GSEB Class 8 Maths Solutions Chapter 3 Understanding Quadrilaterals Exercise 3.3

Question 1. Given a parallelogram ABCD. Complete each statement along with the definition or property used
(i) AD =
(ii) ZDCB =
(iii) OC =
(iv) ∠DAB + mZCDA =
Answer:
(i) \( AD = BC \) [Opposite sides are equivalent]
(ii) \( \angle DCB = \angle DAB \) [Opposite angles are identical]
(iii) \( OC = OA \) [Diagonals divide each other in half]
(iv) \( m\angle DAB + m\angle CDA = 180° \) [Adjacent angles are supplementary]
In simple words: In a parallelogram, sides that are across from each other are the same length. Angles that are across from each other are also the same size. The lines that cut through the middle (diagonals) split each other into two equal parts. Also, angles next to each other add up to 180 degrees.

Exam Tip: Remember the five basic properties of a parallelogram: opposite sides are parallel, opposite sides are equal, opposite angles are equal, adjacent angles are supplementary, and diagonals bisect each other. These properties are fundamental for solving problems involving parallelograms.

Question 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer:
(i) In the given parallelogram:
Opposite angles are equal. The angle \( y \) is opposite to the 100° angle.
Therefore, \( y = 100° \).
Adjacent angles in a parallelogram are supplementary, meaning they sum to 180°.
So, \( x + y = 180° \).
Substituting the value of \( y \): \( x + 100° = 180° \).
\( x = 180° - 100° = 80° \).
Since opposite angles are equal, the angle \( z \) is opposite to angle \( x \).
Therefore, \( z = x = 80° \).
Thus, the values are \( x = 80°, y = 100°, z = 80° \).
In simple words: Look at the first picture. Angle y is across from the 100-degree angle, so it's also 100 degrees. Angle x is next to angle y, so they add up to 180, making x 80 degrees. Angle z is across from angle x, so it's also 80 degrees.

Exam Tip: For problems involving angles in a parallelogram, always start by identifying pairs of opposite and adjacent angles. This simplifies the problem as opposite angles are equal and adjacent angles are supplementary.

Question 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer:
(ii) In the given parallelogram:
Opposite angles are equal. The angle \( z \) is opposite to the 50° angle.
Therefore, \( z = 50° \).
Adjacent angles in a parallelogram are supplementary. So, \( x + 50° = 180° \).
\( x = 180° - 50° = 130° \).
Opposite angles are equal, so \( y = x \).
Therefore, \( y = 130° \).
Thus, the values are \( x = 130°, y = 130°, z = 50° \).
In simple words: Look at the second picture. The angle z is across from the 50-degree angle, so it's also 50 degrees. Angle x is next to the 50-degree angle, so they add up to 180 degrees, making x 130 degrees. Angle y is across from angle x, so it's also 130 degrees.

Exam Tip: Remember two key rules for parallelograms: opposite angles are equal, and adjacent angles sum to 180 degrees. These rules help solve for unknown angles.

Question 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer:
(iii) In the given parallelogram, the diagonals intersect at \( x \). We will assume from the initial solution steps that the diagonals intersect perpendicularly, implying \( x = 90° \). This means the parallelogram is a rhombus.
The angle marked 30° is an angle formed by a diagonal and a side, let's say \( \angle OBC = 30° \).
In triangle BOC, we have \( \angle BOC = x = 90° \) and \( \angle OBC = 30° \).
The sum of angles in a triangle is 180°. So, \( \angle OCB = 180° - 90° - 30° = 60° \).
In a rhombus, diagonals bisect the vertex angles.
Thus, \( y = \angle ABC = 2 \times \angle OBC = 2 \times 30° = 60° \).
And \( z = \angle BCD = 2 \times \angle OCB = 2 \times 60° = 120° \).
So, the values are \( x = 90°, y = 60°, z = 120° \).
In simple words: If diagonals cross at 90 degrees, it's a rhombus. One small angle inside is 30 degrees. This helps us find the other small angle in that triangle is 60 degrees. Because diagonals cut the main angles in half, the top angle (y) is twice 30, so 60 degrees. The bottom right angle (z) is twice 60, so 120 degrees.

Exam Tip: When diagonals are involved, remember their properties: they bisect each other, and in special parallelograms like rhombuses, they are perpendicular and bisect the vertex angles. Identify which angles are given within the triangles formed by the diagonals.

Question 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer:
(iv) In the given parallelogram:
Opposite angles are always equal. So, the angle \( y \) which is opposite to the 80° angle, is \( y = 80° \).
Adjacent angles in a parallelogram are supplementary, meaning they sum up to 180°.
Therefore, \( x + 80° = 180° \).
\( x = 180° - 80° = 100° \).
Since opposite angles are equal, the angle \( z \) which is opposite to angle \( x \), is also \( z = 100° \).
Thus, the values are \( x = 100°, y = 80°, z = 100° \).
In simple words: For this parallelogram, angle y is across from the 80-degree angle, so it's also 80 degrees. Angle x is next to the 80-degree angle, so they add up to 180 degrees, making x 100 degrees. Angle z is across from angle x, so it's also 100 degrees.

Exam Tip: Carefully identify whether angles are opposite or adjacent. Opposite angles are equal, while adjacent angles sum to 180 degrees. This helps avoid confusion with transversal angle properties.

Question 2. Consider the following parallelograms. Find the values of the unknowns x, y, z.
Answer:
(v) In the given parallelogram ABCD, with vertices labeled D (top left), C (top right), B (bottom right), A (bottom left):
1. The angle at vertex B is 112°, i.e., \( \angle ABC = 112° \).
2. Opposite angles in a parallelogram are equal, so \( \angle ADC = \angle ABC = 112° \).
3. The angle at vertex D (\( \angle ADC \)) is split into two parts by the diagonal: 40° and \( x \). So, \( \angle ADC = 40° + x \).
Therefore, \( 40° + x = 112° \).
\( x = 112° - 40° = 72° \).
4. Adjacent angles in a parallelogram are supplementary, so \( \angle DAB + \angle ABC = 180° \).
\( z + 112° = 180° \).
\( z = 180° - 112° = 68° \).
5. Opposite angles are equal, so \( y = \angle DCB = \angle DAB = z \).
Thus, \( y = 68° \).
The values are \( x = 72°, y = 68°, z = 68° \).
In simple words: First, find angle D, which is opposite to the 112-degree angle, so it's also 112 degrees. Angle D is made of 40 degrees and x, so x must be 72 degrees. Next, angle z is next to 112 degrees, so they add to 180, making z 68 degrees. Angle y is across from angle z, so it's also 68 degrees.

Exam Tip: For complex diagrams, first identify all known full vertex angles. Then use opposite and adjacent angle properties to find the full unknown vertex angles. Finally, use any given parts of angles to calculate the remaining unknowns.

Question 3. Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180°?
Answer: If a quadrilateral ABCD has \( \angle D + \angle B = 180° \):
If \( \angle D \) and \( \angle B \) are opposite angles, then for a parallelogram, opposite angles are equal, so \( \angle D = \angle B \).
This would mean \( \angle D + \angle D = 180° \), which simplifies to \( 2\angle D = 180° \), so \( \angle D = 90° \).
If one pair of opposite angles is 90°, then all angles must be 90° (a rectangle), which is a type of parallelogram.
However, if \( \angle D \) and \( \angle B \) are just two angles in a quadrilateral that sum to 180°, without further information about side parallelism or other angles, it could also be a non-parallelogram shape like an isosceles trapezoid.
Thus, the quadrilateral *may* be a parallelogram but not in every instance.
In simple words: If angle D and angle B add up to 180 degrees, it might be a parallelogram (like a rectangle if D and B are opposite angles). But it's not always a parallelogram, as other shapes can have two angles that add up to 180 degrees.

Exam Tip: Carefully distinguish between "must be a parallelogram" and "can be a parallelogram". A single property might allow for a parallelogram but doesn't guarantee it exclusively.

Question 3. Can a quadrilateral ABCD be a parallelogram if
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm?
Answer: For a quadrilateral ABCD to be a parallelogram, both pairs of its opposite sides must have the same length.
Here, we are given \( AB = DC = 8 \text{ cm} \), which satisfies one condition for opposite sides.
However, we are also given \( AD = 4 \text{ cm} \) and \( BC = 4.4 \text{ cm} \).
Since \( AD \neq BC \), the second pair of opposite sides is not equal in length.
Therefore, this quadrilateral cannot be a parallelogram.
In simple words: For a shape to be a parallelogram, the sides that are across from each other must be the same length. Here, one pair of opposite sides is different (4 cm and 4.4 cm), so it can't be a parallelogram.

Exam Tip: Remember that *both* pairs of opposite sides must be equal for a quadrilateral to be a parallelogram. If even one pair is unequal, it's not a parallelogram.

Question 3. Can a quadrilateral ABCD be a parallelogram if
(iii) ∠A = 70° and ∠C = 65°?
Answer: For a quadrilateral ABCD to be a parallelogram, its opposite angles must be equal in measure.
Here, we are given \( \angle A = 70° \) and \( \angle C = 65° \).
Since \( \angle A \neq \angle C \), the condition for opposite angles being equal is not met.
Therefore, this quadrilateral cannot be a parallelogram.
In simple words: If a shape is a parallelogram, the angles across from each other must be exactly the same. Since angle A is 70 degrees and angle C is 65 degrees, they are not the same, so this shape cannot be a parallelogram.

Exam Tip: Always check if opposite angles are equal. If they are not, the figure is definitely not a parallelogram. This is a quick test to rule out parallelograms.

Question 4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Answer: A kite is a quadrilateral that is not a parallelogram but possesses precisely two opposite angles of identical measure. In the accompanying diagram, ABCD represents a kite where the opposing angles \( \angle B \) and \( \angle D \) are equal, but \( \angle A \neq \angle C \).
In simple words: You need to draw a kite. A kite has two pairs of equal-length sides that are next to each other. It also has one pair of opposite angles that are equal, but the other pair is not. This makes it different from a parallelogram.

Exam Tip: When asked to draw a figure, clearly label vertices and any relevant angles or side lengths. For a kite, remember its unique property of having exactly one pair of opposite angles equal.

Question 5. The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measure of each of the angles of the parallelogram.
Answer: Let the parallelogram be ABCD. If two adjacent angles are in the ratio 3 : 2, let these angles be \( \angle A \) and \( \angle B \).
Since adjacent angles in a parallelogram are supplementary, their sum is 180°.
So, we can write \( \angle A = 3x \) and \( \angle B = 2x \).
\( 3x + 2x = 180° \)
\( 5x = 180° \)
\( x = \frac{180°}{5} = 36° \)
Now, we can find the measures of \( \angle A \) and \( \angle B \):
\( \angle A = 3x = 3 \times 36° = 108° \)
\( \angle B = 2x = 2 \times 36° = 72° \)
In a parallelogram, opposite angles are equal.
So, \( \angle C = \angle A = 108° \)
And \( \angle D = \angle B = 72° \)
Therefore, the measures of the angles of the parallelogram are \( \angle A = 108°, \angle B = 72°, \angle C = 108°, \) and \( \angle D = 72° \).
In simple words: When two angles next to each other in a parallelogram are in a 3:2 ratio, we can call them 3x and 2x. They add up to 180 degrees. Solving for x gives 36 degrees. So, the angles are 3 times 36 (108 degrees) and 2 times 36 (72 degrees). The angles opposite these will be the same.

Exam Tip: Remember that adjacent angles in a parallelogram are supplementary (add up to 180°), and opposite angles are equal. This allows you to find all angles once two adjacent angles are known.

Question 6. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Answer: Let's assume the parallelogram is ABCD, and its two adjacent angles \( \angle A \) and \( \angle B \) have equal measure.
We know that adjacent angles in a parallelogram are supplementary, meaning their sum is 180°.
So, \( \angle A + \angle B = 180° \).
Since \( \angle A = \angle B \), we can substitute \( \angle A \) for \( \angle B \):
\( \angle A + \angle A = 180° \)
\( 2\angle A = 180° \)
\( \angle A = \frac{180°}{2} = 90° \)
Therefore, \( \angle B \) is also 90°.
We also know that opposite angles in a parallelogram are equal.
So, \( \angle C = \angle A = 90° \)
And \( \angle D = \angle B = 90° \)
Thus, all angles of the parallelogram are 90°. The parallelogram is a rectangle.
In simple words: If two angles next to each other in a parallelogram are the same size, they must both be 90 degrees because they add up to 180 degrees. Since opposite angles are also the same, all four angles in this parallelogram will be 90 degrees.

Exam Tip: This is a special case: a parallelogram with equal adjacent angles is always a rectangle. Understanding this directly can save time in calculations.

Question 7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Answer: In the parallelogram HOPE, with vertices labeled H (bottom left), E (top left), O (top right), P (bottom right):
1. The exterior angle at vertex P is given as 70°. The interior angle \( \angle OPE \) (which is \( x \)) is supplementary to this exterior angle.
\( x = 180° - 70° = 110° \) (Angles on a straight line are supplementary).
2. In a parallelogram, adjacent angles are supplementary. Angle \( \angle PEH \) (which is \( z \)) is adjacent to \( \angle OPE \) (which is \( x \)).
\( z + x = 180° \)
\( z + 110° = 180° \)
\( z = 180° - 110° = 70° \) (Adjacent angles are supplementary).
3. In a parallelogram, opposite angles are equal. Angle \( \angle HOE \) (which is \( y \)) is opposite to \( \angle PEH \) (which is \( z \)).
\( y = z = 70° \) (Opposite angles are equal).
Thus, the angle measures are \( x = 110°, y = 70°, \) and \( z = 70° \).
In simple words: First, angle x inside the parallelogram is 180 minus the outside 70 degrees, so x is 110 degrees. Then, angle z is next to angle x, so they add up to 180, making z 70 degrees. Angle y is across from angle z, so it's also 70 degrees.

Exam Tip: When dealing with parallelogram problems, always prioritize the fundamental properties: opposite angles are equal, and adjacent angles are supplementary. If a diagram contains conflicting numerical values, use the properties to find a consistent set of answers.

Question 8. figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm.)
Answer:
(i) In the parallelogram GUNS, opposite sides are equal in length.
According to the provided calculations:
One pair of opposite sides gives the equation:
\( 3x = 18 \)
Dividing by 3, we get:
\( x = \frac{18}{3} = 6 \) cm
The other pair of opposite sides gives the equation:
\( 3y - 1 = 26 \)
Adding 1 to both sides:
\( 3y = 26 + 1 \)
\( 3y = 27 \)
Dividing by 3, we get:
\( y = \frac{27}{3} = 9 \) cm
Therefore, the values are \( x = 6 \) cm and \( y = 9 \) cm.
In simple words: For a parallelogram, opposite sides are the same length. We set up two equations from the given labels. Solving the first equation, we find x is 6 cm. Solving the second equation, we find y is 9 cm.

Exam Tip: Always remember that in a parallelogram, both pairs of opposite sides are equal. Use this property to form equations and solve for unknown side lengths.

Question 8. figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm.)
Answer:
(ii) In the parallelogram RUNS, the diagonals bisect each other. This means they cut each other into two equal parts.
From the diagram, the diagonal `RN` is divided into segments `RO` and `ON`.
So, \( RO = ON \).
This gives us the equation: \( x + y = 16 \) (Equation 1)
The diagonal `SU` is divided into segments `OS` and `OU`.
So, \( OS = OU \).
This gives us the equation: \( 7 + y = 20 \)
To find \( y \), subtract 7 from both sides:
\( y = 20 - 7 \)
\( y = 13 \) cm
Now substitute the value of \( y \) into Equation 1:
\( x + 13 = 16 \)
To find \( x \), subtract 13 from both sides:
\( x = 16 - 13 \)
\( x = 3 \) cm
Therefore, the values are \( x = 3 \) cm and \( y = 13 \) cm.
In simple words: For a parallelogram, the lines drawn from corner to corner (diagonals) cut each other in half. So, we make two equations. From the second equation, we find y is 13 cm. Then, we use that to solve the first equation, and find x is 3 cm.

Exam Tip: Remember that in any parallelogram, diagonals always bisect each other. This means the point where they cross divides each diagonal into two equal segments.

Question 9. In the following figure, both RISK and CLUE are parallelograms. Find the value of x.
Answer: Let the intersection point of line segments SK and CE be O.
**For Parallelogram RISK:**
We are given \( \angle K = 120° \).
Adjacent angles in a parallelogram are supplementary. So, \( \angle R + \angle K = 180° \).
\( \angle R = 180° - 120° = 60° \).
Opposite angles in a parallelogram are equal. Thus, \( \angle S = \angle R = 60° \).
**For Parallelogram CLUE:**
We are given \( \angle L = 70° \).
Opposite angles in a parallelogram are equal. So, \( \angle E = \angle L = 70° \).
Now, consider the triangle formed by vertices E, S, and the intersection point O (where x is located). Let's call this triangle \( \triangle EOS \).
The angles inside \( \triangle EOS \) are \( \angle ESO \) (which is \( \angle S \) of parallelogram RISK), \( \angle SEO \) (which is \( \angle E \) of parallelogram CLUE), and \( \angle EOS \) (which is \( x \)).
In this case, \( \angle ESO \) is the full angle \( \angle S \) of parallelogram RISK, which is 60°.
And \( \angle SEO \) is the full angle \( \angle E \) of parallelogram CLUE, which is 70°.
The sum of angles in a triangle is 180°.
\( \angle ESO + \angle SEO + x = 180° \)
\( 60° + 70° + x = 180° \)
\( 130° + x = 180° \)
\( x = 180° - 130° \)
\( x = 50° \)
Therefore, the value of \( x \) is 50°.
In simple words: First, find all the angles for both parallelograms. For RISK, angle S is 60 degrees. For CLUE, angle E is 70 degrees. Then, look at the small triangle made by the crossing lines. The angles in that triangle are angle S, angle E, and angle x. Since angles in a triangle add up to 180, x must be 50 degrees.

Exam Tip: For figures with overlapping shapes, identify common points and form triangles or smaller quadrilaterals. Apply angle properties of parallelograms first, then use triangle angle sum properties.

Question 10. Explain how this figure is a trapezium. Which of its two sides are parallel?
Answer: In the given figure (quadrilateral NMLK):
We are provided with \( \angle M = 100° \) and \( \angle L = 80° \).
Let's sum these two angles: \( \angle M + \angle L = 100° + 80° = 180° \).
Since the sum of consecutive interior angles on the same side of a transversal (ML) is 180°, this indicates that the lines NM and KL are parallel.
A quadrilateral is defined as a trapezium if it has at least one pair of parallel sides.
Therefore, figure NMLK is a trapezium.
The two sides that are parallel are \( \bar{NM} \) and \( \bar{KL} \).
In simple words: When two angles next to each other, like angle M and angle L, add up to 180 degrees, it means the lines connecting them (NM and KL) are parallel. Any four-sided shape with at least one pair of parallel sides is called a trapezium. So, NM and KL are the parallel sides.

Exam Tip: To identify parallel lines, check for supplementary consecutive interior angles. This is a fundamental property that defines trapeziums and parallelograms.

Question 11. Find m∠C in the adjoining figure if \( \bar { AB } || \bar { DC } \)
Answer: In the given figure, ABCD is a trapezium where line segment \( \bar{AB} \) is parallel to line segment \( \bar{DC} \). The line segment \( \bar{BC} \) acts as a transversal.
When two parallel lines are cut by a transversal, the consecutive interior angles (also known as same-side interior angles) are supplementary, meaning they add up to 180°.
Therefore, \( m\angle B + m\angle C = 180° \).
We are given that \( m\angle B = 120° \).
Substituting this value into the equation:
\( 120° + m\angle C = 180° \)
To find \( m\angle C \), subtract 120° from both sides:
\( m\angle C = 180° - 120° \)
\( m\angle C = 60° \)
Thus, the measure of angle C is 60°.
In simple words: Since sides AB and DC are parallel, and BC cuts across them, the angles B and C are "same-side interior angles." These kinds of angles always add up to 180 degrees. So, if angle B is 120 degrees, angle C must be 60 degrees.

Exam Tip: For trapeziums, remember that only one pair of sides is parallel. The angles between a non-parallel side (transversal) and the parallel sides are supplementary.

Question 12. Find the measure of ∠P and ∠S if \( \bar { SP } || \bar { RQ } \) in figure. (If you find m∠R, is there more than one method to find m∠P?)
Answer: In the given trapezium PQRS, we have \( \bar{SP} \parallel \bar{RQ} \).
**Finding \( m\angle P \):**
1. Consider PQ as a transversal intersecting the parallel lines \( \bar{SP} \) and \( \bar{RQ} \).
The consecutive interior angles \( \angle P \) and \( \angle Q \) are supplementary.
So, \( m\angle P + m\angle Q = 180° \).
We are given \( m\angle Q = 130° \).
\( m\angle P + 130° = 180° \)
\( m\angle P = 180° - 130° = 50° \).
**Finding \( m\angle S \):**
2. Consider RS as a transversal intersecting the parallel lines \( \bar{SP} \) and \( \bar{RQ} \).
The consecutive interior angles \( \angle S \) and \( \angle R \) are supplementary.
So, \( m\angle S + m\angle R = 180° \).
From the figure, \( m\angle R \) is indicated by a right angle symbol, so \( m\angle R = 90° \).
\( m\angle S + 90° = 180° \)
\( m\angle S = 180° - 90° = 90° \).
**Alternative method to find \( m\angle P \):**
Yes, if \( m\angle R \) is known, there is another method to find \( m\angle P \).
The sum of all interior angles in any quadrilateral is 360°.
So, \( m\angle P + m\angle Q + m\angle R + m\angle S = 360° \).
Substitute the known and calculated values:
\( m\angle P + 130° + 90° + 90° = 360° \)
\( m\angle P + 310° = 360° \)
\( m\angle P = 360° - 310° = 50° \).
This confirms the value of \( m\angle P \).
In simple words: First, for angles P and Q, because their sides are parallel, they add up to 180 degrees. So P is 50 degrees. Next, for angles S and R, they also add up to 180 degrees, and R is 90 degrees, so S is 90 degrees. Yes, you can also find P by adding all four angles (P, Q, R, S) to 360 degrees and then subtracting the others.

Exam Tip: For trapeziums, focus on the pair of parallel sides. The angles on the same side of a transversal cutting these parallel lines will always be supplementary. The sum of interior angles of any quadrilateral is always 360 degrees, which provides a useful cross-check or alternative method.