CHAPTER−3
UNIT-5
EXERCISE 3.5.2
1. Two angles of a quadrilateral are 70° and 130° and the other two angles are
equal. Find the measure of these two angles.
Ans: Let ∟A and ∟B be x, ∟C=70°
and ∟D = 130°
∟A + ∟B + ∟C + ∟D = 360° (Theorem 7)
x + x + 70° + 130° = 360°
2x = 360° − 200°
X=
160 °
2
= 80°
∴∟A = 80° and ∟B = 80°
2. In the figure suppose ∟P and ∟Q are supplementary angles and
∟R = 125°. Find the measures of ∟S.
S
125°
P
R
Q
Ans: ∟P + ∟Q = 180° (Supplementary angles)
∟P + ∟Q + ∟R + ∟S = 360° (Theorem 7)
180° + 125° + ∟S = 360°
∟S = 360°−(180° + 125°)
∟S = 360°−305°
∟S=55°
3. Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angle is
90°. Find the measures of the other three angles.
Ans: Let the angles be 2x, 3x, 5x.
∟A + ∟B + ∟C + ∟D = 360°
2x + 3x + 5x + 90° = 360°
10x = 360°−90°
x =
270 °
10
x = 27°
∴ The angles are , ∟A = 2x = 2×27 = 54°
∟B = 3x = 3×27 = 81
∟C = 5x = 5×27 = 135°
4.In the adjoining figure, ABCD is a quadrilateral such that ∟D + ∟C =100°.
The bisectors of ∟A and ∟B meet at ∟P. Determine ∟APB.
D
C
P
A
B
Ans: AP and BP are angular bisectors
∟D + ∟C = 100°
To find : ∟APB
Proof : ∟A + ∟B + ∟C + ∟D = 360° (Theorem 7)
∟A + ∟B + 100° = 360°
∟A + ∟B = 360° − 100°
Multipyling by
1
2
1
2
,
1
1
2
2
× ∟A + × ∟B =
× 260
a + b =130°
In Δ APB,
a + b + ∟P = 180°
∟P = 180° − 130°
∟P = 50°
EXERCISE 3.5.3
1. In a trapezium PQRS, PQ || RS, and ∟P = 70° and ∟Q = 80° . Calculate
the measure of ∟S and ∟R.
Ans :
P
Q
70°
80°
S
R
∟P + ∟S = 180° (Supplementary angles)
∟S = 180° − 70° = 110°
∟Q + ∟R = 180° (Supplementary angles)
∟R = 180° −80° = 100°
2. In a trapezium ABCD with AB || CD. It is given that AD is not parallel to
BC. Is ΔABC ≅ Δ ADC? Give reasons.
Ans:
A
B
D
C
In Δ ABC and Δ ADC,
(1) AC = AC (Common side)
(2) ∟BAC = ∟ACD (Alternate angles, AB || CD)
∴ Δ ABC ≅ Δ ADC
It cannot be proved with the help of any beam postulated.
3. In the figure, PQRS is an isosceles trapezium; ∟SRP = 30°, and ∟PQS=40°
Calculate the angles ∟RPQ and ∟RSQ.
Ans:
S
R
30°
P
40°
Q
Data:PQRS is an isosceles trapezium,
PS = RQ and ∟P = ∟Q
∟SRP = 30° and ∟PQS = 40°
To find : ∟RPQ and ∟RSQ
Proof: ∟RPQ = SRP = 30°
(Alternate angles, PQ || SR)
∟RSQ = ∟PRS = 40°
(Alternate angles, PQ || SR)
4. Proof that the base angles of an isosceles trapezium are qual.
Ans:
D
C
A
B
Data: ABCD is an isosceles trapezium.
To Prove: ∟A = ∟B
Construction: Join the diagonal BD and AC.
Proof : In Δ ACB and ΔBDA
(1) BC = AD
Isosceles)
(2)AC = BD
trapezium)
(3) AB = AB (Common side)
∴ ACB ≅ Δ BDA (SSS postualate)
∴ ∟A = ∟B (Conguency property)
5. Suppose in a quadrilateral ABCD, AC= BD and AD = BC. Prove that
ABCD is a trapezium.
Ans:
D
A
Data: ABCD is aquadrilateral,
AC = BD and AD = BC
To prove: ABCD is trapezium
Proof: In Δ ADB and Δ BCA
(1) AD = BC (Data)
(2) AC = BD (Data)
C
B
(3) AB = AB (Common side)
∴ Δ ADB ≅ Δ BCA (SSS postulate)
∴ ∟A = ∟B (Conguency property)
AC = BD (Data), AD = BC (Data)
∴ ABCD is an isosceles trapezium
(Alternate method to be given later)
EXERCISE 3.5.4
1.The adjacent angles of a parallelogram are in the ratio 2:1 Find the
measures of all the angles.
Ans: Let the angles be 2x and x
∟A + ∟B = 2x + x = 180° (Adjecent angleso of percellelograms
ae supplementary)
2x + x = 180°
3x = 180°
X=
180 °
3
= 60°
∴ ∟A = 2x = 60×2 = 120°
∴ ∟B = 60°
∴ ∟C = ∟A = 120°
(Opposite angles of parallelogram)
∟D = ∟B = 60°
(Opposite angles of parallelogram)
2. A field is in the form of a parallelogram, whose perimeter is 450 m and one
of its sides is larger than the other by 75m. Find the lengths of all sides
Ans:
K
L
60°
N
M
Perimeter = AD + DC + CB + BA
450
= x + x + 75 + x + x + 75
450
= 4x + 150
450 – 150 = 4x
300
𝑥
=x
X = 75
∴ Side = 75m(x)
∴ Opposite side = x + 75 = 75 + 75 = 150m
3. In the figure, ABCD is a parallelogram. The diaglonals AC and BD
intersect at O; and ∟DAC = 40°, ∟CAB = 35°, and ∟DOC = 10°.
∟ADC, ∟ACB, and ∟CBD.
D
C
110°
40°
35°
A
B
Ans: Data: ABCD is a parallelogram AD || BC and DC || AB.
The diagonals AC and BD intersect at O
∟DAC = 40°, ∟CAB = 35° and ∟DOC = 110°
To find : (1) ∟ABO (2)∟ADC (3)∟ACB (4) ∟CBD
Proof: ∟DAC + ∟CAB = ∟A
40° + 35 = ∟A
∟A = 75°
 ∟C = ∟A = 75°
(Opposite angles of parallelogram are equal)
 ∟D + ∟A = 180° (Supplementary angles)
 ∟D = 180°−75° = 105°
 ∟B = ∟D = 105° (Opposite angles of pallelgoram are equal)
 ∟DOC = ∟AOB = 110°
(Vertically opposite angles)
 In Δ AOB, ∟A + ∟O + ∟B = 180°
(Sum of all angles of a Δ is 180°)
35° + 110° + ∟B = 180°
∟B = 180°-145°
 (1) ∟ABO = 35°
 (2) ∟ADC = 105° (Proved)
 (3) ∟ACB = ∟CBD = 40°
(Alternate angles, AD || BC)
∟CBD = 105°-35°
(4) ∟CBD = 70°
(1)∟ABO = 35°
(2) ∟ADC = 105°
(3)∟ACB = 40° and (4) ∟CBD = 70°
4. In a parallelogram ABCD, the side DC is produced to E and ∟BCE = 105°.
Calculate and ∟A, ∟B, ∟C and ∟D.
Ans:
A
B
105°
75°
105°
D
75°
105°
C
E
∟BCD + ∟BCE = 180° (Linear pair)
∟BCD = 180° − 105° = 75°
In ABCD,
∟A = ∟C = 75° (Opposite angles of parallelogram)
∟ABC = ∟BCE = 105°
(Alternate angles, AB || DE)
∟D = ∟B = 105°
(Opposite angles of parallelogram)
5. Prove logically the diagonals of a parallelogram bisect each other. Show
conversely that a quadrilateral in which diagonals bisect each other is a
parallelogram.
A
B
Ans:
D
C
ln
Data: ABCD is a parallogram ( ║ )
To Prove : AO = OC and DO = OB
Proof: In ΔABO and DOC
(1)AB = DC (Opposite sides of angles)
(2) ∟AOB = ∟DOC (V.O.A)
(3) ∟ABD = ∟BDC (Alternate angles AB ||
∴ ΔABO ≅ ΔDOC (ASA portulate)
∴ AO = OC & DO = 3O (Conguency properties)
6. In a parallelogram KLMN, ∟K = 60°. Find the measure of all the angles.
Ans:
K
L
60°
N
M
∟M = ∟K = 180° (Supplementary angles)
∟N = 180°−60° = 120°
∟M = ∟K = 60°
Opposite angles of parallelogram
∟L = ∟N = 120°
7. Let ABCD be a quadrilateral in which ∟A = ∟C and ∟B = ∟D. Prove
that ABCD is a parallelogram.
Ans:
A
B
x
y
x
y
D
C
Data: ∟A = ∟C and ∟D = ∟B
To Prove: ABCD is a parallelogram
Proof: ∟A + ∟B + ∟C + ∟D = 360°
x + y + x + y = 360°
2x + 2y = 360°
2(x + y) = 360°
x+ y=
360 °
2
= 180°
Adjacent angles are supplementary
∴ ABCD is a parallelogram.
8. In a quadrilateral ABCD, suppose AB = CD and AD = BC. Prove that
ABCD is a parallelogram.
Ans:
D
C
A
B
Data: ABCD is a quadrilateral,
Ab = DC, AD = BC
To Prove : ABCD is a parallelogram
Construction: Join DB
Proof: In Δ ABD and Δ DBC
(1) DB = DB (Common side)
(2) AD = BC (Data)
(3) AB = DC (Data)
∴ ABD ≅ Δ DBC (SSS postulate)
∴ ∟DBA = ∟BDC (Congruency property)
But they are alternate angles
These we have AB ║ DC
∴ ∟ADB = ∟DBC (C.P)
But they are alternate angles
These we have DA ║ BC
∴ ABCD is a parallelogram.
EXERCISE 3.5.5
1. The sides of the rectangle are in the ratio 2:1. The perimeter is 30 cm.
Calculate the measure of all the sides.
Ans: ABCD is a rectangle
Let AB:BC = 2:1
∴ AB = 2x and BC = x
Perimeter of the rectangle = 2(1+b)
2(1+b) = 30
D
C
2(2x+x) = 30
2×3x = 30
6x
= 30
A
B
x
=
30
6
=5
2x = 2×x = 2×5 = 10
1x = 1×4 = 1×5 = 5
AB = 10m BC = 5cm
CD = 10cm and DA = 5cm
2. In the adjacent rectangle ABCD, ∟OCD = 30°. Caclulate ∟BOC. What
type of triangle is BOC?
ABCD is a rectangle
Diogonals AC and BD bisect each other
At O and AC = BD.
AO = OC = BO = OD
OCD = ODC = 30°
On ΔCOD, ODC + OCD + COD = 1
30° + 30° + COD = 180°
D
C
30°
A
B
60° + COD = 180°
∴ COD
= 180 – 60 =120°
COD + COB = 180
COB
= 180 – 120
COB
= 60°
∴ Δ BOC is an isosceles triangle
3. All rectangle are parallelograms, but all parallelograms are not rectangles.
Justify this statements.
All rectangles have all the properties of parallelograms but a parallelogram
May not have all the properties of a rectangle.
1. In a rectangle all the angles are right angle but in aparallelogram only
opposite angles are equal.
2. In a rectangle the diagonals are equal but in a parallelogram diagnolas are
not equal.
4. Prove logically that the diagonals of a rectangle are equal.
Data : ABCD is a rectangle.
AC and BD are the diagonals.
D
C
To Prove : AC = BD
Proof : In ΔABC and Δ ABD
ABC = BAD [90°]
BC = AD [Opp. Sides]
AB = AB [Common side]
∴ ΔABC ≅ ΔABD [SAS]
∴ AC = BD [CPCT]
5. The sides of a rectangualar park are in the ratio 4:3. If the area is 1728 m2 ,
find the cost of fencing it at the rate of Rs.2.50/m.
Ans: ABCD is a rectangualar park let AB:BC = 4:3
∴ AB = 4x and BC = 3x
D
C
Area of rectangle = 1×b
4x × 3x
= 1728
2
12x
= 1728
2
X
= 144
A
B
X
= 144 = 12
Length = 4x = 4×12 = 48m
Breadth = 3x = 3×12 = 36m
Perimeter of a rectangle = 2(1+b)
2(48 + 36) = 2×84 = 168m
Cost of fencing = perimeter ×Rate
= 168 ×2.50
= Rs.420
6. A rectangualar yard contains two flower beds in the shape of congruent
isosceles right triangle. The remaining portion is a yard of trapezoidal.
Shape (see fig) whose parallel sides have legths 15 m and 25 m. What
fraction of the yard is occupied by the flower bed?
ABCD is a rectangualar yard
AEFB is a trapezium
AB ║ EF
AB = 25 m and EF = 15m
AD = BC = 5m
Area of rectangle = 1×b = 25×5 = 125m2
Area of each flower bed =
=
1
2
×5×5 =
25
2
1
2
D
5m
E
F 5m
5m
5m
A
25m
×b×h
m2
Area of both flower beds = 2×
25
2
Fraction of flower beds to yard =
= 25m2
25
125
C
1
=5
7. In a rhombus ABCD ∟C = 70°. Find the other angle of the rhombus.
ABCD is a rhombus
LC = 70°
But ∟A = ∟C = 70 ∴ ∟A = 70°
B
∟A +∟B =180° (adjacent angles)
D
C
70°
70 + ∟B = 180°
∟B = 180 − 70 = 110°
∟B = ∟D = 110°
∴ ∟A = 70° ∟B = 110° ∟D = 110°
A
B
8. In a rhombus PQRS, ∟SQR = 40° and PQ = 3 cm. Find ∟SPQ, ∟QSR and
the perimeter of the rhombus.
PQRS is a rhombus
∟SQR = 40° and pQ = 3cm
P
S
∟PQS = ∟SQR = 40°
∟PQS = 40° (diagonal bisector angles ) 3 cm
But ∟PQS = ∟QSP [PQ = PS]
∴ ∟QSP = 40°
40°
In Δ PQS, ∟PQS + ∟QSP + ∟SPQ = 180° Q
R
40° + 40° + ∟SPQ = 180
∟SPQ = 180 −80
= 100°
∴ ∟SPQ = 100°
In a rhombus PQ = QR = RS = SP = 3cm
Perimeter of the Rhombus PQRS = 3×4 = 12cm
9. In a rhombus PQRS, if PQ = 3x -7 and QR = x + 3, find PS.
In a rhombus all sides are equal
PQ = QR = RS = SP
∴ PQ = QR
3x -7 = x + 3
3x -x = 3 + 7
2x = 10
X=5
PS = x+3 = 5+3 = 8 cm
10. Let ABCD be a rhombus and ∟ABC = 124°. Calculate ∟A, ∟D and ∟C.
In a rhombus, opposite angles are equal
∟B = ∟D = 124°
∟A + ∟B =180°
Consecutive angles
∟A + 124° = 180°
∟A + 124° = 180°
∟A = 180-124=56°
∴ ∟A = 56° ∟B = 124° and ∟C = 56°
11. Rhombus is a parallelogram: justify.
Rhombus has all the properties of parallelogram i.e
a) Opposite sides are equal and parallel.
b) Opposite angles are equal.
c) Diagonals bisect each other
∴ Rhombus is a parallelogram.
12. In a given square ABCD, if the area of triangle ABD is 36 cm2 , find
(i) the area of triangle BCD;
(ii) the area of the square ABCD.
ABCD is a square
Bd is the diagonal
The diagonal divides the square into two congruent triangles
∴ Δ ABD ≅ Δ BCD
D
C
∴ Area of Δ ABD =area of Δ BCD
Area of Δ ABD= 36 cm2 (given)
∴ Area of Δ BCD = 36 cm2
A
B
Area of the square ABCD = Area of Δ ABD + Area of ΔBCD
=36 cm2 +36 cm2
Area of the squre ABCD =72 cm2.
13. The side of a square ABCD is 5 cm and another square PQRS has
perimeter equal to 40cm. Find the ratio of the area ABCD to the area of
PQRS.
Perimeter of the square ABCD = 4 ×side
= 4×5 = 20
Ratio of
perimeter
of ABCD
Perimeter of PQRS
20
1
= 40 = 2 or 1:2
Area of ABCD = (side)2 = (5)2 = 25cm
Side of PQRS =
perimeter
4
=
40
4
= 10cm
Area of PQRS = (side)2 = (10)2 = 100cm2
Ratio of
Area of ABCD
Area of PQRS
25
1
= 100 = 4 or 1:4
14. A square field has side 20m. Find the length of the wire required to fence it
foure times.
Length of one side of the square = 20 Rs
Length of wire required to fence one round = 4×20
Length of wire required to fence four rounds = 80m
= 4×80m
= 320m
15. List out the differences between square and rhombus.
Square
1. All the angles are equal
2. Diagonals are equal
Rhombus
Opposite angles are equal
Diagonals are unequal
3. Area = side × side=(s)2
Area=
=
1
2
1
2
× Product of diagonals
×d1 ×d2
16. Four congruent rectangles are placed as shown in the figure. Area of the
outer square is 4 times that of the inner square. What is the ratio os length
to breadth of the congruent rectangles?
Let the length of rectangles be „a‟ and breadth be „b‟
Side of outer square –(a+b) units
Side of outer square (a-b) units
Area of outer square = 4 times area of inner square
Area of ABCD = 4(Area of PQRS)
(a + b)2
a+b
a+b
2a – 2b
2a-a
a
a
b
= 4(a-b)
= 2(a-b)
= 2a-2b
= a+b
= b + 2b
= 3b
=
1
3
Ratio : length: headth = 3:1
a
D
b
C
b
S
R
a
p
A
b
q
a
b
B
taking square of both sides
ADDITIONAL PROBLEMS ON “QUADRILATERALS”
I. Complete the following:
a. A Quadrilateral has ………….. sides.(Four 4)
b. A Quadrilateral has…………..diagonals.(2 two)
c. A Quadrilateral in which one pair of sides are paralled to each other
is……. (Trapezium)
d. In an isosceles trapezium, the base angles are ………….(equal)
e. In a Rhombus, diagonals bisect each other in ………angles.(Right)
2. Let ABCD be a parallelogram, What special name will you give it.
a. If AB = BC
A: ABCD will be Rhombus.
D
A
C
B
b.If ∟BAD = 90°
A: ABCD will be Rectangle.
c. If AB = AD and ∟BAD = 90°
A: ABCD will be a Square.
3. A quadrilateral has three acute angles each measuring 70°. What is the
measure of the fourth angle?
A: Sum of 4 angles of a Quadrilateral = 360°
Three of the four angle = 70°
∴ 70 + 70 + 70 + x = 360°
X = 360-210
X= 150°
4. The difference between the two adjacent angles of a parallelogram is 20°.
Find measures of all the angles of the parallelogram.
A: Let ABCD be a 11 grs
D
C
∟A + ∟B = 180° [adjacent of a ║gm is 180°]
and ∟A −∟B = 20° [given]
∴ ∟A = 180−∟B
A
B
∴ 180 − ∟B − ∟B = 20
180 – 20 = 2∟B
160 = 2 ∟B
∴ ∟B = 80°
If ∟B=80°
∟A= 100°
5. The angles of a Quadrilateral are in the ratio 1:2:3:4. Find all the angles of
the quadrilateral.
A. Let the angles of a Quadrilateral
= 1x, 2x, 3x, and 4x.
∴ 1x + 2x + 3x + 4x = 360°
[Sum of the 4 angles of a Quadrilateral is 360°]
10x = 360
X=
360
10
= 36°
∴ angles are 36°, 2×36, 3×36, 4×36
36°, 72°, 108°, 144°
6. Let PQRS be a parallelogram with PQ = 10cm, and QR = 6 cm . Calculate
the measures of the other two sides and the perimeter of PQRS.
A: PQ = SR
S
R
PS = QR [opposite sides of a 11 gm are equal]
∴ PQ = SR = 10cm
6cm
PS = QR = 6 cm
P
Q
10cm
Perimeter = PQ + QR + PS + SR
= 10 + 6 + 6 + 10 = 32 cm.
7. The Perimeter of a Square is 60 cm. Find its side length.
A: Perimeter of a Square = 4a = 60 cm.
a=
60
4
= 15 cm
Length of the side = 15 cm.
8. Let ABCD be a square and Let AC = BD = 10 cm. Let AC and BD intersect
at O. Find OC and OD
A: In a square, diagonals bisect each other.
D
C
∴ AO = OC
BO = OD
AC = BD = 10 cm [given]
A
B
AO = OC =
BO = OD =
1
2
1
2
× 10 = 5 cm
× 10 = 5 cm
9. Let PQRS be a rhombus with PR = 15 cm and QS = 8 cm. Find the area of
the Rhombus.
S
A.: Area of Rhombus =
=
1
2
×PR×QS =
1
2
1
2
d1d2
×15×84
P
R R
= 60 Sq cm.
Q
10. Let ABCD be a parallelogram and suppose the bisectr of ∟A and ∟B
meet at P.Prove that ∟APB = 90°.
A : In the ║gm ABCD.
∟DAB + ∟ABC = 180°
(1)
As AP and BP are angular bisectors.
∟DAP = ∟PAB and ∟ABP = ∟PBC
(2)
∴ In (1) ∟DAP + ∟PAB + ∟ABP + ∟PBC = 180°
From (2) ∟PAB + ∟PAB + ∟ABP + ∟ABP = 180°
2∟PAB + 2∟ABP = 180°
∟PAB + ∟ABP =
180
2
= 90°
C
D
∴ In Δ APB,
P
∟A + ∟P + ∟B = 180°
90°
∟B + ∟A = 90°
90° + ∟P = 180°
∟P =180−90 =90°
A
B
11. Let ABCD be a Square. Locate paints P,Q,R,S on the sides AB, BC, CD,
DA respectively such that CR = BP = CR= DS. Prove that PQRS is a
square.
A:
D
R
S
C
Q
A
P
B
Given:- ABCD is a Square and
AP = BQ = CR = DS = x
T.P.T:- PQRS is a Square or
PQ = QR = RS = PS & ∟P = ∟Q = ∟R = ∟S = 90°
Proof :- ABCD is a Square
∴ AB = BC = CD = AD
AP + BP = BQ + QC = CR + DR = SD + AS
x + BP = x + QC = x + DR = x + AS
⇒ BP = QC = DR = AS
Now In Δs ASP and Δ BPQ
AP = BP
AS = BQ
∟A = ∟B= 90°
∴ Δ ASP ≡ Δ BPQ (SAS Postulate)
∴ PS = PQ
IIIly From ΔDSR, DRCO and ΔBPQ, QR= RS = SP = PQ.
From Δ PQS and Δ RSQ, By SSS Postulate
Δ RQS ≡ Δ PSQ.
∴ ∟R = ∟P IIIly. ∟S = ∟Q
∴ ∟P = ∟R = ∟S =∟Q = 90°.
12. Let ABCD be a rextangle and let P,Q,R,S be the midpoint of AB, BC,CD,
AD respectively. Prove PQRS is a Rhombus.
A:
D
R
S
A
C
Q
P
B
Given :- ABCD is a Rectangle.
AP = BP, BQ = QC, QR= RC, AS = DS
T.P.T:- PQRS is a Rhombus.
Solution:- In ΔASP and Δ BPQ
AP = BP
AS = BQ
∟A = ∟B = 90°
∴ ΔASP ≡ ΔBPQ
⇒ PS = PQ
IIIly From ΔDRS and ΔRCQ, ΔASP and ΔPBQ
PQ = RQ = RS = PS
∴ PQRS is a Rhombus.
13. Let ABCD be a Quadrilateral in which diagonals intersect at O
Perpendicularly.
Prove that AB + BC + CD + DA> AC + BD.
A: Consider the Δs
C
AOB, BOC, COD, DOA
All are right ∟led Δs
D
B
∴ AB, BC,CD and AD are the greatest sides.
i.e AB> AO and OB
BC > BO , OC
A
CD > CO and OD
DA > OD and OA
∴ AB + BC + CD + DA > AO + OB + BO + OC + OC + OD + OA
⇒ AB + BC + CD + DA > AB + CD
14. Let ABCD be a Quadrilateral with diagonals AC and BD. Prove that
(a) AB + BC + CD > AD
D
A: Let us join AC
C
In Δ ABC
AB + BC > AC
In Δ ADC
A
AD + DC > AC
∴ AB + BC + AD + DC > AC
⇒ AB + BC + CD > AD
B
15. Let ABCD be a quadrilateral with diagonals AC and BD. Prove the
following statements.
a) AB + BC + CD > AD
b) AB + BC + CD + AD > 2AC
c) AB + BC + CD + AD >2BD
d) AB + BC + CD + DA > AC + BD
D
C
A
B
A: a) Consder the Δ ADC,
AC + CD > AD …………..(1)
(Sum of two sides of a Δ is greater than III side)
∴ In Δ ABD, AB + BD > AD ………..(2)
In Δ BDC, BC +DC > BD
∴ BC > BD
∴ AB + BC > AD
⇒ AB + BC + CD > AD [From (1), (2) and (3)]
b) In ΔADC , AD + DC > AC………..(1)
In Δ ABC, AB + BC > AC………….(2)
∴ AB + BC + AD + DC > 2AC
c) In Δ ABD , AB + AD > BD……….(1)
In Δ BDC, BC + CD > BD………..(2)
(1) + (2). AB + BC + AD + CD >2 BD
d) In ΔABD, AB + AD > BD
In ΔABC, AB + BC > AC
⇒ AB + BC + CD > AD > AC + BD
In Δ BDC, CD + BD > BD
In ΔADC, AD + CD > AC
b) AB + BC + CD + AD > 2AC
Similar to Previous solution
AB + BC > AC
D
AB + DC > AC
C
∴ AB + BC + AD + DC > 2AC
c) AB + BC + CD + DA > 2 BD
A
In ΔABD, AB + AD > BD
In Δ BDC, BC + CD > BD
B
∴ AB + AD + BC + CD > 2BD
d) AB + BC + CD + AD > AC + BD
Solution of problems (13)
16. Let PQRS be a Kite such that PQ > PS Prove that ∟PQR > ∟PSR.
Solution:
p
Q
S
R
In PSQ ∟PSQ > ∟PQS [angle opposits greater side is greater than the
lesser side]
∟PSQ + ∟QSR > ∟PQS + ∟SQR
⇒ ∟PSR > ∟PQR
17. Let ABCD be a quadrilateral in which AB is the Smallest Side and CD is
the largest side. Prove that ∟A > ∟C and ∟B > ∟D.
B
C
A
In ΔAbc, AB is Smallest side
∴ ∟C is smallest ∟le
∴ ∟B > ∟A and ∟C, ∟B > ∟C
A + B + C = 180 or ∟A > ∟C
In Δ BCD and ABD
AB is smallest side
∴ ∟D is smallest ∟le
CD is largest side
∴ ∟B is largest angle
⇒ ∟B > ∟D
D
18. In a Δ ABC, let D be the mid point of BC. Prove that AB + AC >2AD.
A
B
D
C
In Δ ABD, AB + BD > AD
In Δ ADC, AC + CD > AD
∴ AB + AC + BC > 2 AD
⇒ AB + AC > 2 AD
Properly Δ : Sum of any two sides of Δ is greater than the third side.
19. Let ABCD be a quadrilateral and let P,Q,R and s be the midpoint of AB,
BC , CD and AD respectively. Prove that PQRS is a ║ gm.
C
R
D
S
Q
A
P
B
In Δ ADC,
S and R → Midpoints
∴ SR ║ AC and SR =
1
2
AC
In Δ ABC, P and Q → Midpoints
PQ ║ AC, PQ =
1
2
AC
⇒ SR = PQ and SR ║ PQ
⇒ SRQP is a 11gm
[Midpoint Theorem]