CBSE Class 8 Mathematics Understanding Quadrilaterals Assignments Set 08

VIII - Mathematics Assignment No. 03 - Understanding Quadrilaterals

Question 1. The ratio of the two sides of a parallelogram is \(1:2\) and its perimeter is \(60\text{ m}\). Find the sides of the parallelogram.
Answer: Let the common ratio be \(x\).
The sides of the parallelogram are \(1x\) (or \(x\)) and \(2x\).
Perimeter of a parallelogram = \(2(\text{side}_1 + \text{side}_2)\)
\(2(x + 2x) = 60\)
\(2(3x) = 60\)
\(6x = 60\)
\(x = \frac{60}{6} = 10\text{ m}\)
Therefore, the sides are:
Side 1 = \(10\text{ m}\)
Side 2 = \(2 \times 10 = 20\text{ m}\)
In simple words: Using the given ratio, we set the sides as \(x\) and \(2x\) and used the perimeter formula to find \(x\).

Exam Tip: Always remember that the perimeter of a parallelogram (or rectangle) is twice the sum of its adjacent sides.

Question 2. The sum of the two opposite angles of a parallelogram is \(150^{\circ}\). Find all the angles of the parallelogram.
Answer: Let the two opposite angles be \(\angle A\) and \(\angle C\).
In a parallelogram, opposite angles are equal, so \(\angle A = \angle C\).
Given that the sum of the two opposite angles is \(150^{\circ}\):
\(\angle A + \angle C = 150^{\circ}\)
\(2\angle A = 150^{\circ}\)
\(\angle A = \frac{150^{\circ}}{2} = 75^{\circ}\)
So, \(\angle A = 75^{\circ}\) and \(\angle C = 75^{\circ}\).
The sum of adjacent angles in a parallelogram is \(180^{\circ}\):
\(\angle A + \angle B = 180^{\circ}\)
\(75^{\circ} + \angle B = 180^{\circ}\)
\(\angle B = 180^{\circ} - 75^{\circ} = 105^{\circ}\)
Since opposite angles are equal, \(\angle D = \angle B = 105^{\circ}\).
Therefore, all the angles of the parallelogram are \(75^{\circ}, 105^{\circ}, 75^{\circ},\) and \(105^{\circ}\).
In simple words: Since opposite angles are equal and their sum is \(150^{\circ}\), each is \(75^{\circ}\). Adjacent angles add up to \(180^{\circ}\), giving the other two angles as \(105^{\circ}\).

Exam Tip: Memorize the key properties of parallelograms: opposite angles are equal, and adjacent angles are supplementary (add up to \(180^{\circ}\)).

Question 3. The perimeter of a parallelogram is \(120\text{ cm}\). If one side is greater than the other by \(20\text{ cm}\), find the lengths of the sides of the parallelogram.
Answer: Let the shorter side be \(x\text{ cm}\).
Then the longer side = \((x + 20)\text{ cm}\).
Perimeter = \(2(\text{shorter side} + \text{longer side})\)
\(2(x + x + 20) = 120\)
\(2(2x + 20) = 120\)
\(2x + 20 = \frac{120}{2} = 60\)
\(2x = 60 - 20 = 40\)
\(x = \frac{40}{2} = 20\text{ cm}\)
Shorter side = \(20\text{ cm}\)
Longer side = \(20 + 20 = 40\text{ cm}\)
In simple words: Form a linear equation using the perimeter formula and solve for the shorter side \(x\), then find the longer side.

Exam Tip: Simplify equations by dividing both sides by coefficients early to avoid large numbers.

Question 4. In the parallelogram \(ABCD\), \(\angle DAO = 50^{\circ}\), \(\angle BAO = 45^{\circ}\), \(\angle COD = 95^{\circ}\). Calculate: (i) \(\angle ABO\) (ii) \(\angle ODC\) (iii) \(\angle ACB\) (iv) \(\angle CBD\)
Answer:
(i) In parallelogram \(ABCD\), opposite sides \(AB\) and \(DC\) are parallel, and diagonal \(BD\) acts as a transversal. Therefore, alternate interior angles are equal:
\(\angle ABO = \angle CDO\) (Not directly helpful yet, let us use triangle properties).
Alternatively, consider triangle \(AOB\). We need more angles, or use properties of parallel lines with alternate interior angles:
Since \(AB \parallel DC\) and diagonal \(AC\) is a transversal, \(\angle CAB = \angle ACD\).
Using triangle \(AOB\): Wait, let us use alternate interior angles for parallel lines \(AB \parallel DC\):
\(\angle ODC = \angle ABO\) (alternate interior angles since \(AB \parallel CD\) and transversal \(BD\)).
Let us find angles step by step:
In \(\triangle COD\):
\(\angle COD = 95^{\circ}\).
Since \(AB \parallel CD\) and transversal \(AC\), \(\angle OAB = \angle OCD\) (alternate interior angles). Given \(\angle OAB = 45^{\circ}\), so \(\angle OCD = 45^{\circ}\).
In \(\triangle COD\), the sum of angles is \(180^{\circ}\):
\(\angle ODC + \angle OCD + \angle COD = 180^{\circ}\)
\(\angle ODC + 45^{\circ} + 95^{\circ} = 180^{\circ}\)
\(\angle ODC + 140^{\circ} = 180^{\circ}\)
\(\angle ODC = 180^{\circ} - 140^{\circ} = 40^{\circ}\).
Now, since \(AB \parallel CD\) and transversal \(BD\), alternate interior angles are equal:
\(\angle ABO = \angle ODC = 40^{\circ}\).
(ii) From above, \(\angle ODC = 40^{\circ}\).
(iii) For \(\angle ACB\) (which is same as \(\angle ACB\) or \(\angle ACD\)):
Since \(AB \parallel CD\) and transversal \(AC\), \(\angle ACD = \angle CAB = \angle BAO = 45^{\circ}\).
(iv) For \(\angle CBD\):
\(\angle ABC = \angle ABO + \angle OBC\)... Wait, let us find \(\angle ABC\).
In parallelogram \(ABCD\), adjacent angles are supplementary: \(\angle DAB + \angle ABC = 180^{\circ}\).
\(\angle DAB = \angle DAO + \angle BAO = 50^{\circ} + 45^{\circ} = 95^{\circ}\).
So, \(\angle ABC = 180^{\circ} - 95^{\circ} = 85^{\circ}\).
We know \(\angle ABO = 40^{\circ}\).
\(\angle CBD = \angle ABC - \angle ABO = 85^{\circ} - 40^{\circ} = 45^{\circ}\).
Summary of answers:
(i) \(\angle ABO = 40^{\circ}\)
(ii) \(\angle ODC = 40^{\circ}\)
(iii) \(\angle ACB = 50^{\circ}\) (Correction: using alternate interior angle with \(\angle DAO = 50^{\circ}\), since \(AD \parallel BC\), \(\angle ACB = \angle CAD = \angle DAO = 50^{\circ}\))
(iv) \(\angle CBD = 45^{\circ}\)
In simple words: We used the properties of parallel lines, alternate interior angles, triangle angle sum property (\(180^{\circ}\)), and adjacent supplementary angles of a parallelogram to find each angle step-by-step.

Exam Tip: Look for Z-shapes (alternate interior angles) formed by diagonals and parallel sides to easily transfer angle measurements.

Question 5. In the figure, PQRS is a parallelogram and one of the angle \(\angle P = 80^{\circ}\). Calculate other angles.
Answer: In parallelogram \(PQRS\):
Opposite angles are equal:
\(\angle R = \angle P = 80^{\circ}\)
Adjacent angles are supplementary:
\(\angle P + \angle Q = 180^{\circ}\)
\(80^{\circ} + \angle Q = 180^{\circ}\)
\(\angle Q = 180^{\circ} - 80^{\circ} = 100^{\circ}\)
Opposite angle to \(\angle Q\) is \(\angle S\):
\(\angle S = \angle Q = 100^{\circ}\)
Therefore, the other angles are \(\angle Q = 100^{\circ}\), \(\angle R = 80^{\circ}\), and \(\angle S = 100^{\circ}\).
In simple words: Opposite angles are equal, so \(\angle R = 80^{\circ}\). Adjacent angles add up to \(180^{\circ}\), giving \(\angle Q = 100^{\circ}\) and \(\angle S = 100^{\circ}\).

Exam Tip: For any parallelogram, knowing just one angle allows you to find all other three instantly using equality of opposite angles and supplementary adjacent angles.

Question 6. PQRS is a parallelogram in which \(OQ = 4\text{ cm}\) and \(PR\) is \(6\) more than \(SQ\). Find \(OP\).
Answer: In a parallelogram, diagonals bisect each other.
Since \(OQ = 4\text{ cm}\), the full diagonal \(SQ = 2 \times OQ = 2 \times 4 = 8\text{ cm}\).
Given that diagonal \(PR\) is \(6\text{ cm}\) more than \(SQ\):
\(PR = SQ + 6 = 8 + 6 = 14\text{ cm}\).
Since diagonals bisect each other at point \(O\), \(OP\) is half of diagonal \(PR\):
\(OP = \frac{PR}{2} = \frac{14}{2} = 7\text{ cm}\).
In simple words: First find the length of diagonal \(SQ\) using \(OQ\), add \(6\) to get diagonal \(PR\), and then take half of \(PR\) to find \(OP\).

Question 7. Two adjacent angles of a parallelogram ABCD are in the ratio \(2:3\). Find all the angles of the parallelogram.
Answer: Let the common ratio be \(x\).
The two adjacent angles are \(2x\) and \(3x\).
Since adjacent angles of a parallelogram are supplementary:
\(2x + 3x = 180^{\circ}\)
\(5x = 180^{\circ}\)
\(x = \frac{180^{\circ}}{5} = 36^{\circ}\)
The angles are:
First angle = \(2 \times 36^{\circ} = 72^{\circ}\)
Second angle = \(3 \times 36^{\circ} = 108^{\circ}\)
Since opposite angles are equal, the four angles of the parallelogram are \(72^{\circ}, 108^{\circ}, 72^{\circ},\) and \(108^{\circ}\).
In simple words: Set the adjacent angles as \(2x\) and \(3x\), equate their sum to \(180^{\circ}\) to find \(x\), and then use opposite angle properties.

Exam Tip: Ratio problems are easily solved by introducing a variable multiplier \(x\).

Question 8. PQRS is a parallelogram. The diagonals PR and SQ intersect at O. If \(OQ = (y + 9)\text{ cm}\), \(OS = 13\text{ cm}\), \(OP = 16\text{ cm}\), \(OR = (x + y)\text{ cm}\). Find \(x\) and \(y\).
Answer: In a parallelogram, diagonals bisect each other at the point of intersection \(O\).
Therefore:
1. \(OQ = OS\)
\(y + 9 = 13\)
\(y = 13 - 9 = 4\)

2. \(OP = OR\)
\(16 = x + y\)
Substitute \(y = 4\):
\(16 = x + 4\)
\(x = 16 - 4 = 12\)
Thus, \(x = 12\) and \(y = 4\).
In simple words: Since diagonals bisect each other, equate the two halves of each diagonal (\(OQ = OS\) and \(OP = OR\)) to form equations and solve for \(x\) and \(y\).

Exam Tip: Set up simultaneous equations by equating the segments created by the intersection point on each diagonal separately.