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Joy of Mathematics Class 8 Solutions Chapter 15 Properties of Quadrilaterals
Welcome to NCTB Solution. Here with this post we are going to help 8th class students for the Solutions of Joy of Mathematics Class 8 Book, Chapter 15 Properties of Quadrilaterals. Here students can easily find step by step solutions of all the problems for Properties of Quadrilaterals. Exercise wise proper solutions for every problems. All the problem are solved with easily understandable methods so that all the students can understand easily. Here students will find solutions for Exercise 15.1 and 15.2
Properties of Quadrilaterals Exercise 15.2 Solution :
Question no – (1)
Solution :
(a) From the parallelogram
ABCD,
Length of the CD = 6 cm
(∵ AB || CD) and (AB = 6cm)
(b) From ABCD parallelogram
AD = 4 cm (AD || CB, and CB = 4 cm)
(c) <C = 60° (∵ <A = 60°, since opposite angle are equal)
(d) AB || CD
Now,
<A = 60°
<C = 60°
Let, <B = x°
Now,
we all know that sum of the angle of parallelogram is 360°
<A + <B + <C + <D = 360°
=> 60 + x + 60 + x = 360
=> 2X + 120 = 360
=> 2x = 360 – 120
=> 2x = 240
=> x = 120
∴ <D = 120°
Now,
<A + <D = <60 + <120°
= <180°
Question no – (2)
Solution :
Ratio of two adjacent angles are = 1 : 2
Let, the adjacent is x and 2x
We know,
adjacent angles of parallelogram is 180°
Then, x + 2x = 180
= 3x = 180
= x = 180/3
= x = 60°
Therefore the adjacent angles are
= 60°and 2 × 60
= 120°
Question no – (3)
Solution :
Here, one angle of parallelogram is = 110°
and the opposite angle is 110°
Let, the other two angles are x°
Now,
We know that sum of the angle of a parallelogram are 360°
∴ 110 + 110 + x + x= 360
=> 220 + 2x = 360
=> 2x = 360 – 220
=> 2x = 140
=> x = 70
Therefore the other angles are 70, 110° and 70°
Question no – (4)
Solution :
Here, sum of pair of opposite angels = 200
∴ measure of one angle
= 200/2
= 100°
Let, the other opposite two angles is x°
Now, x° + 100 + 100 + x° = 360°
= 2x° = 360 – 200
= 2x° = 160
= x = 160/2
= 80
= x = 80°
Therefore, the other two angles will be 100° and 80°
Question no – (5)
Solution :
Here , Ratio of two sides of a Parallelogram = 3 : 2
And,
let, the sides be = 3x and 2x,
∴ perimeter of a parallelogram
= 2(3x + 2X)
= 2 × 5x cm
= 10x cm
According to the question,
10x = 40
= x= 40/10
= x = 4
Therefore the length of the two sides are
= (3 × 4)
= 12cm
and (2 × 4)
= 8 cm
Question no – (6)
Solution :
let , the other side of the parallelogram is = x cm
∴ Perimeter of the parallelogram = 2 (x + 10) cm
According to the question,
2(x + 10) = 100
=> 2x + 20 = 100
=> 2x = 100 – 20
=> 2x = 80
=> x = 40
Therefore, the other side of the parallelogram is 40 cm.
Question no – (7)
Solution :
From the figure PQRS
∠p = ∠R (opposite angles of a parallelogram are same)
∴ y + z = 30 + t and ∠Q = ∠S
We know, ∠P + ∠Q + ∠R + ∠S = 360°
=> 30 + t + 110 + 30 + t + 110 = 360°
=> 2t + 280 = 360
=> 2t = 360 – 280
=> 2t = 80
=> t = 80/2
= 40°
∴ ∠R = 30 + t = 30 + 40 = 70
∴ ∠p = 70 = y + z
∴ y = 35°
z = 35°
and x = 110°
Therefore, each of the angle,
x = 110°
y = 35°
z = 35°
t = 40°
Question no – (8)
Solution :
In △BCA and △DAC
Given, ABCD is a parallelogram
AD ≅ BC (opposite side of parallelogram are equal)
AB ≅ DC (opposite side of parallelogram are equal)
AB ≅ DC (common side of both triangle)
So, △BCA ≅ △DAC
(a) Since, ABCD is a parallelogram then
DC = DA (opposite side of the parallelogram are same)
(b) ABCD is a parallelogram then
AB = CD (opposite side of the parallelogram are same)
(c) ∠B = ∠D
ABCD is a parallelogram so opposite angles is same
Question no – (9)
Solution :
In △ADF and △BCE
Given, ABCD is a parallelogram
AD || BC(opposite side of parallelogram)
DC || AB (opposite side of parallelogram)
And, ∠E = ∠F (midpoint)
So, △ADF ≅ △BCE
Then, AF = EC
And FC = DC (since, F and E are the middle point of DC and AB respectively)
So, AECF is a parallelogram. (Proved)
Question no – (10)
Solution :
In the Given figure,
∠C = 3x + 15
and, ∠A = 3x + 15
(Since ∠C and ∠A are opposite angle of parallelogram which is same)
And, ∠D = 2X + 10°
∠B = 2x + 10° {opposite single is same}
We all know that,
<A + <B + <C + <D = 360
=> 3x + 15 + 2x + 10 + 3x + 15 + 2x + 10° = 360
=> 10x + 50 = 360
=> 10x = 360 – 50
=> 10x = 310
=> x = 310/10
=> x = 31
Therefore the required value of x = 31°
(b) In figure,
∠A = 5y, ∠C = 5y
∠B = 3y + 20, ∠D = 3y + 20 (Opposite angle of the parallelogram is same)
Now,
∠A + ∠B + ∠C + ∠D = 360
=> 5y = 3y + 20 + 5y + 3y + 20 = 360
=> 16y + 40 = 360
=> 16y = 360 – 40
=> 16y = 320
=> y = 320/16
=> y = 20
∴ The required value of y = 20
Question no – (11)
Solution :
Given, PQRS is a rectangle of the rectangle are equal
Thus, PR = QS
=> 3y + 5 = 2
=> 3y = 2 – 5
=> 3y = – 3
=> y = – 3/3
= – 1
∴ Thus the required value of y = – 1
We know,
∠S = 90°
=> 30° + x° = 90°
=> x = (90 – 30)°
= 60°
and, ∠Q = 90°
=> 2 × Z = 90°
=> Z = 90/2
= 45°
∴ Thus the value of x and z are 60° and 45° respectively .
Question no – (12)
Solution :
ABCD is a square, and ABO is an equilateral triangle then the angle of the ABO triangle are 60°
∴ ∠X = 90 – 60
= 30°
(a) ∠BCO = 180 – (30 + 90)
= 180 – 120
= 60°
Question no – (13)
Solution :
Given, PQRS
∠P = 90/2
= 45°
∠Q = 45°
∴ Then ∠z = 180 – (45 + 45)
=> ∠z = 180 – (90)
=> z = 90°
and, ∠x = 90/2
= 45° (Since SR ⊥ RQ)
y = 180 – 45
= 135°
Therefore the angles of the square are x = 45°, y = 135°, z = 90°
Question no – (15)
Solution :
Since, ABCD is a rhombus and BD, CA are diagonal.
∠COB = 90°
∠OAB = 30°
∴ ∠OBA = 180 – (90 + 30)
= 180 – 120
= 60°
∴ ∠y = 2 × 60 = 120°
Now, from the triangle △BDC,
∠CDB = 180 – (∠BCD + ∠DBC)
= 180 – (60 + 60)
= 180 – 120
= 60
and, ∠BOC = 60°
∠COB = 90°
∠BCO = 180 – (90 + 60)
= 180 – 150
= 30°
∴ ∠x = 30 × 2
= 60°
∴ ∠z = 2 × 60
= 120°
∴ Thus the required angles are
∠x = 60°
∠y = 120°
∠z = 120°
Question no – (16)
Solution :
Since, ABCD is a rhombus,
(a) △AOB is a right angle triangle
(b) Here,
AO = 8cm
BO = 6cm
From the right angle triangle
AB² = AO2 + BO2
= (8)2 + (6)2
= 64 + 36
= 100
∴ AB = √100
= 10
Therefore, the length of the AB = 10 cm.
(c) Each side of the length of the rhombus is 10 cm
Question no – (17)
Solution :
Since,
TN || RE
TX ⊥ RE
Then, from △RTX we get,
(TR)2 = (TX)2 + (RX)2
=> 72 = 62 + (RX)2
=> (RX)2 = 72 – 62
=> RX = √40 – 36
= √13
Since, it is an isosceles trapezium
So, RX = YE and TN = XY
So, RE = RX + TN + YE
= √13 + 15 + √13
= 15 + 2√13
Question no – (18)
Solution :
Here, ABCD is a rectangle and ∠COD = 120
then, ∠AOB = 120 (Since ∠COD and ∠AOB are
Now,
Let, ∠ABO = x and ∠BAO = x (Since the diagonal of the rectangle is same)
From △AOB
∠OAB + ∠ABO + ∠AOB = 180°
=> x + x + 120 = 180
=> 2x = 180 – 120
=> 2x = 60
=> x = 60/x
= 30
(a) Thus ∠ABO = 30° and ∠OAD = (90 – 30) = 60
(b) ∠ADO = 60° (Since, ∠OAD = ∠ADO)
(c) ∠OBA = 30° and ∠OBC = 90 – 30 = 60°
∴ ∠BCO = 60 (Since ∠OCB = ∠BCO)
Question no – (19)
Solution :
Here,
ABCD is a rhombus AC and BD is a diagonal of the rhombus
OA = OC ….. (i)
OB = OD ….. (ii)
From (ii)
x + 3 = 5
=> x = 5 – 3
=> x = 2
From (i)
12 = x + y
=> 12 = 2 + y
=> y = 12 – 2
=> y = 10
Now, BD = (x + 3) + 5
= (2 + 3) + 5
= 5 + 5
= 10
AC = 12 + (y + x)
= 12 + 10 + 2
= 24
From the △AOB
AB2 = OB2 + OA2
= (5)2 + (12)2
= 25 + 144
= 169
=> AB = √169 = 13
=> z = 13
∴ Perimeter of the rhombus,
= 4 × AB
= 4 × 13
= 52 units
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