CBSE Class 8 Mathematics Understanding Quadrilaterals Assignments Set 05

VIII - Mathematics Assignment No-06 - Understanding Quadrilaterals

M. C. Q. (Choose the correct option from the following)

Q1. Three angles of a quadrilateral are \(45^\circ\), \(76^\circ\) and \(104^\circ\). The fourth angle is
(i) \(135^\circ\)
(ii) \(134^\circ\)
(iii) \(130^\circ\)
(iv) \(145^\circ\)
Answer: (i) \(135^\circ\)
Sum of all four angles of a quadrilateral = \(360^\circ\)
Let the fourth angle be \(x\).
\(45^\circ + 76^\circ + 104^\circ + x = 360^\circ\)
\(225^\circ + x = 360^\circ\)
\(x = 360^\circ - 225^\circ = 135^\circ\)
In simple words: The sum of angles in any four-sided shape is always \(360^\circ\), so we subtract the given three angles from \(360^\circ\) to find the remaining angle.

Exam Tip: Always verify that the sum of all four angles equals \(360^\circ\) after finding your answer.

Q2. The sum of all angles of a quadrilateral is
(i) \(90^\circ\)
(ii) \(360^\circ\)
(iii) \(270^\circ\)
(iv) \(180^\circ\)
Answer: (ii) \(360^\circ\)
By the angle sum property of polygons, the sum of interior angles of an \(n\)-sided polygon is given by \((n - 2) \times 180^\circ\). For a quadrilateral, \(n = 4\), so the sum is \((4 - 2) \times 180^\circ = 2 \times 180^\circ = 360^\circ\).
In simple words: A quadrilateral can be split into two triangles by drawing a diagonal, and since each triangle has a sum of \(180^\circ\), \(2 \times 180^\circ = 360^\circ\).

Exam Tip: Remember the formula \((n - 2) \times 180^\circ\) to quickly find the angle sum for any polygon.

Q3. The angles of a quadrilateral are in the ratio \(1 : 2 : 3 : 4\). The least angle is
(i) \(108^\circ\)
(ii) \(72^\circ\)
(iii) \(36^\circ\)
(iv) \(24^\circ\)
Answer: (iii) \(36^\circ\)
Let the angles be \(1x\), \(2x\), \(3x\), and \(4x\).
Sum of angles = \(1x + 2x + 3x + 4x = 360^\circ\)
\(10x = 360^\circ\)
\(x = \frac{360^\circ}{10} = 36^\circ\)
The least angle is \(1x = 1 \times 36^\circ = 36^\circ\).
In simple words: Use variables for the ratio, add them up to equal \(360^\circ\), solve for \(x\), and then multiply by the smallest ratio number.

Exam Tip: When given a ratio, always attach a common variable \(x\) to each part before setting up the equation.

Q4. The interior angle of a polygon is \(165^\circ\). The number of its sides are
(i) 12
(ii) 15
(iii) 20
(iv) 24
Answer: (iv) 24
Measure of interior angle = \(165^\circ\)
Measure of exterior angle = \(180^\circ - 165^\circ = 15^\circ\)
Number of sides (\(n\)) = \(\frac{360^\circ}{\text{Exterior angle}} = \frac{360^\circ}{15^\circ} = 24\).
In simple words: An interior angle and its corresponding exterior angle always add up to \(180^\circ\). Find the exterior angle first, then divide \(360^\circ\) by it to get the number of sides.

Q5. The measure of an exterior angle of a regular polygon of 10 sides is
(i) \(36^\circ\)
(ii) \(144^\circ\)
(iii) \(40^\circ\)
(iv) \(140^\circ\)
Answer: (i) \(36^\circ\)
Number of sides (\(n\)) = 10
Measure of each exterior angle = \(\frac{360^\circ}{n} = \frac{360^\circ}{10} = 36^\circ\).
In simple words: The sum of all exterior angles in any convex polygon is always \(360^\circ\). Divide that by the number of sides to find one exterior angle.

Exam Tip: Exterior angle of a regular polygon = \(\frac{360^\circ}{n}\).

Q6. The measure of an interior angle of a regular polygon of 20 sides is
(i) \(18^\circ\)
(ii) \(162^\circ\)
(iii) \(160^\circ\)
(iv) \(48^\circ\)
Answer: (ii) \(162^\circ\)
Number of sides (\(n\)) = 20
Exterior angle = \(\frac{360^\circ}{20} = 18^\circ\)
Interior angle = \(180^\circ - \text{Exterior angle} = 180^\circ - 18^\circ = 162^\circ\).
In simple words: Calculate the exterior angle first, then subtract it from \(180^\circ\) to get the interior angle.

Exam Tip: Alternatively, use the interior angle formula \(\frac{(n - 2) \times 180^\circ}{n}\), but finding the exterior angle first involves simpler numbers.

Q7. In a parallelogram ABCD, sum of two adjacent angles are
(i) \(90^\circ\)
(ii) \(360^\circ\)
(iii) \(180^\circ\)
(iv) \(270^\circ\)
Answer: (iii) \(180^\circ\)
Adjacent angles of a parallelogram are supplementary because the opposite sides are parallel, making consecutive interior angles add up to \(180^\circ\).
In simple words: Neighboring angles in a parallelogram always add up to a straight line (\(180^\circ\)).

Exam Tip: Remember: Opposite angles are equal, but adjacent angles are supplementary.

Q8. Two adjacent angles of a parallelogram are in the ratio \(2 : 3\). The largest angle is
(i) \(100^\circ\)
(ii) \(80^\circ\)
(iii) \(104^\circ\)
(iv) \(108^\circ\)
Answer: (iv) \(108^\circ\)
Let the adjacent angles be \(2x\) and \(3x\).
Since adjacent angles are supplementary, \(2x + 3x = 180^\circ\)
\(5x = 180^\circ\)
\(x = \frac{180^\circ}{5} = 36^\circ\)
The largest angle corresponds to \(3x = 3 \times 36^\circ = 108^\circ\).
In simple words: Since adjacent angles add up to \(180^\circ\), divide \(180^\circ\) according to the given ratio to find the angles.

Exam Tip: Do not equate the ratio sum to \(360^\circ\) here; adjacent angles in a parallelogram only add up to \(180^\circ\).

Q9. PQRS is a parallelogram. The value of \(\angle P - \angle R\) is
(i) \(0^\circ\)
(ii) \(180^\circ\)
(iii) \(90^\circ\)
(iv) \(360^\circ\)
Answer: (i) \(0^\circ\)
In a parallelogram PQRS, opposite angles \(\angle P\) and \(\angle R\) are equal (\(\angle P = \angle R\)).
Therefore, \(\angle P - \angle R = \angle P - \angle P = 0^\circ\).
In simple words: Opposite angles of a parallelogram are equal, so subtracting one from the other gives zero.

Exam Tip: Memorize the key properties of parallelograms: opposite sides are equal, opposite angles are equal, and diagonals bisect each other.

Q10. The perimeter of a parallelogram is 50 cm. One side is \(1\frac{1}{2}\) times the other. The smallest side is
(i) 15 cm
(ii) 10 cm
(iii) 5 cm
(iv) 20 cm
Answer: (ii) 10 cm
Let the smaller side be \(x\).
The other side = \(1\frac{1}{2}x = \frac{3}{2}x\)
Perimeter of a parallelogram = \(2 \times (\text{side}_1 + \text{side}_2)\)
\(2 \times \left(x + \frac{3}{2}x\right) = 50\)
\(x + \frac{3}{2}x = \frac{50}{2} = 25\)
\(\frac{5}{2}x = 25\)
\(x = 25 \times \frac{2}{5} = 10\text{ cm}\).
In simple words: Perimeter is twice the sum of adjacent sides. Set up the equation using the given fraction and solve for the smaller side \(x\).

Exam Tip: Perimeter formula for a quadrilateral with two pairs of equal adjacent sides is \(2(a + b)\).