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Maths Wiz Class 8 Solutions Chapter 10 Quadrilaterals
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Wiz Class 8 Math Book, Chapter 10, Quadrilaterals. Here students can easily find step by step solutions of all the problems for Quadrilaterals, Exercise 10A, 10B 10C, 10D and 10E Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 10 solutions. Here in this post all the solutions are based on ICSE latest Syllabus.
Quadrilaterals Exercise 10A Solution :
Question no – (1)
Solution :
Figure – (a)
The angle sum of a quadrilateral is 360°
∴ 125° + 105° + 75° + x° = 360°
= 305 + x° = 360°
= x° = 360° – 305°
= 55°
Figure – (b)
The angle sum of a quadrilateral is 360°
∴ 100° + 70° + 80° + y° = 360°
= 250° + y° = 360°
= y° = 360° – 250°
= 110°
Figure – (c)
The angle of sum of a quadrilateral is 360°
∴ 90° + 140° + a° + 30° = 360°
= 260° + a° = 360°
= a° = 360° – 260°
= 100°
Figure – (d)
The angle sum of a quadrilateral is 360°
∴ 20° + x° + 10° + 40° = 360°
= 70° + x° = 360°
= x° = 360° – 70°
= 290°
Figure – (e)
The angle sum of a quadrilateral is 360°
∴ P° + 70° + 90° + 90° = 360°
= P° + 250° = 360°
= P° = 360° – 250°
= 110°
Figure – (f)
The angle sum of a quadrilateral is 360°
∴ 138° + 90° + x° + 2x° = 360°
= 3x + 228° = 360°
= 3x = 360° – 228°
= x = 132/3
= x = 44
∴ x° = 44
2x° = 2 × 44 = 88°
Figure – (g)
The angle sum of a quadrilateral is 360°
2m + m + m + 2m = 360°
= 6m = 360°
= m = 60°
∴ 2m = 2 × 60° = 120°
m = 60°
m = 60°
2m = 2 × 60° = 120°
Figure – (h)
The angle sum of a quadrilateral is 360°
65° + 240° + 30° + y° = 360°
= 335° + y° = 360°
= y° = 360° – 335°
= 25°
Question no – (2)
Solution :
Figure – (a)
= ∠C = 180° – 60° = 120°
∴ ∠A = 360° – {110°+ 120° + 50}
= 360° – 280°
= 80°
Figure – (b)
∠E = 180° – 35° = 145°
∠D = 180° – 145° = 35°
∠C = x° = 360° – {∠A + ∠B + ∠D}
= 360° – (90° + 136° + 35°)
= 360° – 261°
= 99°
Figure – (c)
3x + 4x + x + 2x = 360°
= 10x = 360°
= x = 36°
∴ x = 36°
2 × x = 2 × 36° = 72°
3x = 3 × 36° = 108°
4x = 4 × 36° = 144°
Figure – (d)
∠D = 180° – 100° = 80°
∠C = 360° – {∠A + ∠b + ∠D}
= 360° – {90° + 70° + 80°}
= 360° – 240°
= 120°
Question no – (3)
Solution :
Let the measures of the angles be
2x, 4x, 6x, 8x
∴ 2x + 4x + 6x + 8x = 360°
= 20x = 360°
= x = 18°
∴ measures of the angles be,
36°, 72°, 108°, 144°
Question no – (4)
Solution :
Sum of the angles of a quadrilateral 360°
x + 2x + 13° + 3x + 10° + x – 6° = 360°
= 7x° + 17° = 360°
= 7x = 360° – 17
= x = 343/7
= x = 49°
Therefore, the value of x will be 49°
Quadrilaterals Exercise 10B Solution :
Question no – (1)
Solution :
(a) A rectangle is a parallelogram.
= always
(b) A rhombus is a square.
= sometimes
(c) A parallelogram is a rhombus.
= sometimes
(d) A rhombus is a rectangle.
= sometimes
(e) A square is a rhombus.
= always
(f) A rectangle is a quadrilateral.
= always
(g) A square is a rectangle.
= always
(h) A rectangle is a square.
= sometimes
(i) A rectangle is a rhombus.
= Never
Question no – (2)
Solution :
A rhombus is equilateral but not equiangular. A rectangle is equiangular but not equilateral.
Quadrilaterals Exercise 10C Solution :
Question no – (1)
Solution :
Figure – (a)
x = 50° since opposite cones are equal
a = 8, b = 5 opposite sides are equal to
Figure – (b)
y = 3, x = 2
Figure – (c)
c = 95°
so, a + b = 360° – (95° + 95°)
= 360° – 190°
= 170°
so, a = 170/2 = 85
b = 170/2 = 85°
Figure – (d)
We knew x ≠ y
so, x + 2x + 2x + x = 360°
= 6x = 360°
= x = 60°
y = 60°
Figure – (f)
x = 6 opposite sides are equal
and 5y = 45°
= y = 15°
Question no – (2)
Solution :
∠AOD + ∠DOC = 230° = 360°
since AO = OD = OC
So, base cone are equal
∠AOC = ∠AOD + ∠DOC
∠AOC = 360 + 230
∵ = 130°
a + A = 130°
= a = 65°
∠DAO + ∠AOD + ∠ADO = 180°
and ∠DOC + ∠DCO + ∠CDO = 180°
∠ADO = ∠DC = 180-65/2 = 575 each one is equal.
and opposite ∠ABC and ∠ADC are equal
So, ∠ADC = ∠ADO + ∠DOC
= 2 × 57.5
= 115°
= x = 115°
Therefore, the value of x will be 115°
Question no – (3)
Solution :
As we know, Same adjacent angle are 180°
∠A + ∠B = 180°
one the cones are 4x and 5x
So, 4x + 5x = 180°
x = 20°
∠A = 4 × 20 = 80°
∠B = 5 × 20 = 100°
similarly opposite cones
∠B + ∠C = 180°
= ∠C = 180 – 100
∠C = 80°
and ∠D = 180° – 80° = 100°
Therefore, the measures of the angles are ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠D 100°
Question no – (4)
Solution :
5x – 21 + x + 75 = 180°
= 6x = 126
= x = 21
one cone ∠A = 5x – 21 = 105 – 21 = 84
∠B = x + 75 = 96
so, ∠C = 180° – 96° = 84°
and ∠D = 180° – 84 = 96°
Question no – (5)
Solution :
Side are 7x and 9x
We know perimeter
= (7x + 9x) = 96
= x = 96/2 × 16
= x = 3
side are = 7 × 3 = 21 cm
and = 9 × 3 = 27 cm
Therefore, the sides of the parallelogram are 21 cm and 27 cm.
Question no – (6)
Solution :
Figure – (a)
It is a Parallelogram
Figure – (b)
It is a Parallelogram
Reason : opposite sides are equal.
Figure – (c)
It is not a Parallelogram
Reason : opposite angles are equal.
Figure – (d)
It is not a parallelogram
Reason : the angle ∠ACB and ∠DAC are not equal.
Figure – (e)
It is not a parallelogram
Figure – (f)
It is not a parallelogram
Reason : opposite angles are not equal.
Question no – (7)
Solution :
Figure – (a)
8b = 10b – 3
= 10b – 8b = 3
= b = 1.5
Now, 5a – 15 = 3a – 5
= 5x – 3a = – 5 + 13
= 2a = 8
= a = 4
Figure – (b)
5b + 6 = 8a – 10
= 8a – 5b =16
= 8a + 16 = 5b
Again, 5b + 6 + 8a – 10 + 2 (4a – 8) = 360
= 5b + 6 + 8a – 10 + 8a – 16 = 360
= 8a + 16 + 6 + 8a – 10 + 8a – 16 = 360
= 24a = 360 + 16 + 10 – 16 – 6
a = 16.5, b = 23.2
Figure – (c)
5b – 7= 3b + 6
= 5b – 3b = 6 + 7
= 2b = 13
= b = 6.5
Now, 3b – 5 =2a
= 3 × 6.5 – 5 = 2a
= 19.5 – 5 = 2a
= 7.25 = a
Quadrilaterals Exercise 10D Solution :
Question no – (2)
Solution :
PQ = 20
Since, = SR = PQ
SK = 35/2 175 cm
Since, SK = SQ/2 = PR/2
Question no – (3)
Solution :
Figure – (a)
x = 7, y = 4
a = 90°,
b = 90°
c = 90°
d = 90°
Figure – (b)
∠x = 55°
∠y = (90° – 55) = 35°
Figure – (c)
15 = 5y
= y = 3
Now, 5x = 90°
x = 18°
Question no – (5)
Solution :
4a + 5 = 15a – 6
= 4a – 15a = – 6 – 5
= 11a = 11
= a = 1
So, sides are = 11 cm
So, BC = 11
Question no – (6)
Solution :
Figure – (a)
∠3 = 28°
∠2 = 28°
So, v5 = 28°
∠1 = 180 – (28 + 28)
= 124°
∠4 = 124°
Figure – (b)
∠4 = 50°
∠1 = 180 – 50/2
= 65
∠2 = 65°
∠3 = 65°
∠1 = 65°
∠5 = 65°
Previous Chapter Solution :
👉 Chapter 1 👈
