Understanding Quadrilaterals - Worksheets

CBSE Worksheet-1
Class 08 - Mathematics (Understanding Quadrilaterals)

---

General Instructions: All questions are compulsory. Q.1 to Q.2 carries one mark each. Q.3 to Q.7 carries two marks each. Q.8 and Q.9 carries three marks each. Q.10 to Q.12 carries four marks each.

---

1. What is the maximum exterior angle possible for a regular polygon?
2. Find the number of sides in a regular polygon when the measure of each exterior angle is 45o.
3. State true or false:
   1. The adjacent angles of a rhombus are complementary.
   2. A polygon in which at least one angle is more than 180° is called a convex polygon.
   3. The angle sum of all interior angles of a convex polygon of sides 7 is 900°.
   4. The angle sum of a convex polygon with number of sides 8 is 1080°.
4. Fill Up the following:
   1. All the sides of a regular polygon are ______ in length.
   2. In a parallelogram, the diagonals are equal and the adjacent sides are unequal. It is a _______.
   3. Each exterior angle of a regular hexagon is of measure _____.
   4. The number of sides in a regular polygon is 15 , then measure of each exterior angle is ______.
5. Match the following:

   Column A	Column B
   1. Rectangle	(a) all side equal
   2. Trapezium	(b) opposite side parallel
   3. Square	(c) opposite side equal
   4. Kite	(d) unequal opposite side

6. Find the measure of each exterior angle of a regular polygon of (i) 9 sides(ii) 15 sides.
7. Each interior angle of a regular polygon are 158o. Can it be an interior angle of a regular polygon? Why?
8. Find the angle measure x in the following figure.
   
9. Take two identical set squares with angles 30° - 60° - 90° and place them adjacently to form a parallelogram as shown in the figure. Justify that the two triangles are congruent.
   
10. ABCD is a rhombus. If AB = 4 cm then what is the perimeter of ABCD?
11. The adjacent angles of a parallelogram are (2x – 4)° and (3x – 1)°. Find the measures of all angles of the parallelogram.
12. Quadrilateral EFGH is a rectangle in which J is the point of intersection of the diagonals. Find the value of x, if JF = 8x + 4 and EG = 24x – 8.
    

CBSE Worksheet-1
Class 08 - Mathematics (Understanding Quadrilaterals)
Solution

---

1. Each exterior angle of an equilateral triangle is 120 degree and hence this the maximum possible value of exterior angle of a regular polygon. This can also be proved by another principle; which states that each exterior angle of a regular polygon is equal to 360 divided by number of sides in the polygon. If 360 is divided by 3, we get 120.
2. If the polygon has n sides,
   Then, we know that; n = 360o/measure of each exterior angle360∘measureofeachexteriorangle$\frac{{360}^{\circ }}{measureofeachexteriorangle}$
   = 360∘45∘$\frac{{360}^{\circ }}{{45}^{\circ }}$
   = 8
   Therefore, the regular polygon has 8 sides.
3. 1. False
   2. False
   3. True
   4. True
4. 1. equal
   2. rectangle
   3. 60°
   4. 24°
5. 1. (c)
   2. (b)
   3. (a)
   4. (d)
6. (i) 9 sides
   Size of each exterior angle =360∘9=40∘$=\frac{{360}^{\circ }}{9}={40}^{\circ }$
   (ii) 15 sides
   Size of each exterior angle =360∘15=24∘$=\frac{{360}^{\circ }}{15}={24}^{\circ }$
7. No, because each exterior angles 180° – 22° = 158°, which is not a divisor of 360°.
8. x + 30° + x + (180° – 70°) + (180° – 60°) = (5 – 2) × 180° [ By angle sum property of a pentagon]
   ∴ 2x + 30° + 110° + 120° = 540°
   ∴ 2x + 260° = 540°
   ∴ 2x = 540° – 260°
   ∴ 2x = 280°
   ∴ x=280∘2$x=\frac{{280}^{\circ }}{2}$
   ∴ x = 140°
9. We can further strengthen this idea through a logical argument also.
   Consider a parallelogram ABCD. Draw any one diagonal, say AC¯¯¯¯¯¯¯¯$\overline{AC}$. Looking at the angles,
   ∠1 = ∠2 and ∠3 = ∠4
   Since in triangles ABC and ADC, ∠1 = ∠2, ∠3 = ∠4 and AC¯¯¯¯¯¯¯¯$\overline{AC}$ is common so, by ASA congruency condition,
   ΔABC≅ΔCDA$\mathrm{\Delta }ABC\cong \mathrm{\Delta }CDA$
   This gives AB = DC
   and BC = AD [CPCT]
   
10. ∴$\therefore$ A rhombus is a quadrilateral with sides of equal length.
    The perimeter of ABCD = addition of all four sides
    The perimeter of ABCD = 4 + 4 + 4+ 4
    The perimeter of ABCD = 16 cm
11. Since, the adjacent angles of a parallelogram are supplementary.
    (2x – 4)° + (3x – 1)° = 180°
    ⇒ 2x – 4+ 3x – 1 = 180°
    ⇒ 5x – 5 = 180°
    ⇒ 5x = 180° + 5
    ⇒ 5x = 185°
    ⇒ x = 185°/5
    ⇒ x = 37
    Thus, the adjacent angles are
    2x – 4 = 2 ×$\times$ 37° - 4 = 74 – 4 = 70
    and 3x – 1 = 3 ×$\times$ 37° - 1 = 111 – 1 = 110°
    Hence, the angles are 70°, 110°, 70°, 110° [∵ Opposite angles in a parallelogram are equal]
12. Given, EFGH is a rectangle in which diagonals are intersecting at the point J.
    We know that, the diagonals of a rectangle insect each other and are equal
    then, EG = 2 ×$\times$ JF
    ⇒ 24x – 8 = 2 (8x + 4)
    ⇒ 24x – 8 = 16x + 8
    ⇒ 24x – 16x = 8 + 8
    ⇒ 8x = 16
    ⇒ x = 2