RBSE Solutions for Class 8 Maths Chapter 6 Polygons Ex 6.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Exercise 6.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 6 |
| Chapter Name | Polygons |
| Exercise | Exercise 6.2 |
| Number of Questions | 8 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 6 Polygons Ex 6.2
Question 1.
Fill in the blanks choosing the right option.
(i) Adjacent angles of a parallelogram are ……… (equal/supplement)
(ii) Diagonals of a rectangle are ……… (equal/perpendicular bisect)
(iii) In any trapezium AB || CD, If A = 100° then the value of ∠D will be ……(100/80°)
(iv) If in any quadrilateral, diagonal bisect each other on right angle then, it is called ……(parallelogram/rhombus)
(v) All squares are……(congruent/similar)
Solution
(i) supplementary
(ii) equal
(iii) 80°
(iv) rhombus
(v) similar.
Question 2.
In figure, BEST is a parallelogram. Find the values of x, y and z.
Solution
∵ BEST is a parallelogram.
∴ ∠B + x° = 180°
Adjacent angles of a parallelogram are supplementary
⇒ 110° + x° = 180°
⇒ x° = 180° – 110°
⇒ x° = 70°
⇒ x = 70
∵ Opposite angles of a parallelogram are equal
∴ x° = y°
⇒ 70° = y°
⇒ y = 70
and z° = 110°
⇒ z = 110
Question 3.
In following parallelograms, find the unknown value of x, y, z.
Solution
(i) x° = 90°
=> x = 90
∆OAD in,
OA = OD | given that
∴ ∠OAD = ∠ODA (= y°)
and also we take
∠AOD + ∠OAD + ∠ODA = 180°
∵ Sum of angles of a triangle is 180°
⇒ 90° + y° + y° = 180°
⇒ 90° + 2y° = 180°
⇒ 2y° = 180° – 90° = 90°
⇒ y° = \(\frac { 90 }{ 2 }\) = 45°
⇒ y = 45
∵ diagonals of a parallelogram bisect to each other.
∴ OA = OC
∵ OA = OD
∴ OC = OD
∴ ∠DOC = 90°
∴ As above
z° = 45°
⇒ z = 45
(ii) y° = 112° ∵ Opposite angles of a parallelogram are equal
⇒ y = 112
40° + z° + y° = 180°
∵ Adjacent angles of a parallelogram are supplementary
⇒ 40° + z° + 112° = 180°
⇒ z° + 152° = 180°
⇒ z° = 180° – 152° = 28°
⇒ z = 28
In end,
x° + y° + 40° = 180°
∵ Sum of angles of a triangle is 180°
⇒ x° + 112° + 40° = 180°
⇒ x° + 152° = 180°
⇒ x° = 180° – 152°
⇒ x° = 28°
⇒ x = 28
Question 4.
In any parallelogram, the ratio of two adjacent angles is 1 : 5. Find the value of all angles of parallelogram.
Solution
Let ABCD is a parallelogram According to question
∠A : ∠B = 1 : 5
∠A + ∠B = 180°
∵Sum of two adjacent angles of a parallelogram is 180°
Sum of Ratio = 1 + 5 = 6
∴∠A = \(\frac { 1 }{ 6 }\) x 180° = 30°
∠B = \(\frac { 5 }{ 6 }\) x 80° = 150°
∵ Opposite angles of a parallelogram are equal
∠C = ∠A = 30°
and ∠D = ∠B = 150°
∴Angles of a parallelogram are 30°, 150°, 30° and 150°.
Question 5.
In the following figures RISK and STEW are parallelograms, find the values of x and y (length in cm)
Solution
(i) ∵ Opposite sides of a parallelogram are equal.
∴3x = 18
and 3y – 1 = 26
⇒ x = \(\frac { 18 }{ 3 }\) = 6
and 3y = 26 + 1
⇒ 3y = 27
⇒ y = \(\frac { 27 }{ 2 }\) = 9
So, x = 6 cm and y = 9 cm
(ii) ∵ Diagonals of a parallelogram bisect each other
x + y = 16….(1)
and y + 7 = 20….(2)
From equation (2),
y = 20 – 7 = 13
We put y = 13 in equation (1)
x + 13 = 16
=> x = 16 – 13 = 3
So, x = 3 cm and y = 13 cm
Question 6.
HOPE is a rectangle.Its diagonals intersect each other at S.Find the value of x if SH = 2x + 4 and SE = 3x + 1.
Solution
∵ Diagonal of a rectangle bisect each other
∴ SE = SO
and SH = SP
∴ EO = 2SE
= 2 (3x + 1)
and HP = 2 SH = 2 (2x + 4)
∵ Diagonal of a rectangle are equal.
∴ EO = HP
⇒ 2(3x + 1) = 2(2x + 4)
⇒ 3x + 1 = 2x + 4
⇒ 3x – 2x = 4 – 1
⇒ x = 3
Question 7.
PEAR is a rhombus. Find the value of x, y and z, also write the causes.
Solution
∵ Diagonals of a rhombus bisect each other.
∴ OA = OP
⇒ x = 5 cm. ∵ OP = 5 cm. (given that)
and OE = OR
⇒ y = 12 cm. ∵ OR = 12 cm. (given that)
∵ sides of a rhombus are equal.
∴ EP = PR
⇒ z = 13 cm.
Question 8.
In a trapezium PQRS, PQ || SR, find the value of ∠x and ∠y.
Solution
∵ PQ || SR
∴ x + 90° = 180°
⇒ x = 180° – 90° = 90°
Again ∴ PQRS is a quadrilateral
∴ x + y + 130° + 90° = 360°
∵ We know that sum of all angles of a quadrilateral = 360°
⇒ 90° + y + 130° + 90° = 360°
⇒ y + 310° = 360°
⇒ y = 360° – 310°
⇒ y = 50°
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