8.213° {SInce OS=OQ=Radius of circle}
∠SOQ = 180 -∠OSQ-∠OQS = 180-8.213-8.213 = 163.574°
Red Area = Area of ΔOSQ = ½×OS×OQ×sin (∠SOQ) = ½×OS×OQ×sin (∠SOQ)
Red Area = ½×2√3×2√3×sin 163.574° = 1.697 cm²
Yellow Area = Area of sector SOQ = (163.574/360)π(2√3)² = 17.129 cm
Blue area = 17.129-1.697 = 15.432 cm²
A
C
B
R
Q
P
O 2
√3
 cm
2√3 cmS
6 cm
6 cm
4 cm
4 c
m
10 
cm
10 cm
Let AM = MD = DN = ND = x
AX = CY {Due to symmetry}
From figure
MN² = DM²+DN² = x²+x² ⇒ MN = x√2
MN = DZ = x√2 {Diameter of circle}
So, radius of circle = ½x√2
BD² = (2x)²+(2x)² = 8x² ⇒ BD = 2x√2
BG = x√2
OB = 2x√2 - ½x√2 = 1.5x√2
BQ² = OB²-OQ²
BQ² = (1.5x√2)²-(½x√2)² ⇒ BQ = 2x
BP = BQ = 2x A B
CD
M
N
P
Q
X
Y
x
x
x
x
2x
2x
O
½x√2
½x
√2
½x
√2
G
H
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 4
Blue area = Yellow area - Red area
SOLUTION 029
From ΔOPB
Area = ½×OP×BP = ½×½x√2×2x = ½x²√2
Also, Area = ½×HP×OB = ½×HP×1.5x√2 = ½x²√2 ⇒ HP = ⅔x
PQ = 2×⅔x = (4/3)x
From ΔBPH
BH² = BP²-HP² = (2x)²-(⅔x)² = 4x²-(4/9)x² 
BH² = (32/9)x² ⇒ BH = (4/3)√2 x
From ΔBXY & ΔBPQ
PQ/XY = BH/BG ⇒ PQ×BG = BH×XY 
(4/3)x×x√2 = (4/3)√2 x×XY ⇒ XY = x
AX+XY+YB = 2x√2
2AX + x = 2x√2
AX = YB = (√2-½)x
AX:XY:YB = (√2-½)x : x : (√2-½)x
AX:XY:YB = √2-½ : 1 : √2-½
AX:XY:YB = 2√2-1 : 2 : 2√2-1
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 5
Let x is the side of blue square 
From figure
OD = ON = x {∠OND = 45°}
MB = MN = x {∠BNM = 45°}
BD² = AB²+AD² = (2x)²+(2x)²
BD² = 8x² ⇒ BD = 2x√2
Blue area = x²
Let y is the side of red square
DS = RS = y {∠OND = 45°}
PB = PQ = y {∠BNM = 45°}
BD = OS+SP+PB = y+y+y = 3y = 2x√2
 ⇒ y = ⅔x√2
Red area = y²
Blue area : Red area = x² : y² = x² : (⅔x√2)² = 1 : 8/9 = 9 : 8
Blue area : Red area = 9 : 8
SOLUTION 030
A
C
B
O
16 cm
14 
cm
10 cmR
R
From figure
AC² = BA²+BC²-2×BA×BC cos B
14² = 16²+10²-2×16×10 cos B
cos B = ½ ⇒ ∠B = 60°
∠AOC = 2×60 = 120° {Inscribed angle}
∠OAC = ∠OCA {OA = OC}
∠OAC + ∠OCA + ∠AOC = 180°
∠OAC + ∠OCA = 180-120 = 60°
∠OAC = ∠OCA = 30°
2×R = 14/sin 60 ⇒ R = ⅓×14√3 cm
Yellow area = ⅓πR² = ⅓π×(⅓×14√3)² = ⅑×196π cm²
Red area = ½×R×R×sin 60 = ½×⅓×14√3×⅓×14√3×sin 60 = ⅓×49√3 cm²
Blue area = ⅑×196π - ⅓×49√3 cm²
A B
CD
S
R
P
Q
O
N
Mx
x
x x
x
x
y
y
yy
y
y
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 4 6
SOLUTION 031
Blue circle is circumcircle of ΔPQR
From ΔAPC & ΔBPC
PC² = AC²-AP² = BC²-PB²
98²-AP² = 70²-(AB-AP)² = 70²-(112-AP)²
9604-AP² = 4900-12544+224×AP-AP² ⇒ AP = 77 cm
PB = 112 - 77 = 35 cm
From ΔARB & ΔCRB
BR² = AB²-AR² = BC²-RC²
112²-AR² = 70²-(AC-AR)² = 70²-(98-AR)²
12544-AR² = 4900-9604+196×AR-AR² ⇒ AR = 88 cm
CR = 98 - 88 = 10 cm
From ΔABQ & ΔCQA
AQ² = AB²-BQ² = AC²-CQ²
112²-BQ² = 98²-(BC-BQ)² = 98²-(70-BQ)²
12544-AR² = 9604-4900+140×BQ-AR² ⇒ BQ = 56 cm
CQ = 70 - 56 = 14 cm
From ΔABC
BC² = AB²+AC²-2×AB×AC cos (∠CAB)
70² = 112²+98²-2×112×98×cos (∠CAB)  ⇒ cos (∠CAB) = 11/14
AC² = AB²+BC²-2×AB×BC cos (∠ABC)
98² = 112²+70²-2×112×70 cos (∠ABC) ⇒ cos (∠ABC) = ½
AB² = AC²+BC²-2×AC×BC cos (∠ACB)
112² = 98²+70²-2×98×70 cos (∠ACB) ⇒ cos (∠ACB) = ⅐
PR² = AR²+AP²-2×AR×AP cos (∠CAB)
PR² = 88²+77²-2×88×77×11/14 = 3025 ⇒ PR = 55 cm
QR² = CQ²+CR²-2×CQ×CR cos (∠ACB)
QR² = 14²+10²-2×14×10×⅐ = 256 ⇒ QR = 16 cm
PQ² = BP²+BQ²-2×BP×BQ cos (∠ABC)
PQ² = 35²+56²-2×35×56×½ = 2401 ⇒ PQ = 49 cm
A P
C
B
R
Q
35 cm
88
 cm
77 cm10 cm
56 cm
14 
cm
airesshahin@gmail.com 04 Jun 2021
SOLUTION 032
Let ∠BAD = θ
From ΔABD
∠BAD = ∠ADB = θ {AB = DB}
∠ABD = 180-∠BAD-∠ADB = 180-2θ
From ΔABC
∠CAB = ∠CBA {AC = BC}
15-θ = 180-2θ = 3θ = 165 ⇒ θ = 55°
∠CBA = 180-2θ = 180-110 = 70°
∠CAB = ∠CBA = 70°
∠ACB = 180-∠CAB-∠CBA = 180-140
∠ACB = 40°
AC/sin (∠ABC) = 2R = 2×6 = 12
AC/sin 70 = 12 ⇒ AC = 12×sin 70
AC = BC = 12×sin 70
Area of ΔABC = ½×AC×BC×sin 40 = ½×12×sin 70×12×sin 70×sin 40
Area of ΔABC ≈ 40.867 cm²
M a t h s s o l u t i o n s 4 7
From ΔPQR
Let a = PR = 55 cm, b = RQ = 16 cm, c = QP = 49 cm
Δ = area of triangle & 2s = perimeter of triangle
2s = 55+16+49 = 120 ⇒ s = 60 cm
Δ² = s(s-a)(s-b)(s-c) {Heron's formula}
Δ² = 60(60-55)(60-16)(60-49)
Δ² = 145200 ⇒ Δ = 220√3 cm²
also Δ = abc/(4R) = 220√3
55×16×49/(4R) = 220√3 ⇒ R = ⅓×49√3 cm
Radius of circle = ⅓×49√3 cm
A P
C
B
R
Q
35 cm
88
 cm
77 cm
10 cm
56 cm
14 
cm
16 cm
55 cm
49
 c
m
A
C
B
D
15°
θ
θ
18
0-
2θ
airesshahin@gmail.com 04 Jun 2021
∠ACB = 90° {ΔABC is right triangle}
∠RPC = ∠PCQ = ∠RQC = 90° {AC & BC are tangent of semicircle}
so, area of blue triangle = ½×R²
ΔABC & ΔARP are similar triangles
R/6 = (8-R)/8 ⇒ 8R = 6(8-R)
8R = 48-6R ⇒ R = 24/7 cm
Area of blue triangle = ½×(24/7)²
Area of blue triangle = 5.878 cm²
From ΔABC 
BC² =AB²+AC²-2×AB²×AC²×cos (∠BAC)
10² = 16²+14²-2×16×14×cos (∠BAC) ⇒ cos (∠BAC) = 11/14 
∠BAC ≈ 38.21°
∠CAP = ∠BAP = 38.21/2 = 19.105°
From ΔABQ
AQ = AC/2 = 14/2 = 7 cm
BQ² = AQ²+AB²-2×AQ×AB cos (∠BAC)
BQ² = 7²+16²-2×7×16×11/14 = 129 ⇒ BQ = √129 cm
AQ² = AB²+BQ²-2×AB²×BQ²×cos (∠ABQ)
7² = 16²+129-2×16×√129×cos (∠ABQ) ⇒ ∠ABQ ≈ 22.41°
∠AXB = 180 - ∠ABQ -∠BAP = 180-22.41-19.105 = 138.485°
Area of ΔABX = ½×AX×AB×sin (∠BAX)
AB/sin (∠AXB) = AX/sin (∠ABX) = BX/sin (∠BAX) {sine rule}
16/sin 138.485 = AX/sin 22.41 = BX/sin 19.105
AX = (16×sin 22.41)/sin 138.485 = 9.203 cm
Area of ΔABX = ½×9.203×16×sin (∠BAX) = ½×9.203×16×sin 19.105 = 24.098 cm²
Area of ΔABX = ½×AB×XY = ½×16×XY = 24.098
XY = 3.012 cm
M a t h s s o l u t i o n s 4 8
SOLUTION 034
SOLUTION 033
A B
C
P
Q
R
R
R
R
8-
R
R
6-R
A B
C
PQ
X
Y
7 c
m
7 c
m
16 cm
10 cm
airesshahin@gmail.com 04 Jun 2021
From figure
CD² = AC²-AD² = BC²-BD²
8²-AD² = 6²-(AB-AD)² = 6²-(10-AD)²
64-AD² = 36-100+20×AD-AD²
AD = 6.4 cm
BD = 10-6.4 = 3.6 cm
Area of ΔABC = ½×AC×BC = ½×8×6 = 24 cm²
Area of ΔABC = ½×AB×CD = ½×10×CD = 24
CD = 4.8 cm
From ΔABD
BD² = AB²+AD² = (2R)²+(2R)²
BD² = (2R)²+(2R)²=4R²+4R² = 8R² ⇒ BD = 2R√2 cm
From ΔAOP
(R+r)² = R²+(2R-r)²
R²+2Rr+r² = R²+4R²-4Rr+r² ⇒ r = ⅔R
From ΔPXD
∠PDX = 45° then ∠PXD = 45°
so, a² = R²+R² = 2R² ⇒ a = R√2 cm
From ΔOBY
∠OBY = 45° then ∠OYB = 45°
so, c² = r²+r² = 2r² ⇒ c = r√2 cm
c = ⅔R√2 cm
b = BD-a-c = 2R√2 - R√2 - ⅔R√2 = ⅓R√2 cm
a:b:c = R√2 : ⅓R√2 : ⅔R√2 = 3 : 1 : 2
a:b:c = 3 : 1 : 2
A B
X
Y
P
Q
b
2R-r
M a t h s s o l u t i o n s 4 9
SOLUTION 035
SOLUTION 036
CD
c
a
C
BA D
R
R
r r
r
R
RP
O
r
8 c
m
6 cm
6.4 cm 4.8 cm
airesshahin@gmail.com 04 Jun 2021
From ΔPQR
PQ² = PR²+QR² = (x-y)²+(x+y)²
PQ² = x²-2xy+y² + x²+2xy+y²
PQ² = 2x²+2y² ⇒ PQ = √2 √(x²+y²)
From ΔXYZ
XY² = XZ²+YZ² = x²+y² ⇒ XY = √(x²+y²)
PQ : XY = √2 √(x²+y²) : √(x²+y²) = √2 : 1
PQ : XY = √2 : 1
SOLUTION 037 P
x
x+y
R
From ΔACD
Area of ΔACD = (6.4/10)×24 = 15.36 cm²
Perimeter of ΔACD = 8+4.8+6.4 = 19.2 cm
Radius of circle = (2×15.36)/19.2 = 1.6 cm
From ΔBCD
Area of ΔBCD = 24-15.36 = 8.64 cm²
Perimeter of ΔBCD = 6+4.8+3.6 = 14.4 cm
Radius of circle = (2×8.64)/14.4 = 1.2 cm
Red area : Blue area = π(1.2)² : π(1.6)² = 1.44 : 2.56
Red area : Blue area = 9 : 16
Let Q is center of circle and r is the radius of circle
AB = AO+BO = 4+3 = 7 cm
AQ = AB/2 = ⁷/₂ cm
From figure
AO×BO = CO×DO ⇒ 4×3 = 2×DO ⇒ DO = 6 cm
CD = CO+DO = 2+6 = 8 cm 
RC = ⁸/₂ = 4 cm
PQ = RO = RC-CO = 4-2 = 2 cm
M a t h s s o l u t i o n s 5 0
Q
Y
X
y
y
x
x-
y
Z
SOLUTION 038
3 cm
2 cm
A
D
C
B
O
P
r
Q
R
⁷/₂ cm
airesshahin@gmail.com 04 Jun 2021
Blue area = Area of kite ADXG - Area of sector GAD
From figure
∠EAB = 120° & ∠BAD = 90°
∠EAD = ∠EAB-∠BAD = 120-90 = 30°
∠EAD = ∠BAG = 30° {symmetry}
∠DAG = ∠EAB-∠EAD-∠BAG = 120-30-30
∠DAG = 60°
∠DAX = ∠GAX = 30° {symmetry}
Area of kite ADXG = DX×AD = AD×tan 30 ×AD
Area of kite ADXG = 4×tan 30 ×4 = ⅓×16√3 cm²
Area of sector GAD = ⅙×π×4² = ⅓×8π cm²
Blue area = ⅓×16√3 - ⅓×8π cm²
Blue area = ⅓×8(2√3 - π) cm²
From ΔAPQ
AP² = AQ²+PQ²
r² = (⁷/₂)²+2² = ⁶⁵/₄  ⇒ r = ½√65 cm
Radius = ½√65 cm
From ΔDCQ
DC = 2 cm & CQ = 1 cm
sin (∠CDQ) = ½ ⇒ ∠CDQ = 30°
∠CBR = ∠CDQ = 30°
From ΔBCD
∠BCD = 108°
∠BCD = ∠BCD {BC = DC}
∠CBD+∠CDB = 180-108 = 72 ⇒ ∠CBD = ∠CDB = 36°
∠PBD = ∠PDB = 36+30 = 66° ⇒ ∠P = 180-66-66
∠P = 48°
M a t h s s o l u t i o n s 5 1
SOLUTION 039
SOLUTION 040
A
E
B
CD
G
F
X
4 cm 4
 c
m
4 cm
4 c
m
A
B
C
D
E
P
Q
1 
cm
R
airesshahin@gmail.com 04 Jun 2021
Area of blue triangle = ½×BC×BQ×sin (∠CBQ)
EB² = AE²+AB²-2×AE×AB×cos (∠A)
EB² = 10²+10²-2×10×10×cos 108 = 50√5 + 150
EB ≈ 16.180 cm
From ΔABE
∠AEB =∠ABE = ½(180-108) = 36°
∠BED = 108-∠AEB = 108-36 = 72°
From ΔDER
DR = DE×sin (∠BED) = 10×sin 72 ≈ 9.510 cm
DR = PQ = 9.510 cm 
Let radius of quarter circle = 2x, then
Radius of semicircle = 2x/2 = x
Area of semicircle BPO = a+c = ½πx²
Area of quarter circle = 2a+b+c = ¼π(2x)² = πx²
a+c = ½πx².........................eq(1)
2a+b+c = πx²......................eq(2)
From eq(1) & eq(2)
a+b+½πx² = πx² 
a+b = ½πx².........................eq(3)
From eq(1) & eq(3)
a+b = ½πx² = a+c
a+b = a+c ⇒ b = c
Red Area : Blue Area = c:b = 1:1
Red Area : Blue Area = 1:1
SOLUTION 042
M a t h s s o l u t i o n s 5 2
SOLUTION 041
O A
B
a
a
b
c
P
A
F
E
D
C
B
G
H
P
Q
R
10 cm
10 cm
airesshahin@gmail.com 04 Jun 2021
From ΔBPQ
∠PBQ = 45°
PB = PQ = 9.510 cm
BQ² = 9.510²+9.510² ⇒ BQ = 13.449 cm
∠BED = ∠EBC = 72°
∠CBQ = ∠EBC-∠PBQ = 72-45 = 27°
Area of Blue triangle = ½×10×13.449×sin 27 ≈ 30.53 cm²
Area of Blue triangle ≈ 30.53 cm²
From figure
AD = BC = 6 cm 
Let OY = x then, from ΔROZ
OZ² = (2+x)²-(2-x)² = 8x
OZ = XY = √(8x)
BC = BX+XY+YC
6 = 2 + √(8x) + x
√(8x) = 4-x
8x = 16-8x+x²
x²-16x+16 = 0 ⇒ x = 8±4√3
x is less than 6 so, x = 8-4√3 cm
Radius of smallest circle = 8-4√3 cm
P
R
A
2-x
M a t h s s o l u t i o n s 5 3
SOLUTION 043
6 
cm
Q
B
CD
Y
XZ
O x
x
2 
cm
2 
cm
x
SOLUTION 044
A
D
C
B
X
P
O
Let ∠CDX = x then ∠O = 2x {∠O is central angle}
∠P = x {Inscribed angle from ∠O}
∠CXD = 90° {Inscribed in a semicircle}
∠DCX = 90-x {From ΔCXD}
We know CX = DX {symmetry}
so ∠CDX = ∠DCX = x = 90-x ⇒ x = 45° or ∠P = 45°
airesshahin@gmail.com 04 Jun 2021
From figure ΔABC is right triangle 
 {6, 8 & 10 pythagorean triples}
so AB = diameter of circle = 10 cm
so LP = 5 cm {Radius of circle}
MB = MC = 3 cm {CB = 6 cm}
XM×MQ = MB×MC
(10-y)y = 3×3 
10y-y² = 9 ⇒ y = MQ = 1 & 9
MQ is less than 5 (radius) then MQ = 1 cm
NA = NC = 4 cm {AC = 8 cm}
YN×NR = NA×NC
(10-x)x = 4×4
10x-x² = 16 ⇒ x = NR = 2 & 8
RN is less than 5 (radius) then RN = 2 cm
LP + NR + MQ = 5 + 2 +1 = 8 cm
LP+NR+MQ = 8 cm
SOLUTION 045
From figure
sin (∠ABQ) = AX/AB = 3/9 = ⅓
sin (∠BAQ) = BY/AB = 6/9 = ⅔
From ΔABQ
∠AQB = 180-∠BAQ-∠ABQ = 180-sin⁻¹(⅔) - sin⁻¹(⅓)
sin ∠AQB = sin (180-sin⁻¹(⅔) - sin⁻¹(⅓)) = sin(sin⁻¹(⅔) + sin⁻¹(⅓))
sin ∠AQB = ⅑(√5 + 4√2)
AB/sin (∠AQB) = AQ/sin(∠ABQ) = BQ/sin (∠BAQ) {sine rule}
AQ = AB × sin(∠ABQ)/sin (∠AQB) = 9×⅓/⅑(√5 + 4√2) = 4√2-√5 cm
Area of ΔABQ = ½×AB×AQ×sin (∠BAQ) = ½×9×(4√2-√5)×⅔ = 12√2 - 3√5 cm²
M a t h s s o l u t i o n s 5 4
SOLUTION 046
A B
C
L
P
MN
Q
R
X
5 cm 5 cm
5 
cm
Y
5 cm5-x
x
y
5-y
5 c
m
3 c
m
3 c
m
4 cm
4 cm
A B
X
Y
P
Q
3 cm 6 cm
6 cm
3 
cm
airesshahin@gmail.com 04 Jun 2021
SOLUTION 048
M a t h s s o l u t i o n s 5 5
SOLUTION 047
From fig(2)
OI = QI = OJ = PJ = QH = PH = 4√3 / 2 = 2√3 cm
∠IOJ =∠HQI = ∠JPH = 60°
so ΔOJI, ΔIQH & ΔPJH are a equilateral triangle
That is, IJ = JH = HI = 2√3 cm
Area of ΔHIJ = (2√3)²×√3 / 4 = 3√3 cm²
Area of ΔHIJ = 3√3 cm²
Let radius of circle = 2x & AB = 2y then,
From ΔOXR
OX²+RX² = (2x)² = 4x²
2 OX = 4x² {OX = RX because OY ⟂ RS}
OX = x√2 
From ΔOCY
OY²+CY² = OC²
(x√2 + 2y)² + (y)² = (2x)²
2x² + 4xy√2 + 4y² + y² = 4x²
5y² + 4xy√2 + - 2x² = 0 ⇒ y = ⅕(-2x√2 ± 3x√2)
y is not become negative so y = ⅕x√2
From fig(1)
OH = cos 30 × OQ
6 = ½√3 ×OQ ⇒ OQ = 4√3 cm
OQ = OP {symmetry}
OQ = OP = PQ = 4√3 cm {OQ = OP & ∠POQ = 60°}
O
B
A
P
Q
H
I
J
6 cm30°
30°
O
J
I
H
P
A
B Q
60°
2√
3 c
m
2√3 cm
fig(1)
fig(2)
P Q
RS
A B
CD
2x
O
X
Y
2y
y
x√
2 
2x
airesshahin@gmail.com 04 Jun 2021
From ΔPQR
PQ²+RQ² = PR²
2 PQ² = (4x)²
PQ² = 16x²/2 = 8x²
PQ = 2x√2
AB : PQ = 2y : 2x = 2×⅕x√2 : 2x√2 = 1:5
AB:PQ = 1:5
M a t h s s o l u t i o n s 5 6
QUESTION 050
From ΔMQC
MC/XC = MQ/XB = CQ/BC
CQ² = MQ²-MC² = 6²-2² = 36-4 = 32
CQ = 4√2 cm
From figure ΔMQC & ΔBXC are similar 
 {BX || MQ}
MC/XC = MQ/XB = CQ/BC
MC/XC = CQ/BC
2/4 = 4√2 / BC ⇒ BC = 8√2 cm
BC = BA = 8√2 cm
sin (∠CBX) = sin (∠CQM) = 2/6 = ⅓
cos (∠CBX) = cos (∠CQM) = 4√2/6 = ⅔√2
∠ABC = 2×∠CBX 
sin (∠ABC) = sin (2×∠CBX) = 2 sin (∠CBX) cos (∠CBX)
sin (∠ABC) = 2×⅓×⅔√2 = 4√2 / 9
Area of ΔABC = ½×BA×BC×sin (∠ABC) = ½×8√2 × 8√2 × 4√2 / 9 = ⅑×256√2 cm²
Area of ΔABC = ⅑×256√2 cm²
A
C
BP
R
Q
X Y
M
2 
cm2 
cm
2 
cm
6 cm
6 cm
4√2 cm
airesshahin@gmail.com 04 Jun 2021
M a t h s s o l u t i o n s 5 7
SOLUTION 050
From ΔACQ
∠A+∠C+∠AQC = 180°..............eq(1)
From ΔBDP
∠B+∠D+∠BPD = 180°..............eq(2)
From ΔCET
∠C+∠E+∠CTE = 180°..............eq(3)
From ΔBER
∠B+∠E+∠BRE = 180°..............eq(4)
From ΔADS
∠A+∠D+∠ASD = 180°..............eq(5)
Add eq(1), eq(2), eq(3), eq(4) & eq(5) then
(∠A+∠C+∠AQC) + (∠B+∠D+∠BPD) +(∠C+∠E+∠CTE) + 
 (∠B+∠E+∠BRE) + (∠A+∠D+∠ASD) = 180+180+180+180+180 = 900
2∠A + 2∠B + 2∠C + 2∠D + 2∠E + (∠AQC + ∠BPD + ∠CTE + ∠BRE + ∠ASD) = 900
∠AQC + ∠BPD + ∠CTE + ∠BRE + ∠ASD = 540 {sum of interior 
 angles of the pentagon}
2(∠A + ∠B + ∠C + ∠D + ∠E) + 540 = 900
2(∠A + ∠B + ∠C + ∠D + ∠E) = 360
∠A + ∠B + ∠C + ∠D + ∠E = 180°
A
E
D
C
B
RQ
P
T
S
airesshahin@gmail.com 04 Jun 2021