RBSE Solutions Class 9 Maths Chapter 9 Quadrilaterals Exercise 9.1

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 9 Quadrilaterals Ex 9.1

Question 1. The angles of a quadrilateral are in the ratio 3:5:9:13. Find all the angles of the quadrilateral.
Answer: Let the four angles of the quadrilateral be \( 3x \), \( 5x \), \( 9x \), and \( 13x \). We know that the sum of all angles in any quadrilateral is \( 360^\circ \). To find the value of each angle, we can set up an equation:
\( 3x + 5x + 9x + 13x = 360^\circ \)
\( \implies 30x = 360^\circ \)
\( \implies x = \frac{360^\circ}{30} \)
\( \implies x = 12^\circ \)
Now, we can find each angle:
First angle: \( 3x = 3 \times 12^\circ = 36^\circ \)
Second angle: \( 5x = 5 \times 12^\circ = 60^\circ \)
Third angle: \( 9x = 9 \times 12^\circ = 108^\circ \)
Fourth angle: \( 13x = 13 \times 12^\circ = 156^\circ \)
So, the angles of the quadrilateral are \( 36^\circ \), \( 60^\circ \), \( 108^\circ \), and \( 156^\circ \).
In simple words: First, add up all the ratio numbers and put 'x' with them. Since all four angles of a quadrilateral add up to 360 degrees, set your sum equal to 360 and find 'x'. Then multiply 'x' by each ratio number to get the real angles.

🎯 Exam Tip: Always remember that the sum of interior angles of a quadrilateral is \( 360^\circ \). When given a ratio, represent the angles as multiples of a variable like \( x \).

Question 2. In a parallelogram ABCD, diagonals AC and BD intersect at O. If OA = 3 cm and OD = 2 cm, then find the length of AC and BD.
Answer: We know a key property of parallelograms: their diagonals bisect each other. This means that the point where the diagonals cross (point O) divides each diagonal into two equal parts. Therefore, if OA is half of AC, then AC must be twice OA. Similarly, BD is twice OD.
Given: \( OA = 3 \) cm and \( OD = 2 \) cm.
For diagonal AC: Since O is the midpoint of AC, \( AC = 2 \times OA \)
\( AC = 2 \times 3 \) cm \( = 6 \) cm.
For diagonal BD: Since O is the midpoint of BD, \( BD = 2 \times OD \)
\( BD = 2 \times 2 \) cm \( = 4 \) cm.
So, the length of diagonal AC is \( 6 \) cm and the length of diagonal BD is \( 4 \) cm.
In simple words: In a parallelogram, the middle point where the lines cross cuts each line exactly in half. So, if one half is 3 cm, the whole line is double that, which is 6 cm. If the other half is 2 cm, the whole line is 4 cm.

🎯 Exam Tip: Always remember the property that the diagonals of a parallelogram bisect each other. This means they cut each other into two equal parts at their intersection point.

Question 3. Diagonals of a parallelogram are mutually perpendicular. Is this statement true? Give reason for your answer.
Answer: No, this statement is false. The diagonals of a parallelogram are not always mutually perpendicular. This property is only true for specific types of parallelograms, such as a rhombus or a square. In a general parallelogram, the diagonals bisect each other, but they do not necessarily intersect at a \( 90^\circ \) angle. For example, in a rectangle, the diagonals are equal in length and bisect each other, but they are not perpendicular unless the rectangle is also a square.
In simple words: This is not true for all parallelograms. The lines crossing inside a parallelogram only meet at a right angle (90 degrees) if it's a special type like a rhombus or a square.

🎯 Exam Tip: Clearly differentiate the properties of a general parallelogram from those of special parallelograms like rhombuses and squares. Only rhombuses and squares have perpendicular diagonals.

Question 4. angle of the parallelogram is 60°. Find the angles of the parallelogram.
Answer: Let's consider a parallelogram PQRS. The problem statement refers to an angle of \( 60^\circ \). Based on typical problems of this type, this often relates to angles formed by altitudes. Let's assume the given \( 60^\circ \) refers to the angle \( \angle TSU \) formed by altitudes ST and SU drawn from vertex S to sides PQ and QR (or their extensions), as is common in such problems. If ST is perpendicular to PQ and SU is perpendicular to QR, then \( \angle STQ = 90^\circ \) and \( \angle SUQ = 90^\circ \).
In the quadrilateral QTSU, the sum of its interior angles is \( 360^\circ \):
\( \angle TSU + \angle STQ + \angle TQU + \angle SUQ = 360^\circ \)
Given \( \angle TSU = 60^\circ \)
\( 60^\circ + 90^\circ + \angle TQU + 90^\circ = 360^\circ \)
\( 240^\circ + \angle TQU = 360^\circ \)
\( \implies \angle TQU = 360^\circ - 240^\circ \)
\( \implies \angle TQU = 120^\circ \)
Since \( \angle TQU \) is the same as \( \angle PQR \) (interior angle of the parallelogram), then \( \angle PQR = 120^\circ \).
In a parallelogram, opposite angles are equal:
\( \angle PSR = \angle PQR = 120^\circ \)
Adjacent angles in a parallelogram are supplementary (sum to \( 180^\circ \)):
\( \angle QPS + \angle PQR = 180^\circ \)
\( \angle QPS + 120^\circ = 180^\circ \)
\( \implies \angle QPS = 180^\circ - 120^\circ \)
\( \implies \angle QPS = 60^\circ \)
Again, opposite angles are equal:
\( \angle QRS = \angle QPS = 60^\circ \)
Therefore, the angles of the parallelogram are \( 120^\circ \), \( 60^\circ \), \( 120^\circ \), and \( 60^\circ \).
In simple words: Imagine a parallelogram. If two lines (altitudes) drop from one corner, making right angles with the opposite sides, and the angle between these two dropped lines is 60 degrees, we can find the parallelogram's angles. By using the property that a small quadrilateral's angles add up to 360 degrees, we find one angle of the parallelogram is 120 degrees. Then, because opposite angles are equal and nearby angles add to 180 degrees, all the other angles can be figured out.

🎯 Exam Tip: When dealing with altitudes in a parallelogram, remember that they form a quadrilateral (like QTSU here) where two angles are \( 90^\circ \). Use the angle sum property of quadrilaterals to find the third angle, which often corresponds to an interior angle of the parallelogram.

Question 5. Can all the four angles a quadrilateral are obtuse? Give reason for your answer.
Answer: No, all four angles of a quadrilateral cannot be obtuse. An obtuse angle is an angle greater than \( 90^\circ \) but less than \( 180^\circ \). If all four angles of a quadrilateral were obtuse, their sum would be greater than \( 4 \times 90^\circ = 360^\circ \). However, the sum of the interior angles of any quadrilateral is exactly \( 360^\circ \). Since the sum would exceed \( 360^\circ \), it is not possible for all four angles to be obtuse.
In simple words: No, a four-sided shape cannot have all its angles bigger than 90 degrees. If they were all big, the total sum of the angles would be more than 360 degrees, but a quadrilateral's angles must always add up to exactly 360 degrees.

🎯 Exam Tip: Define "obtuse angle" and state the angle sum property of a quadrilateral clearly in your reasoning to get full marks.

Question 6. One angle of a quadrilateral is 108° and the other three angles are equal. Find the value of each of the equal angle.
Answer: Let the three equal angles of the quadrilateral be represented by \( x \). We know that the sum of the interior angles of a quadrilateral is \( 360^\circ \).
Given one angle is \( 108^\circ \).
So, we can set up the equation:
\( 108^\circ + x + x + x = 360^\circ \)
\( 108^\circ + 3x = 360^\circ \)
Now, subtract \( 108^\circ \) from both sides to find \( 3x \):
\( 3x = 360^\circ - 108^\circ \)
\( 3x = 252^\circ \)
To find \( x \), divide \( 252^\circ \) by 3:
\( x = \frac{252^\circ}{3} \)
\( x = 84^\circ \)
Therefore, each of the three equal angles is \( 84^\circ \). This problem is a good example of applying algebraic methods to geometric properties.
In simple words: If one angle is 108 degrees and the other three are the same, subtract 108 from 360 (the total for a quadrilateral). This leaves 252 degrees for the three equal angles. Then, divide 252 by 3 to find out what each of those equal angles is.

🎯 Exam Tip: Always start by clearly stating the known sum of angles for the polygon (e.g., \( 360^\circ \) for a quadrilateral) and then form an algebraic equation to solve for the unknown angles.

Question 7. ABCD is a trapezium in which AB || DC and ∠A = ∠B = 45°, then find the values of the angles C and D.
Answer: In a trapezium ABCD, if AB is parallel to DC (\( AB \parallel DC \)), then the adjacent angles between the parallel sides and a transversal (a non-parallel side) are supplementary. This means they add up to \( 180^\circ \).
Given \( \angle A = 45^\circ \) and \( \angle B = 45^\circ \).
For side AD (which is a transversal cutting parallel lines AB and DC):
\( \angle A + \angle D = 180^\circ \) (Sum of interior angles on the same side of a transversal)
\( 45^\circ + \angle D = 180^\circ \)
\( \implies \angle D = 180^\circ - 45^\circ \)
\( \implies \angle D = 135^\circ \)
For side BC (which is another transversal cutting parallel lines AB and DC):
\( \angle B + \angle C = 180^\circ \) (Sum of interior angles on the same side of a transversal)
\( 45^\circ + \angle C = 180^\circ \)
\( \implies \angle C = 180^\circ - 45^\circ \)
\( \implies \angle C = 135^\circ \)
So, \( \angle C = 135^\circ \) and \( \angle D = 135^\circ \). This type of trapezium, where non-parallel sides have equal base angles, is called an isosceles trapezium.
In simple words: In a trapezium, when two sides are parallel, the angles on the same side between these parallel lines always add up to 180 degrees. Since angles A and B are 45 degrees, angles D and C must each be 180 minus 45 degrees, which is 135 degrees.

🎯 Exam Tip: The key to solving problems involving trapeziums with parallel sides is to use the property of co-interior angles, which states that angles on the same side of a transversal add up to \( 180^\circ \).

Question 9. In a parallelogram ABCD, E and F are the points on the diagonals AC in such a way that AE = CF then show that BFDE is a parallelogram.
Answer: To prove that BFDE is a parallelogram, we need to show that its opposite sides are parallel and equal, or that its diagonals bisect each other. Let's use properties of parallelograms and triangle congruence.
Consider \( \triangle ABE \) and \( \triangle CDF \):
1. \( AB = CD \) (Opposite sides of parallelogram ABCD)
2. \( AE = CF \) (Given)
3. \( \angle BAE = \angle DCF \) (Alternate interior angles, since \( AB \parallel CD \) and AC is a transversal)
Therefore, by SAS (Side-Angle-Side) congruence rule, \( \triangle ABE \cong \triangle CDF \).
From congruence, \( BE = DF \) (Corresponding Parts of Congruent Triangles).
Similarly, consider \( \triangle ADE \) and \( \triangle CBF \):
1. \( AD = CB \) (Opposite sides of parallelogram ABCD)
2. \( AE = CF \) (Given)
3. \( \angle DAE = \angle BCF \) (Alternate interior angles, since \( AD \parallel BC \) and AC is a transversal)
Therefore, by SAS congruence rule, \( \triangle ADE \cong \triangle CBF \).
From congruence, \( DE = BF \) (Corresponding Parts of Congruent Triangles).
Since both pairs of opposite sides of quadrilateral BFDE are equal ( \( BE = DF \) and \( DE = BF \)), BFDE is a parallelogram. A quadrilateral is a parallelogram if its opposite sides are equal in length.
In simple words: We have a big parallelogram and two points on its main diagonal. We show that two pairs of triangles are identical (congruent). Because these triangles are identical, their matching sides are equal. This proves that the new shape formed by these points also has opposite sides that are equal, which means it is a parallelogram.

🎯 Exam Tip: The most common ways to prove a quadrilateral is a parallelogram are to show that: (a) both pairs of opposite sides are parallel, (b) both pairs of opposite sides are equal, (c) both pairs of opposite angles are equal, or (d) diagonals bisect each other. Choose the method that best fits the given information.

Question 10. In figure, ABCD is a parallelogram and lines AQ and CP are the bisector of ∠A – ∠C respectively. Prove the APCQ is a parallelogram.
Answer: Given that ABCD is a parallelogram, we know that opposite angles are equal, so \( \angle A = \angle C \).
Also, AQ is the bisector of \( \angle A \), meaning \( \angle PAQ = \frac{1}{2} \angle A \).
And CP is the bisector of \( \angle C \), meaning \( \angle PCQ = \frac{1}{2} \angle C \).
Since \( \angle A = \angle C \), it follows that \( \frac{1}{2} \angle A = \frac{1}{2} \angle C \).
So, \( \angle PAQ = \angle PCQ \).
Now, in parallelogram ABCD, \( AB \parallel DC \). Because of this, \( AP \parallel CQ \) (since P lies on AB and Q lies on DC).
If a pair of opposite sides of a quadrilateral are parallel and equal, then it is a parallelogram. Here, we have \( AP \parallel CQ \). Also, we can use the property that if opposite angles are equal and one pair of opposite sides are parallel, it can lead to the conclusion. A simpler way is to consider that since \( \angle A = \angle C \), and \( AB \parallel DC \), then \( \angle DAB + \angle ADC = 180^\circ \). When bisected, and considering the parallel lines, we can show that \( AQ \parallel PC \).
Let's refine the proof by showing that one pair of opposite sides (AQ and PC) are parallel.
Since \( AB \parallel DC \) (opposite sides of parallelogram), and AQ and PC are segments from these sides, then \( AQ \parallel PC \) if we can show that alternate interior angles formed by a transversal are equal. This is complex. Let's use the property that if opposite sides are parallel and equal, or if diagonals bisect each other. We already showed \( \angle PAQ = \angle PCQ \).
Also, \( AB \parallel DC \), so \( AP \parallel CQ \).
Consider \( \triangle APQ \) and \( \triangle CQP \).
Since \( AB \parallel DC \), and AQ and CP are transversals, this gives issues directly. Let's use another method.

Alternatively, we know that \( AQ \parallel PC \) if their alternate interior angles with respect to diagonal AC are equal. Let's reconsider.
Since \( ABCD \) is a parallelogram, \( AB \parallel DC \).
Since \( AQ \) bisects \( \angle A \) and \( CP \) bisects \( \angle C \), and \( \angle A = \angle C \), it implies \( \angleDAQ = \angle BC P \).
Also, \( AD \parallel BC \), so \( \angle DAQ = \angle CQA \) (alternate interior angles).
Similarly, \( \angle BCP = \angle APC \) (alternate interior angles).
Therefore, \( \angle CQA = \angle APC \). This implies \( AQ \parallel PC \).
We already know \( AP \parallel QC \) (as AB is parallel to DC and P, Q are points on them).
Since both pairs of opposite sides are parallel ( \( AP \parallel QC \) and \( AQ \parallel PC \) ), the quadrilateral APCQ is a parallelogram. A parallelogram has both pairs of opposite sides parallel to each other.
In simple words: We start with a parallelogram. Since opposite angles are equal, and the lines AQ and CP cut these angles in half, the halves are also equal. Because the big parallelogram's sides are parallel, it makes the lines AQ and CP parallel to each other. Also, since AP is part of AB and CQ is part of DC, and AB is parallel to DC, then AP is parallel to CQ. Since both pairs of opposite sides (AP and CQ, and AQ and CP) are parallel, the inner shape APCQ is also a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram, look for ways to show that both pairs of opposite sides are parallel. Using angle bisector properties along with alternate interior angles from the parent parallelogram is often effective.

Question 12. ABCD is a parallelogram and AP and CQ are the perpendiculars from A and C on its diagonal BD, respectively. Prove that AP = CQ.
Answer: To prove \( AP = CQ \), we can use the concept of the area of triangles within the parallelogram. In a parallelogram, the diagonal divides it into two congruent triangles, meaning they have equal areas. So, the area of \( \triangle ABD \) is equal to the area of \( \triangle CDB \).
Area of \( \triangle ABD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times AP \)
Area of \( \triangle CDB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times BD \times CQ \)
Since Area \( (\triangle ABD) = \) Area \( (\triangle CDB) \),
\( \frac{1}{2} \times BD \times AP = \frac{1}{2} \times BD \times CQ \)
We can cancel \( \frac{1}{2} \times BD \) from both sides:
\( \implies AP = CQ \)
Hence proved. This shows that the perpendicular distances from opposite vertices to a diagonal in a parallelogram are equal. This is a simple and elegant proof.
In simple words: In a parallelogram, the line going from one corner to the opposite (the diagonal) splits it into two triangles that are the exact same size. If you draw straight lines down from the other two corners onto this diagonal, these lines are the 'heights' of those two triangles. Since the triangles are the same size and share the same base (the diagonal), their heights must also be equal.

🎯 Exam Tip: Remember that the area of a triangle can be expressed as \( \frac{1}{2} \times \text{base} \times \text{height} \). If two triangles share the same base and have equal areas, their corresponding heights must be equal. This property is very useful for proofs.

Question 14. In figure. ABCD is a parallelogram. P and Q are the mid-points of opposite sides AB and DC of a parallelogram ABCD. Prove that PRQS is a parallelogram.
Answer: Given that ABCD is a parallelogram, we know that \( AB \parallel DC \) and \( AB = DC \).
Since P is the midpoint of AB, \( AP = PB = \frac{1}{2} AB \).
Since Q is the midpoint of DC, \( DQ = QC = \frac{1}{2} DC \).
Because \( AB = DC \), it follows that \( \frac{1}{2} AB = \frac{1}{2} DC \).
So, \( AP = QC \) and \( PB = DQ \).
Consider quadrilateral APCQ: Since \( AB \parallel DC \), then \( AP \parallel QC \). We also have \( AP = QC \). A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram. Therefore, APCQ is a parallelogram.
From APCQ being a parallelogram, we get \( AQ \parallel PC \). Thus, \( SQ \parallel PR \) (as S and R lie on AQ and PC respectively).

Now, consider quadrilateral PBQD:
Since \( AB \parallel DC \), then \( PB \parallel DQ \).
We also have \( PB = DQ \).
Therefore, PBQD is a parallelogram.
From PBQD being a parallelogram, we get \( PD \parallel BQ \). Thus, \( SP \parallel RQ \) (as S and R lie on PD and BQ respectively).

In quadrilateral PRQS, we have shown:
1. \( SQ \parallel PR \)
2. \( SP \parallel RQ \)
Since both pairs of opposite sides are parallel, PRQS is a parallelogram. This means that connecting midpoints of opposite sides in a parallelogram creates another parallelogram inside.
In simple words: We have a parallelogram ABCD. P is the middle of side AB, and Q is the middle of side DC. Since AB is parallel and equal to DC, and P and Q are midpoints, this means AP is parallel and equal to QC. So, APCQ is a parallelogram. This makes AQ parallel to PC. Similarly, PBQD is also a parallelogram, which means PD is parallel to BQ. Inside the big parallelogram, we have a new shape PRQS. Because AQ is parallel to PC (so SQ parallel to PR) and PD is parallel to BQ (so SP parallel to RQ), both pairs of opposite sides of PRQS are parallel. Therefore, PRQS is also a parallelogram.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram, showing that both pairs of opposite sides are parallel is a very effective method. Use the properties of the larger parallelogram (e.g., opposite sides are parallel and equal) to establish the conditions for the inner quadrilateral.