RBSE Solutions Class 9 Maths Chapter 9 Quadrilaterals Important Questions

Multiple Choice Questions

Question 1. In the given figure, ABCD is a parallelogram in which \( \angle ADC = 130^\circ \), then \( \angle CBE \) is equal to:
(a) 60°
(b) 130°
(c) 50°
(d) 70°
Answer: (c) 50°
In simple words: The angle \( \angle ADC \) is 130 degrees. In a parallelogram, consecutive angles add up to 180 degrees, so \( \angle DAB \) is 50 degrees. Since AB is parallel to DC, and AE is a transversal, \( \angle DAB \) and \( \angle CBE \) are corresponding angles if extended or using properties of exterior angles, leading to \( \angle CBE \) being 50 degrees.

🎯 Exam Tip: Remember that opposite angles in a parallelogram are equal, and consecutive angles are supplementary (add up to 180°).

Question 2. In the given figure, PQR is a triangle in which X and Y are the mid-point of the sides PR and QR respectively. If PQ = 6 cm, QR = 7 cm and PR = 8 cm, then XY is equal to:
(a) 12 cm
(b) 3.5 cm
(c) 3 cm
(d) 4 cm
Answer: (c) 3 cm
In simple words: X and Y are the middle points of two sides of the triangle. According to the Midpoint Theorem, the line segment connecting these two midpoints (XY) will be half the length of the third side (PQ). Since PQ is 6 cm, XY will be 3 cm.

🎯 Exam Tip: The Midpoint Theorem is key here: the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Question 4. If an angle of a parallelogram is two-third of its adjacent angle, then the smallest angle of the parallelogram is:
(a) 54°
(b) 72°
(c) 84°
(d) 108°
Answer: (b) 72°
In simple words: Let one angle be x and the adjacent angle be y. We know y = (2/3)x and also that adjacent angles in a parallelogram add up to 180 degrees (x + y = 180). By solving these two equations, we can find the values of x and y. The smaller of the two angles will be 72 degrees.

🎯 Exam Tip: For problems involving ratios or relationships between adjacent angles of a parallelogram, always use the property that their sum is 180 degrees.

Question 5. The ratio of the side and diagonal of a square is:
(a) \( 1:\sqrt{2} \)
(b) \( 3:\sqrt{2} \)
(c) \( \sqrt{2}:1 \)
(d) \( \sqrt{2}:3 \)
Answer: (a) \( 1:\sqrt{2} \)
In simple words: If a square has a side length of 's', then its diagonal can be found using the Pythagorean theorem as \( s\sqrt{2} \). The ratio of the side to the diagonal is then \( s : s\sqrt{2} \), which simplifies to \( 1 : \sqrt{2} \).

🎯 Exam Tip: Remember the special relationship in a square: the diagonal is always \( \sqrt{2} \) times the side length. This is a common geometric fact.

Question 6. In the given figure, ABCD is a rhombus. If \( \angle OAB = 35^\circ \), then the value of x is:
(A) 25°
(B) 35°
(C) 55°
(D) 70°
Answer: (C) 55°
In simple words: In a rhombus, the diagonals cross each other at a right angle (90 degrees). So, in the triangle AOB, angle AOB is 90 degrees. We are given angle OAB as 35 degrees. Since the angles in a triangle add up to 180 degrees, we can find angle ABO (which is x) by subtracting 90 and 35 from 180.

🎯 Exam Tip: Key properties of a rhombus include all sides being equal, opposite angles being equal, and diagonals bisecting each other at right angles.

Question 7. If the two adjacent angles of a parallelogram are \( (3x - 20^\circ) \) and \( (50^\circ - x) \), then the value of x is:
(A) 55°
(B) 75°
(C) 20°
(D) 80°
Answer: (B) 75°
In simple words: Adjacent angles in a parallelogram always add up to 180 degrees. So, we can set up an equation where \( (3x - 20) + (50 - x) = 180 \). By solving this equation, we will find that the value of x is 75.

🎯 Exam Tip: Remember the key property: the sum of any two consecutive (adjacent) angles of a parallelogram is 180 degrees.

Question 8. In the given figure, ABCD is a quadrilateral and P, Q, R and S are the mid-points of the sides AB, BC, CD, DA respectively. If BD =12 cm, then length of QR is:
(A) 6 cm
(B) 8 cm
(C) 3 cm
(D) 4 cm
Answer: (A) 6 cm
In simple words: Since Q and R are the mid-points of sides BC and CD respectively, the line segment QR connects these mid-points in triangle BCD. According to the Midpoint Theorem, QR will be parallel to the third side BD and exactly half its length. Since BD is 12 cm, QR will be 6 cm.

🎯 Exam Tip: Apply the Midpoint Theorem to the relevant triangle (BCD in this case) to find the relationship between the segment connecting midpoints and the third side.

Question 10. In a quadrilateral PQRS, \( \angle P, \angle Q, \angle R \) and \( \angle S \) are interior angles. If \( \angle P:\angle Q:\angle R:\angle S = 1:2:3:4 \), then which angle is equal to 144°.
(A) \( \angle P \)
(B) \( \angle Q \)
(C) \( \angle R \)
(D) \( \angle S \)
Answer: (D) \( \angle S \)
In simple words: The sum of all angles in any quadrilateral is 360 degrees. We are given the ratio of the angles as 1:2:3:4. Let the angles be \( x, 2x, 3x, \) and \( 4x \). If we add these up and set them equal to 360, we find \( 10x = 360 \), which means \( x = 36 \). Then, \( \angle S = 4x = 4 \times 36 = 144 \) degrees.

🎯 Exam Tip: Always remember that the sum of interior angles of a quadrilateral is 360 degrees. Use this fact when dealing with ratios of angles.

Very Short Answer Type Questions

Question 1. Two consecutive angles of a parallelogram are in the ratio 1 : 3. Find the smallest angle.
Answer: Let the two consecutive angles be \( x \) and \( 3x \).
Since the sum of two consecutive angles of a parallelogram is 180°, we have:
\( x + 3x = 180^\circ \)
\( \implies 4x = 180^\circ \)
\( \implies x = \frac{180^\circ}{4} \)
\( \implies x = 45^\circ \)
The angles are \( 45^\circ \) and \( 3 \times 45^\circ = 135^\circ \). The smallest angle is 45°. This helps understand how angles are distributed in such a shape.
In simple words: In a parallelogram, two angles next to each other always add up to 180 degrees. If their ratio is 1:3, then the angles are 45 degrees and 135 degrees. The smaller one is 45 degrees.

🎯 Exam Tip: When given a ratio for angles, represent them as \( x, 2x, 3x \), etc., and use the properties of the shape (e.g., sum of angles, parallel lines) to form an equation.

Question 3. The three angles of a quadrilateral measure 56°, 100° and 88°. Find the measure of the fourth angle.
Answer: Let the measure of the fourth angle be \( x \).
As we know that sum of the angles of a quadrilateral is 360°.
\( 56^\circ + 100^\circ + 88^\circ + x = 360^\circ \)
\( \implies 244^\circ + x = 360^\circ \)
\( \implies x = 360^\circ - 244^\circ \)
\( \implies x = 116^\circ \)
Hence, the measure of the fourth angle is 116°. Every four-sided shape follows this fundamental rule.
In simple words: A four-sided shape's angles always add up to 360 degrees. If you have three angles, add them up and then subtract that total from 360 to find the missing fourth angle.

🎯 Exam Tip: Always recall the sum of interior angles for common polygons: triangle (180°), quadrilateral (360°), pentagon (540°), etc.

Question 4. In a quadrilateral ABCD, \( \angle A = 3x, \angle B = 5x, \angle C = 20x \) and \( \angle D = 8x \). Find the value of x.
Answer: The sum of the angles of a quadrilateral is 360°.
\( \implies \angle A + \angle B + \angle C + \angle D = 360^\circ \)
\( \implies 3x + 5x + 20x + 8x = 360^\circ \)
\( \implies 36x = 360^\circ \)
\( \implies x = \frac{360^\circ}{36} \)
\( \implies x = 10^\circ \)
Therefore, the value of x is 10 degrees. This constant sum helps us find unknown angle parts.
In simple words: All the angles inside a four-sided shape add up to 360 degrees. If each angle is given as a multiple of 'x', you can add up all the 'x' values and set them equal to 360 to find what 'x' is.

🎯 Exam Tip: When angles are expressed in terms of a variable, add them up and equate to the total angle sum of the polygon to solve for the variable.

Question 5. An angle is inclined to one side of the rectangle at 25°. Then find the acute angle between the diagonals.
Answer: Let's consider a rectangle with diagonals intersecting at O. If one angle formed by a diagonal and a side is 25°, for example, \( \angle ODC = 25^\circ \).
In a rectangle, diagonals are equal and bisect each other. So, \( OD = OC \).
This means \( \triangle OCD \) is an isosceles triangle.
Therefore, \( \angle OCD = \angle ODC = 25^\circ \).
Now, in \( \triangle OCD \), the sum of angles is 180°:
\( \angle ODC + \angle OCD + \angle COD = 180^\circ \)
\( \implies 25^\circ + 25^\circ + \angle COD = 180^\circ \)
\( \implies 50^\circ + \angle COD = 180^\circ \)
\( \implies \angle COD = 180^\circ - 50^\circ \)
\( \implies \angle COD = 130^\circ \)
The acute angle between the diagonals is the smaller of the two angles formed by their intersection. Since angles on a straight line add up to 180°, the acute angle \( \angle DOA \) is:
\( \angle DOA = 180^\circ - \angle COD \)
\( \implies \angle DOA = 180^\circ - 130^\circ \)
\( \implies \angle DOA = 50^\circ \)
The acute angle between the diagonals is 50°. The isosceles triangle property is crucial here.
In simple words: In a rectangle, when a diagonal meets a side, it forms an angle (here, 25 degrees). Since the diagonals cut each other in half and are equal, this creates a special triangle where two sides are the same length. This means two angles in that triangle are also 25 degrees. The third angle in that triangle can be found, and then the smaller angle formed where the diagonals cross will be 50 degrees.

🎯 Exam Tip: Remember that the diagonals of a rectangle are equal and bisect each other. This creates isosceles triangles at the intersection, which is key to solving angle problems.

Question 6. In the given figure, BCPQ and BCDA are two parallelograms on the same base BC. Find the value of x + y.
Answer: Given that BCDA is a parallelogram.
In a parallelogram, opposite angles are equal.
So, \( \angle A = \angle BCD \). From the figure, \( \angle A = y \).
Thus, \( y = 100^\circ \). The sum of angles in a parallelogram is 360 degrees.
Also, BCPQ is a parallelogram sharing the same base BC.
In parallelogram BCDA, consecutive angles sum to 180°, so \( \angle ABC + \angle BCD = 180^\circ \).
Since \( \angle BCD = 100^\circ \), then \( \angle ABC = 80^\circ \).
In parallelogram BCPQ, \( \angle CBQ = 120^\circ \).
\( \angle ABQ = \angle ABC + \angle CBQ = 80^\circ + 120^\circ = 200^\circ \). (This is not directly useful here, focus on exterior angle property)
The angle \( \angle PC D \) in parallelogram BCDA is \( \angle C \). Also, \( \angle BCP \) in parallelogram BCPQ is opposite \( \angle BQP \).
Let's use the exterior angle property for a triangle formed by extending a side or the sum of angles on a straight line.
From the figure, at point C, the angles \( 100^\circ \) and \( x \) are involved.
Since BCDA is a parallelogram, \( BC || AD \). So \( \angle ABC \) and \( \angle DAB \) are consecutive angles.
In parallelogram BCPQ, \( BC || PQ \).
Angles \( \angle BCD = 100^\circ \) (from the figure). In parallelogram BCDA, \( y = \angle A = 100^\circ \) (opposite angles).
Now, consider point C in the figure. We have \( \angle BCD = 100^\circ \). Also, there is an angle \( 120^\circ \) at Q.
Let's assume the angle \( \angle BC P \) in parallelogram BCPQ is part of the 120-degree angle shown at Q or that \( \angle BQC \) is \( 120^\circ \).
If \( \angle BCD = 100^\circ \), then \( \angle CBD \) is not directly given. However, the angles \( 100^\circ \) and \( 120^\circ \) are depicted within a combined shape.
Let's re-examine the image interpretation. The \( 100^\circ \) is \( \angle D \) in ABCD and \( y \) is \( \angle A \). In parallelogram BCDA, \( \angle A = \angle C_{BCD} = y = 100^\circ \). So \( \angle BCD = 100^\circ \).
The angle \( 120^\circ \) appears to be \( \angle CBQ \), or an angle at Q. Assuming it's \( \angle CBQ = 120^\circ \).
Then in parallelogram BCPQ, \( \angle CQP \) should be \( 120^\circ \).
The angle x is shown as \( \angle CPQ \).
Let's assume the angles shown as \( 100^\circ \) and \( 120^\circ \) are \( \angle ADC \) and \( \angle CBQ \) respectively.
From BCDA (parallelogram): \( \angle A = \angle BCD \) (opposite angles), so \( y = \angle BCD \).
Also, consecutive angles sum to 180°, so \( \angle ADC + \angle BCD = 180^\circ \). If \( \angle ADC = 100^\circ \), then \( \angle BCD = 80^\circ \). This implies \( y = 80^\circ \).
From BCPQ (parallelogram): \( \angle CBQ = 120^\circ \) (from diagram). Then \( \angle CPQ = x = 120^\circ \) (opposite angles).
So \( x = 120^\circ \) and \( y = 80^\circ \).
Then \( x + y = 120^\circ + 80^\circ = 200^\circ \).
Let's re-interpret the OCR's provided "Solution".
"BCDA is a parallelogram \( \angle A = y = 100^\circ \) (opposite angles of a parallelogram are equal)". This means \( \angle D = 100^\circ \). So \( y = 100^\circ \).
"Also \( x + 100^\circ = 120^\circ \) (exterior angle property)". This line implies that \( 120^\circ \) is an exterior angle and \( x \) and \( 100^\circ \) are interior angles of a triangle. This structure is not clear from the diagram itself.
Let's follow the solution text exactly: if \( y = 100^\circ \) and \( x + 100^\circ = 120^\circ \), then \( x = 20^\circ \).
Then \( x + y = 20 + 100 = 120^\circ \).
The diagram in the OCR is difficult to precisely map to the values and properties in the solution. Assuming the solution's logic for \( x \) and \( y \): \( \angle A = y = 100^\circ \) (opposite to the given 100° in the diagram, which is probably \( \angle D \)). And \( x \) is an angle related to \( 120^\circ \) by exterior angle property. The 120° in the diagram is \( \angle QCB \). If \( \angle QCB = 120^\circ \), then in parallelogram BCPQ, \( \angle BPQ = 120^\circ \) and \( \angle BPC = 60^\circ \).
If \( y = 100^\circ \), it implies \( \angle BCD = 100^\circ \). The exterior angle property \( x+100=120 \) suggests x and 100 are interior angles of a triangle and 120 is the exterior. This doesn't seem to fit the parallelograms easily.
Let's use the interpretation from the OCR solution: \( y = 100^\circ \). The angle \( 120^\circ \) is shown as an interior angle at B or Q. If \( \angle QBC = 120^\circ \), then \( \angle QPC = 120^\circ \). If \( \angle BQC \) or \( \angle PQC \) is \( 120^\circ \).
Let's assume the angles as interpreted by the solution: \( \angle A = y \) and \( \angle D = 100^\circ \). Then in parallelogram BCDA, \( y = 100^\circ \).
Now consider the triangle formed by line segment CD (from BCDA) and PQ (from BCPQ). The angle labeled \( 120^\circ \) appears to be \( \angle QCB \).
The angle \( x \) is \( \angle CPD \).
If \( y = 100^\circ \) (angle A or BCD), and \( 120^\circ \) is \( \angle BQP \), then \( x \) is a part of \( \angle BQP \).
The provided solution ( \( x+100 = 120 \)) suggests that \( x \) is an angle in a triangle, and \( 100 \) is another, and \( 120 \) is the exterior. This implies \( \angle QCP = 100^\circ \), and \( \angle CQD = x \), and \( \angle PQC \) is exterior to triangle CDQ.
This is confusing. Let's assume the OCR solution implies \( \angle A = y = 100^\circ \). And perhaps the \( 120^\circ \) is an external angle to a triangle where \( x \) and \( 100^\circ \) are internal. This is not obvious from the figure. Without a clearer diagram or explanation of the 120-degree angle and x, it's hard to verify the solution's steps.
However, following the provided steps: \( y = 100^\circ \). And \( x + 100^\circ = 120^\circ \), which means \( x = 20^\circ \). Then \( x + y = 120^\circ \). This logic must be coming from a specific interpretation of angle positions in the original full diagram.
Let's present the steps exactly as given in the solution, maintaining the numbers.
Given: BCDA is a parallelogram.
\( \angle A = y = 100^\circ \) (opposite angles of a parallelogram are equal).
Also, \( x + 100^\circ = 120^\circ \) (exterior angle property).
\( \implies x = 120^\circ - 100^\circ \)
\( \implies x = 20^\circ \).
Therefore, \( x + y = 20^\circ + 100^\circ = 120^\circ \). This shows how properties of angles help in finding combined values.
In simple words: We have two parallelograms sharing a side. In a parallelogram, opposite angles are equal, so one angle (y) is 100 degrees. Another angle (x) can be found using the exterior angle rule which states an exterior angle equals the sum of the two opposite interior angles. If an exterior angle is 120 degrees and one interior angle is 100 degrees, then x must be 20 degrees. So, x plus y is 120 degrees.

🎯 Exam Tip: Pay close attention to how angles are labeled and their relationship (e.g., opposite, consecutive, interior, exterior) in diagrams involving multiple connected shapes.

Question 7. "The opposite sides of a parallelogram are equal”. Prove it.
Answer: Given: A parallelogram ABCD.
To prove: \( AB = CD \) and \( BC = DA \).
Construction: Join diagonal BD.
Proof: In parallelogram ABCD, we know that \( AB || DC \) and \( AD || BC \).
Consider \( \triangle ABD \) and \( \triangle CDB \).
Since \( AB || DC \), and BD is a transversal, \( \angle ABD = \angle CDB \) (alternate interior angles).
Since \( AD || BC \), and BD is a transversal, \( \angle ADB = \angle CBD \) (alternate interior angles).
BD = BD (common side).
Therefore, by Angle-Side-Angle (ASA) congruence rule, \( \triangle ABD \cong \triangle CDB \).
Since the triangles are congruent, their corresponding parts are equal (CPCTC):
\( AB = CD \)
And \( AD = BC \).
Hence, the opposite sides of a parallelogram are equal. This proof relies on fundamental geometric properties of parallel lines and triangles.
In simple words: To show that opposite sides of a parallelogram are equal, we can draw a line (diagonal) across it. This splits the parallelogram into two triangles. Because the sides are parallel, we can use a rule about alternate angles being equal. Then, using triangle congruence rules, we can prove that these two triangles are exactly the same, which means their matching sides are also equal.

🎯 Exam Tip: For proofs, clearly state the given, what to prove, construction (if any), and then logically deduce the conclusion using theorems like alternate interior angles and congruence rules.

Question 8. In figure, D and E are the mid-points of the sides AB and AC respectively of \( \triangle ABC \). If BC = 6.4 cm, find DE.
Answer: Given that D and E are mid-points of sides AB and AC respectively in \( \triangle ABC \).
According to the Mid-point Theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of its length.
Therefore,
\( DE = \frac{1}{2} BC \)
\( DE = \frac{1}{2} \times 6.4 \text{ cm} \)
\( \implies DE = 3.2 \text{ cm} \)
Thus, DE is 3.2 cm. This theorem provides a direct way to find the length of this segment.
In simple words: If you connect the middle points of two sides of a triangle, that new line will be half as long as the third side of the triangle. Here, the third side (BC) is 6.4 cm, so the line connecting the mid-points (DE) will be 3.2 cm.

🎯 Exam Tip: The Mid-point Theorem is a fundamental concept for problems involving midpoints of triangle sides; apply it directly to find unknown lengths.

Question 9. In the given figure, PQRS is a rectangle. If \( \angle RPQ = 30^\circ \) then find the value of \( (x + y) \).
Answer: In a rectangle PQRS, the diagonals PR and QS are equal and bisect each other. Let them intersect at O.
So, \( PO = OQ = OR = OS \).
Given \( \angle RPQ = 30^\circ \).
Consider \( \triangle POQ \). Since \( PO = OQ \), \( \triangle POQ \) is an isosceles triangle.
In \( \triangle PQR \), \( \angle PQR = 90^\circ \) (angle of a rectangle).
So, \( \angle PRQ = 90^\circ - \angle RPQ = 90^\circ - 30^\circ = 60^\circ \).
Since \( SR || PQ \) (opposite sides of a rectangle), and PR is a transversal, \( \angle SRP = \angle RPQ = 30^\circ \) (alternate interior angles).
Similarly, \( \angle QSP = \angle PRS = 30^\circ \) and \( \angle RQS = \angle PRQ = 60^\circ \).
Now, in \( \triangle SOR \), \( OS = OR \), so it's an isosceles triangle.
\( y = \angle SRO = \angle RSO = 30^\circ \). (The y shown in the diagram is \( \angle SOR \)).
Let's re-evaluate the angles x and y from the diagram. It shows x as \( \angle POQ \) and y as \( \angle SRO \).
Since \( \angle SRO = \angle RPQ = 30^\circ \) (alternate interior angles if SP || RQ and PR is transversal, which is not the case for SRO).
However, \( \angle SRQ = 90^\circ \). If \( \angle SRO \) is y, then in \( \triangle SOR \), \( OS=OR \), so \( \angle OSR = \angle ORS = y \).
\( \angle QPR = 30^\circ \). In \( \triangle PQR \), \( \angle PQR = 90^\circ \). So \( \angle PRQ = 90^\circ - 30^\circ = 60^\circ \).
Now, \( PQ || SR \), so \( \angle QPR = \angle PRS = 30^\circ \) (alternate interior angles). So \( y = 30^\circ \).
In \( \triangle POQ \), \( PO = OQ \), so \( \angle OPQ = \angle OQP = 30^\circ \).
Therefore, \( x = \angle POQ = 180^\circ - (\angle OPQ + \angle OQP) = 180^\circ - (30^\circ + 30^\circ) = 180^\circ - 60^\circ = 120^\circ \).
So, \( x = 120^\circ \) and \( y = 30^\circ \).
Then \( x + y = 120^\circ + 30^\circ = 150^\circ \). This combines properties of rectangles and triangles effectively.
*The source's solution for Question 9 is missing (only shows the image). My derivation above uses standard rectangle/triangle properties.*
In simple words: In a rectangle, the diagonal lines are equal and cut each other in half. This creates special triangles where two sides are equal. If one angle formed by a diagonal and a side is 30 degrees, we can use triangle rules and parallel line rules to find the other angles. The angle 'y' (formed by a diagonal and a side) will be 30 degrees. The angle 'x' (where the diagonals cross) will be 120 degrees. Adding them gives 150 degrees.

🎯 Exam Tip: Remember that diagonals of a rectangle are equal and bisect each other, forming isosceles triangles. Use alternate interior angles with parallel sides and the sum of angles in a triangle.

Question 10. In the given figure, AD and BE are medians of \( \triangle ABC \) and BE || DF. Prove that \( CF = \frac{1}{4}AC \).
Answer: Given: AD and BE are medians of \( \triangle ABC \). So D is the mid-point of BC, and E is the mid-point of AC.
Also given: \( BE || DF \).
To prove: \( CF = \frac{1}{4} AC \).
Proof:
1. Consider \( \triangle BEC \).
D is the mid-point of BC (since AD is a median).
\( DF || BE \) (given).
By the converse of the Mid-point Theorem, if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side.
Therefore, DF bisects EC. This means F is the mid-point of EC.
So, \( CF = FE = \frac{1}{2} EC \).
2. We know that BE is a median to side AC, so E is the mid-point of AC.
This means \( AE = EC = \frac{1}{2} AC \).
3. Now, substitute the value of EC from step 2 into the equation from step 1:
\( CF = \frac{1}{2} EC \)
\( \implies CF = \frac{1}{2} \left( \frac{1}{2} AC \right) \)
\( \implies CF = \frac{1}{4} AC \).
Hence, it is proved that \( CF = \frac{1}{4} AC \). This demonstration links medians, midpoints, and parallel lines in a systematic way.
In simple words: In triangle ABC, AD and BE are lines that go from a corner to the middle of the opposite side (medians). We are told that line DF is parallel to line BE. Because D is the middle of BC and DF is parallel to BE, F must be the middle of EC. Since E is the middle of AC, it means EC is half of AC. So, if CF is half of EC, and EC is half of AC, then CF must be one-quarter of AC.

🎯 Exam Tip: Problems involving medians and parallel lines often use the Mid-point Theorem or its converse. Clearly identify the triangles and midpoints to apply the theorem correctly.

Long Answer Type Questions

Question 1. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Answer: Given: ABCD is a quadrilateral in which diagonals AC and BD are equal (AC = BD) and bisect each other at right angles (intersect at O such that \( AO=OC \), \( BO=OD \) and \( \angle AOB = \angle BOC = \angle COD = \angle DOA = 90^\circ \)).
To prove: ABCD is a square.
Proof:
1. Since the diagonals bisect each other, ABCD is a parallelogram.
2. Since the diagonals of the parallelogram ABCD bisect each other at right angles, ABCD is a rhombus. This is a special property where all sides become equal.
Thus, \( AB = BC = CD = DA \).
3. Now, we need to show that one angle is 90°. Consider \( \triangle ABC \) and \( \triangle BAD \).
\( AB = BA \) (Common side).
\( BC = AD \) (Sides of a rhombus are equal).
\( AC = BD \) (Given that diagonals are equal).
Therefore, by Side-Side-Side (SSS) congruence rule, \( \triangle ABC \cong \triangle BAD \).
By Corresponding Parts of Congruent Triangles (CPCTC), \( \angle ABC = \angle BAD \).
4. In a parallelogram, consecutive angles are supplementary (add up to 180°).
So, \( \angle ABC + \angle BAD = 180^\circ \).
Since \( \angle ABC = \angle BAD \), we have \( 2 \angle BAD = 180^\circ \).
\( \implies \angle BAD = 90^\circ \).
5. Since ABCD is a rhombus (all sides equal) and one of its angles is 90°, it means all its angles are 90° (opposite angles are equal, consecutive angles are supplementary).
Therefore, ABCD is a square. This step-by-step process shows how specific diagonal properties lead to a square.
In simple words: If a four-sided shape has diagonals that are equal, cut each other in half, and cross at a perfect right angle, then that shape is a square. We first prove it's a parallelogram because diagonals bisect each other. Then, because they meet at right angles, it's a rhombus (all sides equal). Finally, since the diagonals are equal, we can show that one of its corner angles is 90 degrees, which makes the rhombus a square.

🎯 Exam Tip: To prove a quadrilateral is a square, you must demonstrate it has properties of both a rhombus (all sides equal) and a rectangle (all angles 90°). Diagonals are crucial for this.

Question 2. In a trapezium ABCD, E is the mid-point of AD and EF || AB. Prove that \( EF = \frac{1}{2} (AB + DC) \).
Answer: Given: ABCD is a trapezium with \( AB || DC \). E is the mid-point of AD, and \( EF || AB \).
To prove: \( EF = \frac{1}{2} (AB + DC) \).
Construction: Join B to D, intersecting EF at G.
Proof:
1. In \( \triangle DAB \):
E is the mid-point of AD (given).
\( EG || AB \) (since \( EF || AB \), then part EG is also parallel to AB).
By the converse of the Mid-point Theorem, G is the mid-point of BD.
Also, \( EG = \frac{1}{2} AB \) (Mid-point Theorem). Call this Equation (i). This shows the relationship within the first triangle.
2. In \( \triangle BCD \):
G is the mid-point of BD (proved above).
Since \( EF || AB \) and \( AB || DC \), it follows that \( EF || DC \). Therefore, \( GF || DC \).
By the converse of the Mid-point Theorem, F is the mid-point of BC.
Also, \( GF = \frac{1}{2} DC \) (Mid-point Theorem). Call this Equation (ii). This shows the relationship within the second triangle.
3. Now, add Equation (i) and Equation (ii):
\( EG + GF = \frac{1}{2} AB + \frac{1}{2} DC \)
\( \implies EF = \frac{1}{2} (AB + DC) \).
Hence, proved. This demonstrates how to break down a complex shape into simpler triangles to apply known theorems.
In simple words: Imagine a trapezoid (a shape with two parallel sides). If you draw a line from the middle of one non-parallel side, parallel to the two main sides, that line will go to the middle of the other non-parallel side. The length of this middle line (EF) is exactly half the sum of the two parallel sides (AB and DC). We prove this by adding a diagonal line inside the trapezoid, which creates two triangles, and then using the Midpoint Theorem on each triangle.

🎯 Exam Tip: When dealing with trapeziums and midpoints, adding a diagonal often helps split the trapezium into two triangles, allowing you to apply the Mid-point Theorem effectively.

Question 3. In the given figure, PQRS is a parallelogram in which PQ is produced to T such that QT = PQ. Prove that ST bisects RQ.
Answer: Given: PQRS is a parallelogram. PQ is produced to T such that \( QT = PQ \).
To prove: ST bisects RQ (meaning the intersection point of ST and RQ is the midpoint of RQ). Let's call the intersection point O.
Proof:
1. Since PQRS is a parallelogram, \( PQ || SR \) and \( PQ = SR \).
2. We are given \( QT = PQ \). Combining with \( PQ = SR \), we get \( QT = SR \).
3. Consider \( \triangle QOT \) and \( \triangle ROS \).
Since \( PQ || SR \), and RQ is a transversal, \( \angle OQT = \angle ORS \) (alternate interior angles).
Also, \( PQ || SR \), and ST is a transversal, \( \angle QTO = \angle RSO \) (alternate interior angles).
We have \( QT = SR \) (from step 2).
Therefore, by Angle-Side-Angle (ASA) congruence rule, \( \triangle QOT \cong \triangle ROS \).
4. Since the triangles are congruent, their corresponding parts are equal (CPCTC):
\( QO = RO \).
This means O is the mid-point of RQ.
Thus, ST bisects RQ. This proof effectively uses properties of parallelograms and triangle congruence.
In simple words: We have a parallelogram PQRS. One side (PQ) is extended to T, making the extension (QT) the same length as PQ. We need to show that the line ST cuts the line RQ exactly in half. We do this by looking at two triangles formed by the lines, \( \triangle QOT \) and \( \triangle ROS \). Since PQRS is a parallelogram, opposite sides are parallel and equal. This lets us use rules about parallel lines and alternate angles. By showing these two triangles are identical, we can prove that the line ST cuts RQ at its midpoint.

🎯 Exam Tip: When a problem asks to prove bisection, aim to prove congruence of triangles such that the bisected segment's parts are corresponding sides of the congruent triangles.

Question 1. Show that the line segment joining the mid-points of the opposite sides of a quadrilateral bisect each other.
Answer: Given: ABCD is a quadrilateral. P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. PR and SQ are the line segments joining the mid-points of opposite sides.
To prove: PR and SQ bisect each other.
Construction: Join PQ, QR, RS, SP, and AC.
Proof:
1. In \( \triangle ABC \):
P is the mid-point of AB and Q is the mid-point of BC.
By the Mid-point Theorem, \( PQ || AC \) and \( PQ = \frac{1}{2} AC \). Call this Equation (i).
2. In \( \triangle ADC \):
S is the mid-point of AD and R is the mid-point of CD.
By the Mid-point Theorem, \( SR || AC \) and \( SR = \frac{1}{2} AC \). Call this Equation (ii).
3. From Equation (i) and Equation (ii):
We have \( PQ || AC \) and \( SR || AC \), so \( PQ || SR \).
We also have \( PQ = \frac{1}{2} AC \) and \( SR = \frac{1}{2} AC \), so \( PQ = SR \).
4. Since one pair of opposite sides of quadrilateral PQRS (PQ and SR) are both parallel and equal, PQRS is a parallelogram.
5. We know that the diagonals of a parallelogram bisect each other. Here, PR and SQ are the diagonals of parallelogram PQRS.
Therefore, PR and SQ bisect each other. This shows a property of quadrilaterals based on their midpoints.
In simple words: If you take any four-sided shape and find the middle points of all its sides, then connect the middle points of opposite sides (drawing two lines), those two connecting lines will always cut each other exactly in half. We prove this by drawing another line (a diagonal) across the original shape, which helps to form a new parallelogram by connecting the midpoints. Since the diagonals of any parallelogram cut each other in half, the proof is complete.

🎯 Exam Tip: The Mid-point Theorem is essential here; remember it states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

Question 4. In triangle ABC, P, Q and R are respectively the mid-points of sides BC, CA and AB. PR and BQ meet at X. CR and PQ meet at Y. Prove that \( XY = \frac{1}{4} BC \).
Answer: Given: \( \triangle ABC \) with P, Q, R as mid-points of BC, CA, AB respectively. PR and BQ meet at X. CR and PQ meet at Y.
To prove: \( XY = \frac{1}{4} BC \).
Proof:
1. In \( \triangle ABC \):
R is the mid-point of AB, and P is the mid-point of BC.
By the Mid-point Theorem, \( RP || AC \) and \( RP = \frac{1}{2} AC \).
Also, Q is the mid-point of AC. So, \( RQ || BC \) and \( RQ = \frac{1}{2} BC \).
2. Consider the quadrilateral RPQC.
Since R is midpoint of AB and Q is midpoint of AC, \( RQ || BC \) (from midpoint theorem). Also P is on BC.
This makes RPQC a trapezium (or even a parallelogram if RP=QC, which is true as RP=1/2AC and QC=1/2AC).
\( R \) is mid-point of AB, \( Q \) is mid-point of AC. So \( RQ || BC \).
Also \( P \) is mid-point of BC, so \( BP = PC \).
In \( \triangle ABQ \), R is the mid-point of AB. X is the intersection of BQ and PR.
In \( \triangle BQC \), P is the mid-point of BC.
Consider \( \triangle ABQ \). R is the mid-point of AB. RP is parallel to AQ (a part of AC). This is incorrect as RP is parallel to AC (whole side).
Let's focus on \( \triangle BPQ \).
P is the mid-point of BC.
Q is the mid-point of AC.
R is the mid-point of AB.
From Midpoint Theorem, \( PQ || AB \Rightarrow PQ || AR \). And \( PR || AC \Rightarrow PR || AQ \).
Therefore, quadrilateral ARP Q is a parallelogram.
Its diagonals RQ and AP bisect each other.
Also, \( RQ || BC \). Since \( P \) is the midpoint of BC, \( RQ || BP \).
And \( RQ = \frac{1}{2} BC = BP \).
So, BPQR is a parallelogram.
Its diagonals PR and BQ bisect each other. Thus, X is the mid-point of PR.
Similarly, PCQR is a parallelogram (since \( PQ || RC \) and \( PQ = RC \) because \( PQ = \frac{1}{2} AB \) and \( RC = \frac{1}{2} AB \)).
Its diagonals CR and PQ bisect each other. Thus, Y is the mid-point of PQ.
3. Now, consider \( \triangle RPQ \).
X is the mid-point of PR.
Y is the mid-point of PQ.
By the Mid-point Theorem in \( \triangle RPQ \):
\( XY || RQ \) and \( XY = \frac{1}{2} RQ \).
4. We already know that \( RQ = \frac{1}{2} BC \) (from step 1).
Substitute this into the equation from step 3:
\( XY = \frac{1}{2} \left( \frac{1}{2} BC \right) \)
\( \implies XY = \frac{1}{4} BC \).
Hence, proved. This demonstrates a recursive application of the midpoint theorem.
In simple words: In a triangle, P, Q, and R are the middle points of the three sides. When we connect these midpoints in certain ways, we form new lines and triangles. By using the Midpoint Theorem multiple times, we can show that the line segment XY is half the length of RQ. Since RQ is itself half the length of BC, it means XY is one-quarter the length of BC.

🎯 Exam Tip: For complex midpoint problems, first establish parallelograms or apply the Mid-point Theorem to larger triangles, then apply it again to smaller triangles formed by the midpoints.