CBSE Class 8 Mathematics Understanding Quadrilaterals Assignments Set 07

VIII - Mathematics Assignment No. 04 - Understanding Quadrilaterals

Question 1. Two adjacent angles of a parallelogram are \((2x + 10)^\circ\) and \((3x - 40)^\circ\). Find the measure of all angles of the parallelogram.
Answer: Let the two adjacent angles be \(A\) and \(B\).
We know that adjacent angles of a parallelogram are supplementary (their sum is \(180^\circ\)).
Therefore, \((2x + 10)^\circ + (3x - 40)^\circ = 180^\circ\)
\(5x - 30 = 180\)
\(5x = 210\)
\(x = 42\)
First adjacent angle = \(2(42) + 10 = 84 + 10 = 94^\circ\)
Second adjacent angle = \(3(42) - 40 = 126 - 40 = 86^\circ\)
Since opposite angles of a parallelogram are equal, the four angles of the parallelogram are \(94^\circ\), \(86^\circ\), \(94^\circ\), and \(86^\circ\).
In simple words: Adjacent angles add up to \(180^\circ\). Find \(x\) first, then use the property that opposite angles are equal.

Exam Tip: Always remember that consecutive angles in any parallelogram sum to \(180^\circ\).

Question 2. Two adjacent sides of a parallelogram are such that one is twice the other. If the perimeter of the parallelogram is 120 m, find the length of its adjacent sides.
Answer: Let the shorter adjacent side be \(x\) m.
Then the longer adjacent side = \(2x\) m.
Perimeter of a parallelogram = \(2 \times (\text{sum of adjacent sides})\)
\(2(x + 2x) = 120\)
\(2(3x) = 120\)
\(6x = 120\)
\(x = 20\)
Shorter side = 20 m
Longer side = \(2 \times 20 = 40\) m
In simple words: Use the perimeter formula for a rectangle or parallelogram, which is 2 times the sum of length and breadth.

Exam Tip: Pay close attention to perimeter formulas; perimeter is the total distance around the outside of the shape.

Question 3. An angle of a parallelogram is \(60^\circ\) more than its adjacent angle. Find the measure of all its angles.
Answer: Let the measure of one adjacent angle be \(x\^\circ\).
The other adjacent angle = \((x + 60)^\circ\).
Since adjacent angles of a parallelogram are supplementary:
\(x + (x + 60) = 180\)
\(2x + 60 = 180\)
\(2x = 120\)
\(x = 60\)
First angle = \(60^\circ\)
Second angle = \(60^\circ + 60^\circ = 120^\circ\)
Since opposite angles of a parallelogram are equal, the four angles are \(60^\circ\), \(120^\circ\), \(60^\circ\), and \(120^\circ\).
In simple words: Set one angle as \(x\) and the other as \(x + 60\), then solve for \(x\) using the \(180^\circ\) sum property.

Question 4. In the adjoining figure, \(ABCD\) is a parallelogram in which \(BE\) and \(CE\) are the bisectors of \(\angle B\), \(\angle C\) respectively. Find the measure of \(\angle E\).
Answer: In parallelogram \(ABCD\), adjacent angles \(\angle B\) and \(\angle C\) are supplementary:
\(\angle B + \angle C = 180^\circ\)
Since \(BE\) is the bisector of \(\angle B\):
\(\angle EBC = \frac{1}{2} \angle B\)
Since \(CE\) is the bisector of \(\angle C\):
\(\angle ECB = \frac{1}{2} \angle C\)
In triangle \(EBC\), the sum of angles is \(180^\circ\):
\(\angle E + \angle EBC + \angle ECB = 180^\circ\)
\(\angle E + \frac{1}{2} \angle B + \frac{1}{2} \angle C = 180^\circ\)
\(\angle E + \frac{1}{2}(\angle B + \angle C) = 180^\circ\)
Substitute \(\angle B + \angle C = 180^\circ\):
\(\angle E + \frac{1}{2}(180^\circ) = 180^\circ\)
\(\angle E + 90^\circ = 180^\circ\)
\(\angle E = 180^\circ - 90^\circ = 90^\circ\)
In simple words: Angle bisectors cut angles in half. Half of angle B plus half of angle C equals \(90^\circ\), leaving \(90^\circ\) for angle E in the triangle.

Exam Tip: Angle bisectors of adjacent angles of a parallelogram always intersect at a right angle (\(90^\circ\)).

Question 5. The diagonals of a rhombus are 10 cm and 24 cm. Find the length of a side of the rhombus.
Answer: Let the diagonals of the rhombus be \(d_1 = 10\) cm and \(d_2 = 24\) cm.
We know that the diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Half of the first diagonal = \(\frac{10}{2} = 5\) cm.
Half of the second diagonal = \(\frac{24}{2} = 12\) cm.
The side of the rhombus forms the hypotenuse of a right-angled triangle with these half-diagonals as its other two sides.
Using Pythagoras theorem:
\(\text{Side} = \sqrt{5^2 + 12^2}\)
\(\text{Side} = \sqrt{25 + 144}\)
\(\text{Side} = \sqrt{169} = 13\) cm.
In simple words: Diagonals cut each other at \(90^\circ\) in half. Use Pythagoras theorem on the four right triangles formed inside the rhombus.

Exam Tip: Memorize common Pythagorean triplets like \((5, 12, 13)\) to save calculation time in exams.

Question 6. In the adjoining figure, \(SPQ\) is a right-angled triangle and \(O\) is the mid-point on the hypotenuse \(SQ\). Explain why \(O\) is equidistant from \(P\), \(Q\), \(S\).
Answer: Produce \(PO\) to a point \(R\) such that \(PO = OR\), and join \(R\) to \(S\) and \(Q\) to form rectangle \(SPQR\).
Since \(SPQ\) is a right-angled triangle at \(\angle P\) and diagonals of a rectangle are equal and bisect each other:
\(SQ = PR\)
Since \(O\) is the midpoint of the diagonal \(SQ\) and diagonal \(PR\):
\(OP = OQ = OS = OR\)
Thus, point \(O\) is equidistant from \(P\), \(Q\), and \(S\).
In simple words: Completing the right triangle into a rectangle shows that the midpoint of the hypotenuse is equidistant from all three vertices.

Exam Tip: State clearly that the midpoint of the hypotenuse of a right-angled triangle is equidistant from all vertices as a standard geometric property.

Question 7. Find the length of the diagonal of a rectangle whose sides are 5 cm and 12 cm.
Answer: Let the length of the rectangle be \(l = 12\) cm and breadth be \(b = 5\) cm.
The diagonal forms the hypotenuse of a right-angled triangle with sides \(l\) and \(b\).
Using Pythagoras theorem:
\(\text{Diagonal} = \sqrt{l^2 + b^2}\)
\(\text{Diagonal} = \sqrt{12^2 + 5^2}\)
\(\text{Diagonal} = \sqrt{144 + 25}\)
\(\text{Diagonal} = \sqrt{169} = 13\) cm.
In simple words: The diagonal divides the rectangle into two right-angled triangles. Apply Pythagoras theorem: \(\text{base}^2 + \text{height}^2 = \text{hypotenuse}^2\).

Exam Tip: Every angle of a rectangle is \(90^\circ\), making Pythagoras theorem directly applicable to find diagonals.

Question 8. \(PQRS\) is a rectangle in which the diagonals \(PR\) and \(QS\) intersect each other at point \(O\). If \(OP = 2x + 7\) and \(OQ = 5x + 4\), find the value of \(x\).
Answer: We know that the diagonals of a rectangle are equal in length and they bisect each other.
Therefore, all four half-diagonals are equal: \(OP = OQ = OR = OS\).
Given \(OP = 2x + 7\) and \(OQ = 5x + 4\):
\(2x + 7 = 5x + 4\)
\(7 - 4 = 5x - 2x\)
\(3 = 3x\)
\(x = 1\)
In simple words: Diagonals of a rectangle are equal and bisect each other, which means \(OP\) equals \(OQ\). Set them equal to each other and solve for \(x\).

Exam Tip: Always quote the property "diagonals of a rectangle are equal and bisect each other" before equating expressions.