Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(A) are available in PDF format to boost the exam preparation of students. The method of finding the angles of a quadrilateral is the main concept talked about under this exercise. Multiple solved examples are present to help students grasp the various tricks of solving complex problems with ease. Students can find Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(A) PDF, from the links provided here.
Selina Solutions Concise Mathematics Class 6 Chapter 27: Quadrilateral Exercise 27(A) Download PDF
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Access Selina Solutions Concise Mathematics Class 6 Chapter 27 Quadrilateral Exercise 27(A)
Exercise 27(A)
1. Two angles of a quadrilateral are 890 and 1130. If the other two angles are equal; find the equal angles.
Solution:
Let us consider the other angle as x0
As per the question, we have
890 + 1130 + x0 + x0 = 3600
2x0 = 3600 – 2020
2x0 = 158
x0 = 158 / 2
We get,
x = 790
Therefore, the other two equal angles are 790 each.
2. Two angles of a quadrilateral are 680 and 760. If the other two angles are in the ratio 5: 7; find the measure of each of them.
Solution:
Given
Two angles are 680 and 760
Let us consider the other two angles as 5x and 7x
Hence,
680 + 760 + 5x + 7x = 3600
12x + 1440 = 3600
12x = 3600 – 1440
12x = 2160
x = 2160 / 12
We get,
x = 180
Now, the other angles is calculated as below
5x = 5 × 180 = 900
7x = 7 × 180 = 1260
Therefore, the values of the other angles are 900 and 1260
3. Angles of a quadrilateral are (4x)0, 5(x+2)0, (7x – 20)0 and 6(x + 3)0. Find
(i) the value of x.
(ii) each angle of the quadrilateral.
Solution:
Given
The angles of quadrilateral are,
(4x)0, 5(x + 2)0, (7x – 20)0 and 6(x + 3)0
We know that the sum of angles in a quadrilateral is 3600
Hence,
(4x)0 + 5(x + 2)0 + (7x – 20)0 + 6(x + 3)0 = 3600
4x + 5x + 100 + 7x – 200 + 6x + 180 = 3600
22x + 80 = 3600
22x = 3600 – 80
22x = 3520
x = 3520 / 22
We get,
x = 160
Hence, the value of x is 160
Therefore, the angles are,
(4x)0 = (4 × 16)0
= 640
5(x + 2)0 = 5(16 + 2)0
= 900
6(x + 3)0 = 6(16 + 3)0
= 1140
And,
(7x – 20)° = (7×16 – 20)°
= 92°
4. Use the information given in the following figure to find:
(i) x
(ii) ∠B and ∠C
Solution:
Here, given that,
∠A = 900
∠B = (2x + 4)0
∠C = (3x – 5)0
∠D = (8x – 15)0
We know that,
All the angles in a quadrilateral is 3600
So,
∠A + ∠B + ∠C + ∠D = 3600
900 + (2x + 4)0 + (3x – 5)0 + (8x – 15)0 = 3600
On further calculation, we get
900 + 2x + 40 + 3x – 50 + 8x – 150 = 3600
740 + 13x = 3600
13x = 3600 – 740
13x = 2860
x = 2860 / 13
We get,
x = 220
The value of x is 220
Now,
∠B = 2x + 4 = 2 × 220 + 4
= 480
∠C = 3x – 5 = 3 × 220 – 5
= 610
Therefore, ∠B = 480 and ∠C = 610
5. In quadrilateral ABCD, side AB is parallel to side DC. If ∠A: ∠D = 1: 2 and ∠C: ∠B = 4: 5
(i) Calculate each angle of the quadrilateral.
(ii) Assign a special name to quadrilateral ABCD.
Solution:
Given
∠A: ∠D = 1: 2
Let us consider ∠A = x and ∠D = 2x
∠C: ∠B = 4: 5
Let us consider ∠C = 4y and ∠B = 5y
Also, given
AB || DC and the sum of opposite angles of quadrilateral is 1800
So,
∠A + ∠D = 1800
x + 2x = 1800
3x = 1800
We get,
x = 600
Therefore, ∠A = 600
∠D = 2x
= 2 × 600
= 1200
Therefore, ∠D = 1200
Now,
∠B + ∠C = 1800
5y + 4y = 1800
9y = 1800
We get,
y = 200
Now,
∠B = 5y = 5 × 200
= 1000
∠C = 4y = 4 × 200
= 800
Therefore, ∠A = 600; ∠B = 1000; ∠C = 800 and ∠D = 1200
6. From the following figure find:
(i) x,
(ii) ∠ABC,
(iii) ∠ACD.
Solution:
(i)We know that,
In quadrilateral the sum of angles is equal to 3600
Hence,
x + 4x + 3x + 4x + 480 = 3600
12x = 3600 – 480
12x = 312
We get,
x = 260
Hence, the value of x is 260
(ii) ∠ABC = 4x
4 × 260 = 1040
Therefore, ∠ABC = 1040
(iii) ∠ACD = 1800 – 4x – 480
= 1800 – 4 × 260 – 480
= 1800 – 1040 – 480
We get,
= 280
Therefore, ∠ACD = 280
7. Given: In quadrilateral ABCD; ∠C = 640, ∠D = ∠C – 80; ∠A = 5(a + 2)0 and ∠B = 2(2a + 7)0.
Solution:
Given
∠C = 640
∠D = ∠C – 80
= 640 – 80
We get,
∠D = 560
∠A = 5 (a + 2)0
∠B = 2(2a + 7)0
We know that, sum of all the angles in a quadrilateral = 3600
So,
∠A + ∠B + ∠C + ∠D = 3600
5(a + 2)0 + 2(2a + 7)0 + 640 + 560 = 3600
On further calculation, we get
5a + 100 + 4a + 140 + 640 + 560 = 3600
9a + 1440 = 3600
9a = 3600 – 1440
9a = 2160
We get,
a = 240
∠A = 5(a + 2)
= 5(24 + 2)
We get,
= 1300
8. In the given figure
∠b = 2a + 15
And ∠c = 3a + 5; find the values of b and c
Solution:
∠b = 2a + 15 and
∠c = 3a + 5
Sum of angles of a quadrilateral = 3600
700 + ∠a + ∠b + ∠c = 3600
700 + a + (2a + 15) + (3a + 5) = 3600
700 + a + 2a + 15 + 3a + 5 = 3600
6a + 900 = 3600
6a = 3600 – 900
6a = 2700
We get,
a = 450
Hence, ∠a = 450
b = 2a + 15 = 2 × 450 + 15
= 900 + 15
= 1050
c = 3a + 5 = 3 × 450 + 5
= 1350 + 5
= 1400
Therefore, ∠a = 450; ∠b = 1050 and ∠c = 1400
9. Three angles of a quadrilateral are equal. If the fourth angle is 690; find the measure of equal angles.
Solution:
Given that,
Three angles of a quadrilateral are equal
Let us consider each angle as x0
Hence,
x0 + x0 + x0 + 690 = 3600
3x = 3600 – 690
3x = 2910
x = 2910 / 3
We get,
x = 970
Therefore, the measure of all the equal angles is 970
10. In quadrilateral PQRS, ∠P: ∠Q : ∠R: ∠S = 3: 4: 6: 7.
Calculate each angle of the quadrilateral and then prove that PQ and SR are parallel to each other. Is PS also parallel to QR?
Solution:
Given
∠P: ∠Q: ∠R: ∠S = 3: 4: 6: 7
Let ∠P = 3x
∠Q = 4x
∠R = 6x and
∠S = 7x
Hence,
∠P + ∠Q + ∠R + ∠S = 3600
3x + 4x + 6x + 7x = 3600
20x = 3600
x = 3600 / 20
We get,
x = 180
So,
∠P = 3x = 3 × 180
= 540
∠Q = 4x = 4 × 180
= 720
∠R = 6x = 6 × 180
= 1080
∠S = 7x = 7 × 180
= 1260
Now, adding two adjacent angles, we get
∠Q + ∠R = 720 + 1080
= 1800 and
∠P + ∠S = 540 + 1260
= 1800
Therefore, PQ || RS
Since,
∠P + ∠Q = 540 + 720
= 1260
Which is not equal to1800
Therefore, PS and QR are not parallel
11. Use the information given in the following figure to find the value of x.
Solution:
Given
A, B, C and D are the vertices of quadrilateral and BA is produced to E
Here,
∠EAD = 700
Hence,
∠DAB = 1800 – 700 [By straight line]
∠DAB = 1100
Hence,
∠EAD + ∠DAB = 1800
The sum of angles of a quadrilateral is 3600
1100 + 800 + 560 + 3x – 60 = 3600
3x = 3600 – 1100 – 800 – 560 + 60
3x = 3600– 2400
3x = 1200
x = 1200 / 3
We get,
x = 400
Therefore, the value of x is 400
12. The following figure shows a quadrilateral in which sides AB and DC are parallel. If ∠A: ∠D = 4: 5, ∠B = (3x – 15)0 and ∠C = (4x + 20)0, find each angle of the quadrilateral ABCD.
Solution:
Let us consider ∠A = 4x and
∠D = 5x
Since, AB || DC
So,
∠A + ∠D = 1800
Substituting the value of angle A and D, we get
4x + 5x = 1800
9x = 1800
x = 200
Now,
∠A = 4x = 4 × 200
= 800
∠D = 5x = 5 × 200
= 1000
Similarly since, AB || DC
∠B + ∠C = 1800
3x – 150 + 4x + 200 = 1800
7x + 50 = 1800
7x = 1800 – 50
7x = 1750
We get,
x = 250
∠B = 3x – 150 = 3 × 250 – 150
= 750 – 150
= 600 and
∠C = 4x + 200 = 4 × 250 + 200
= 1000 + 200
= 1200
