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Maths Ace Class 8 Solutions Chapter 11 Quadrilaterals and its Basics
Welcome to NCTB Solutions. Here with this post we are going to help 8th class students for the Solutions of Maths Ace Prime Class 8 Math Book, Chapter 11, Quadrilaterals and its Basics. Here students can easily find step by step solutions of all the problems for Quadrilaterals and its Basics, Exercise 11.1, 11.2, 11.3 and 11.4 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 11 solutions.
Quadrilaterals and its Basics Exercise 11.1 Solution :
Question no – (1)
Solution :
Figure – (a)
= Let, ∠1 = x
8 ∠1 + ∠2 = 180°
= ∠2 = (180° – x)
Now, ∠3 = 90° and,
∠3 + ∠4 = 180°
∠4 = 180° – ∠3
= 180° – 90°
= 90°
We know sum of interior angles of pentagon= 540°
Now, 180°- x + 90°+ 2x – 10 + 85° + 120° = 540°
= 270° – 10 + 205 + x = 540°
= x = 540° – 465°
= 75°
∴ (2x – 10°) = (2 × 75) – 10
= 150 – 10
= 140°
Figure – (b)
We know, sum of all exterior angles = 3600
∴ (4x – 5 + 4x + 20 + 5x – 5 + x) = 360°
= 14x – 5 + 20 – 5 = 360°
= 14x + 10 = 360°
= 14x = 36° – 10
= 350/14
= 25°
∴ (4x – 5) = (4 × 25) – 5 = 95°
(4x + 20) = (4 × 25) +20 = 120°
and, 5x – 5
= (5 × 25) – 5
= (125 – 5°)
= 120°
Figure – (c)
From given figure,
45 + 25° + 20° + x = 360°
= x = 360° – 90°
= 270°
Figure – (d)
Here, 84 + 96° + 2x + x = 360°
= 3x = 360° – 180°
= 180°
= x = 180/3 = 60°
Now, 2x = (2 × 60°)
= 120°
Question no – (2)
Solution :
(a) We know, 10 right angles,
= (10 × 90°)
= 900°
∴ (x – 2) 180 = 900
= 180x – 360 = 900
= 180x = 900 + 360
= x = 1260/180
= 7 sides
Therefore, the number of sides of the polygon is 7.
(b) Sum of interior angles = 720°
∴ (x – 2) × 180 = 720
= x – 2 = 720/180
= 4
= x = 4 + 2
= 6 sides
Hence, the number of sides of a polygon is 6.
(c) Sum of interior angles = 1620°
∴ (x – 2) 180 = 1620°
= x – 2 = 1620°/180°
= 9
= x = 9 + 2
= 11 sides
Therefore, the number of sides of the polygon is 11.
(d) Given, Sum of interior angles = 540°
∴ (x – 2) × 180 = 540
= x – 2 = 540/180
= 3
= x = 3 + 2
= 5 sides
Therefore, the number of sides of the polygon is 5.
Question no – (3)
Solution :
(a) Nonagon,
= Sum of interior angles in nonagon is,
= 180° × 7
= 1260°
Measure of each angle,
= 1260/9
= 140°
(b) Decagon,
Sum of interior angles of decagon,
= 180° × 8
= 1440°
Measure of each angle,
= 1440/10
= 144°
(c) 24-gon,
= Sum of interior angle in 24 sided polygon,
= 180 × 22
= 3960°
Measure of each angle,
= 3960/24
= 165°
(d) Hexagon,
= Sum of interior angle of Hexagon,
= 180 × 4
= 720°
Measure of each angle,
= 720/6
= 120°
Question no – (4)
Solution :
(a) We know, sum of exterior angle is = 360°
∴ 12 sides
= 360/12
= 30°
Thus, the measure of each exterior angle will be 30°
(b) We know, Sum of exterior angle is = 360°
∴ For 30 sides,
= 360/30
= 12°
Hence, the measure of each exterior angle will be 12°
(c) We know, Sum of exterior angle is = 360°
∴ For 9 sides
= 360°/9
= 40°
Hence, the measure of each exterior angle will be 40°
(d) We know, Sum of exterior angle is = 360°
∴ For 18 side
= 360°/18
= 20°
Therefore, the measure of each exterior angle will be 20°
Question no – (5)
Solution :
(a) Each interior angle is = 144°
Let, side of polygon = ‘x’
∴ (x – 2) × 180/x = 144°
= 180x – 360 = 144x
= 180x – 144x = 360°
= 36x = 360°
= x = 360/36
= 10
Therefore, the number of sides are 10.
(b) Each interior angle is = 160°
∴ (x – 2) × 180/x = 160
= 180x – 360 = 160x
= 180x – 160x = 360°
= 20x = 360°
= x = 360/20
= 18 sides
Therefore, the number of sides are 18.
(c) Each interior angle is = 156°
= (x – 2) × 180/x = 156
= 180x – 360 = 156x
= 180x – 156x = 360°
= 24x = 360°
= x = 360/24
= 15 sides
Hence, the number of sides are 15.
(d) Each interior angle is = 120°
∴ (x – 2) × 180/x = 120°
= 180x – 360° = 120x
= 180x – 120x = 360°
= 60x = 360°
= x = 360/60
= 6 sides
Therefore, the number of sides are 6.
Question no – (6)
Solution :
It is not possible to have a regular polygon with an exterior angle of measure 250 because, the formula 3600/x 360 is not divisible by 25.
Question no – (7)
Solution :
As per the question,
Polygon having 7 side
∴ Sum of interior angles,
= (x – 2) × 180°
= (7 – 2) × 180°
= 5 × 180°
= 900°
Therefore, the sum of the interior angle will be 900°
Quadrilaterals and its Basics Exercise 11.2 Solution :
Question no – (1)
Solution :
(a) 45° + 90° + 45° + 90°
= 270 It the sum should be 360°, then it is correct.
Hence, 270° ≠ 360°, its incorrect
(b) 90° + 75° + 120° + 75°
= 360°, therefore its correct
(c) 118° + 120° + 60° + 92°
= 390° ≠ 360°
Thus, its incorrect
(d) 110° + 60° + 120° + 70°
= 360°
Hence, its correct.
Question no – (2)
Solution :
(a) Given ratio = 1 : 3 : 5 : 6
Let, the angles, x, 3x, 5x, 6x
∴ x + 3x + 5x + 6x = 360°
= 15x = 360°
= x = 360°/15
= 24°
Measure of each angle,
∴ x = 24°
∴ 3x = 3 × 24° = 72°,
∴ 5x = 5 × 24° = 120°,
∴ 6x = 6 × 24 = 144°
(b) Given ratio, 7 : 8 : 9 : 12
Let, angles are 7x, 8x, 9x, 12x
∴ 7x + 8x + 9x + 12x = 360°
= 36x = 360°
= x = 360/36
= 10°
Now, measure of each angle.
∴ 7x = (7 × 10) = 70°
∴ 8x = (8 × 10) = 80°
∴ 9x = (9 × 10) = 90°
∴ 12x = (12 × 10) = 120°
(c) Given ratio, 3 : 4 : 5 : 6
Let, angles 3x, 4x, 5x, and 6x
∴ 3x + 4x + 5x + 6x = 360°
= 18x = 360°
= x = 360/18
= 20°
Now, measure of each angle.
∴ 3x = (3 × 20) = 60°
∴ 4x = (4 × 20) = 80°
∴ 5x = (5 × 20) = 100°
∴ 6x = (6 × 20) = 120°
(d) Given ratio, 4 : 5 : 5 : 6
= Let, angles 4x, 5x, 5x, 6x
∴ 4x + 5x + 6x + 6x = 360°
= 20x = 360°
= x = 360/20
= 18°
Now, the measure of each angle,
∴ 4x = (4 × 18) = 72°
∴ 5x = (5 × 18) = 90°
∴ 5x = (5 × 18) = 90°
∴ 6x = (6 × 18) = 108°
Question no – (3)
Solution :
Let, x be the equal angle,
∴ x + x + 280 = 360°
= 2x = 360° – 280°
= 80°
= x = 80/2
= 40°
Therefore, equal angles are 40°, 40°
Question no – (4)
Solution :
According to the given question,
Three angles of a quadrilateral are 75°, 110° and 60°
Let, the fourth angle is x
∴ 75° + 110° + 60° + x = 360°
= x + 245° = 360°
= x = 360° – 245°
= 115°
Thus, the measure of the fourth angle 115°
Question no – (5)
Solution :
Four angles of a quadrilateral are equal.
Let, the equal angle be x
∴ x + x + x + x + = 360°
= 4x = 360°
= x = 360/4
= x = 90°
Therefore, the measure of each angle will be 90°.
Quadrilaterals and its Basics Exercise 11.3 Solution :
Question no – (2)
Solution :
We know, In parallelogram, opposite angles are equal
∴ Angle ∠B = ∠D = 120°
Sum of interior angles is 360°
∴ 120 + A + 120 + C = 360°
= A + C = 360 – 240°
= 120°
∵ ∠A = ∠C
∴ 2C = 120°
= C = 120/2 = 60°
∴ ∠C = 60°,
∴ ∠D = 120°
Therefore, the measure of ∠C is 60°, and ∠D is 120°.
Question no – (3)
Solution :
As we know, In a parallelogram opposite side are equal, therefore, other two side is same 18 cm and 25 cm.
Question no – (4)
Solution :
From figure,
PR = 20 cm
QS = 28 cm
Now,
PO = 10 cm,
QO = 14 cm
Question no – (5)
Solution :
Let, the angle be 4x and 5x
4x + 5x = 180°
= 9x = 180°
= x = 180/9
= 20°
Now, ∠1 = 4 × 20° = 80°
∴ ∠2 = 5 × 20 = 100°
∴ ∠1 = ∠3 = 80°
∴ ∠2 = ∠4 = 100°
[∵ Opposite angles of parallelogram are equal]
Question no – (6)
Solution :
Let, the side A and B
∴ A = 5B
Now, 2(A + B) = 240
= 2(5B + B) = 240
= 2 × 6B = 240
= 12B = 240
= B = 240/12 = 20
Now, A = (5 × 20) = 100
Therefore, the length of the parallelogram will be 100 cm, 20 cm, 100 cm, and 20 cm.
Question no – (7)
Solution :
Given, Sum of two opposite angle of parallelogram is 300°
We know, Sum of all angles is = 360°
∴ (360° – 300°)
= 60°
∴ Other opposite angles total,
= 60°
∴ One angle value,
= 60/2
= 30°
Therefore, other two angles are 30°, 60°
Question no – (8)
Solution :
As We know, opposite angles are equal,
∴ (3x + 4) = (5x – 2)
= 3x – 5x = -2 -4
= -2x = -6
= x = 3
∴ 1st angle
= 3 × 3 + 4
= 13°
∴ 2nd angle
= 5 × 3 – 2
= 13°
Now, sum of adjacent angle is 180°
Let, the adjacent angle be ‘y’
∴ y + 13 = 180°
= y = 180 – 13
= 167°
∴ ∠2 = ∠4 = 167°
Question no – (9)
Solution :
From the given figure,
ABCD is a parallelogram
BE = EC
Prone, DF = 2DC
Now, AB = CD (opposite side of || gm are equal)
AB 11 CD
So, AB 2|| DF (∵ AB || DC)
∠ABE = ∠ECF (Alternate angles, AB || DF)
In △ ABE and △ ECF,
BE = EC (given)
∠AEB = ∠CEF
∠ABE = ∠ECF [ AB || DF]
∴ △ ABE ≅ △ ECF (A – S – A)
∴ AB = CF ………… (1)
AB = CD (opposite side of 11 gm are equal) …………(2)
∴ CF = CD (From (1) and (2)
∴ DF = DC = CF
∴ DF = DC + DC
= 2DC…(Proved)
Quadrilaterals and its Basics Exercise 11.4 Solution :
Question no – (1)
Solution :
Given, Rectangle breadth and length are 8 cm and 15 cm
Length of the diagonals.
= √82 + 152
= √64 + 225
= √289
= 17 cm
Thus, the length of the diagonal of a rectangle 17 cm.
Question no – (2)
Solution :
From question,
2x + 6 = 3x + 4
= 2x – 3x = 4 – 6
= -x = -2
= x = 2
∴ Diagonals,
= 2 × 2 + 6 = 10
= 3 × 2 + 4 = 10
Question no – (3)
Solution :
Given,
1st diagonals AC = 24 cm
BD = 10 cm
∴ AO = 1/2 × 24 = 12
BO = 1/2 × 10 = 5 cm
∴ AB2 = AO2 + BO2
= (12)2 + 52
= 144 + 25
= 169
∴ AB = √169
= 13 CM
Therefore, the length of each side will be 13 cm.
Question no – (5)
Solution :
In the question,
rhombus whose perimeter is = 60 cm.
Let each side be ‘x’
∴ x + x +x + x + x = 60
= 4x = 60
= x = 15 cm
Therefore, the length of each side of the rhombus is 15 cm.
Question no – (6)
Solution :
One angle of a rhombus will be 56°
∴ Another angle
= (180 – 56°) = 124°
∴ ∠1 = ∠3 = 56°
∠2 = ∠4 = 124°
Therefore, the measure of the other angles will be 124°.
Question no – (7)
Solution :
Here, AB = BC = CD = AD = BD
△ABD is an Equilateral angle
∴ ∠BAD = 60°
= ∠BCD = 60° [Opposite angles of 11 cm are equal]
∴ ∠BAD + ∠ABC = 180°
= ∠ABC = 180° – 60°
= 120°
Therefore, Angle value will be 60°, 120°
Question no – (8)
Solution :
Let, the angles x and 5x
∴ x + 5x = 180°
= 6x = 180°
= x = 30°
∴ One angle is 30°
∴ Another angle is,
= (30 × 5)
= 150°
Therefore, the angles of the rhombus are 30° and 150°
Question no – (9)
Solution :
lengths of the diagonals of a rhombus are 16 cm and 12 cm.
Here, d1 = 16 cm d2 = 12 cm
∴ a2 = (d1/2)2 + (d2/2)2
= (16/2)2 + (12/2)2
= 82 + 62
= 64 + 36
= a2 = 100
= a = 10 cm
Therefore, the side of the rhombus is 10 cm.
Next Chapter Solution :
👉 Chapter 2 👈
