CBSE Class 9 Mathematics Quadrilaterals Worksheet Set 04

Question 1. From the figure parallelogram PQRS, the values of \( \angle SQP \) and \( \angle QSP \) are
(a) 45°
(b) 60°, 45°
(c) 45°, 45°.
(d) 60°, 60°.
Answer: (b) 60°, 45°
From the given parallelogram \( PQRS \), the angle \( \angle SQP \) is directly indicated as \( 60^\circ \). In triangle \( PQS \), the sum of all interior angles equals \( 180^\circ \):
\( \angle P + \angle SQP + \angle QSP = 180^\circ \)
\( 75^\circ + 60^\circ + \angle QSP = 180^\circ \)
\( 135^\circ + \angle QSP = 180^\circ \)
\( \implies \angle QSP = 180^\circ - 135^\circ = 45^\circ \)
Thus, \( \angle SQP = 60^\circ \) and \( \angle QSP = 45^\circ \).
In simple words: The triangle contains angles adding to 180 degrees, so subtracting 75 and 60 from 180 gives the missing angle of 45 degrees.

Exam Tip: Be sure to write the angles in the exact sequence requested by the question to avoid selecting reversed values.

Question 2. Which of the following statements is true ?
(a) The diagonals of a rectangle are perpendicular.
(b) The diagonals of a rhombus are equal.
(c) Every square is not a rhombus.
(d) None of these
Answer: (d) None of these
A rectangle has equal diagonals, but they are generally not perpendicular. A rhombus possesses perpendicular diagonals, yet they are typically unequal in length. Furthermore, every square fulfills all requirements of a rhombus because all four of its sides are equal in length, making statement (c) false as well. Therefore, none of the choices provided is correct.
In simple words: None of the first three claims is correct, which means the right choice is none of these.

Exam Tip: Keep the specific diagonal properties of rectangles, rhombuses, and squares distinct to answer classification questions accurately.

Question 3. Choose the correct statement.
(a) The diagonals of a parallelogram are equal
(b) The diagonals of a rectangle are perpendicular to each other
(c) If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus.
(d) Every quadrilateral is either a trapezium or a parallelogram or a kite.
Answer: (c) If the diagonals of a quadrilateral intersect at right angles, it is not necessarily a rhombus.
Perpendicular diagonals alone do not guarantee a rhombus; for instance, a kite or an irregular quadrilateral can have diagonals crossing at right angles without having four equal sides or bisecting each other. Statements (a), (b), and (d) are incorrect because parallelograms usually have unequal diagonals, rectangle diagonals are not necessarily perpendicular, and many quadrilaterals belong to none of those specific categories.
In simple words: Diagonals meeting at ninety degrees can occur in shapes like kites, so the quadrilateral does not have to be a rhombus.

Exam Tip: Remember that a quadrilateral is a rhombus only when its diagonals are perpendicular bisectors of one another, not merely perpendicular.

Question 4. In a trapezium ABCD, AB || CD. If \( \angle A = 95^\circ \) then \( \angle D = \)?
(a) 110°
(b) 85°
(c) 70°
(d) 310°
Answer: (b) 85°
Because line segment \( AB \) is parallel to \( CD \), the side \( AD \) serves as a transversal. Consecutive interior angles on the same side of a transversal are supplementary:
\( \angle A + \angle D = 180^\circ \)
\( 95^\circ + \angle D = 180^\circ \)
\( \implies \angle D = 180^\circ - 95^\circ = 85^\circ \)
In simple words: The two angles between the parallel sides along the same leg must sum up to 180 degrees.

Exam Tip: Identify which sides are parallel so that you apply the supplementary angle rule to the correct pair of vertices.

Question 5. RSTU is a parallelogram as shown in the figure below. Then the shown angles x and y are related as
(a) x = y.
(b) x < y.
(c) x > y.
(d) cannot be determined from given data.
Answer: (d) cannot be determined from given data.
In an arbitrary parallelogram, a diagonal does not necessarily bisect the vertex angle unless the adjacent sides are equal in length, as in a rhombus. Since no lengths of the sides \( RS \) and \( ST \) are provided, the relative sizes of angles \( x \) and \( y \) cannot be established.
In simple words: Without knowing the side lengths of the parallelogram, we cannot tell which of the two split angles is bigger.

Exam Tip: A diagonal bisects vertex angles only in rhombuses and squares; do not assume angle bisection holds in general parallelograms.

Question 6. In a rhombus ABCD, the diagonals intersect at O. If AB = 21 cm, diagonal BD = 16 cm, then the length of the diagonal AC is …….
(a) 38.82 cm
(b) 40 cm
(c) 50 cm
(d) none
Answer: (a) 38.82 cm
The diagonals of a rhombus bisect each other at right angles. Hence, in right triangle \( AOB \):
\( OB = \frac{BD}{2} = \frac{16}{2} = 8\text{ cm} \)
Applying the Pythagorean theorem:
\( OA^2 + OB^2 = AB^2 \)
\( OA^2 + 8^2 = 21^2 \)
\( OA^2 + 64 = 441 \)
\( OA^2 = 441 - 64 = 377 \)
\( \implies OA = \sqrt{377} \approx 19.416\text{ cm} \)
Therefore, the full diagonal \( AC \) has length:
\( AC = 2 \times OA = 2 \times 19.416 \approx 38.82\text{ cm} \)
In simple words: Halve the known diagonal, use Pythagoras' theorem on the right triangle to find half of the other diagonal, and double it.

Exam Tip: Remember to multiply the calculated semi-diagonal by 2 to state the full diagonal length.

Question 7. In a parallelogram ABCD, AB = 9 cm and BC = 7 cm. Each of its diagonals is less than ............
(a) 3 cm
(b) 4 cm
(c) 7 cm
(d) 16 cm
Answer: (d) 16 cm
By the triangle inequality theorem, the sum of any two sides of a triangle must strictly exceed the third side. In triangle \( ABC \), we have:
\( AC < AB + BC \)
\( AC < 9\text{ cm} + 7\text{ cm} = 16\text{ cm} \)
Similarly, in triangle \( ABD \):
\( BD < AB + AD = 9\text{ cm} + 7\text{ cm} = 16\text{ cm} \)
Consequently, both diagonals must each measure less than \( 16\text{ cm} \).
In simple words: Any diagonal forms a triangle with two adjacent sides, so its length must be smaller than the sum of those sides.

Exam Tip: Use the triangle inequality \( a + b > c \) whenever bounding the possible lengths of quadrilateral diagonals.

Question 8. In a parallelogram ABCD \( \angle B = 160^\circ \) ,then find the measurement of \( \angle C \).
(a) 20°
(b) 65°
(c) 90°
(d) 75°
Answer: (a) 20°
Consecutive angles of any parallelogram add up to \( 180^\circ \):
\( \angle B + \angle C = 180^\circ \)
\( 160^\circ + \angle C = 180^\circ \)
\( \implies \angle C = 180^\circ - 160^\circ = 20^\circ \)
In simple words: Neighboring angles in a parallelogram always combine to make 180 degrees.

Exam Tip: Do not confuse consecutive angles (supplementary) with opposite angles (equal) in a parallelogram.

Question 9. In a parallelogram ABCD, if AB = 3x - 4, CD = y - 1, AD = y + 5 and BC = 5x - 8 ,then ratio of AB : BC is
(a) 71 : 21
(b) 12 : 11
(c) 11 : 17
(d) 4 : 7
Answer: (c) 11 : 17
Opposite sides of a parallelogram are equal in length:
\( AB = CD \implies 3x - 4 = y - 1 \implies y = 3x - 3 \quad \text{--- (1)} \)
\( AD = BC \implies y + 5 = 5x - 8 \implies y = 5x - 13 \quad \text{--- (2)} \)
Equating both expressions for \( y \):
\( 3x - 3 = 5x - 13 \)
\( 2x = 10 \implies x = 5 \)
Now, calculating the side lengths:
\( AB = 3(5) - 4 = 15 - 4 = 11 \)
\( BC = 5(5) - 8 = 25 - 8 = 17 \)
Thus, the ratio is:
\( AB : BC = 11 : 17 \)
In simple words: Set opposite sides equal to solve for x, then substitute x back to find the side lengths and their ratio.

Exam Tip: Double check your solution by verifying that \( AD \) equals \( BC \) and \( AB \) equals \( CD \) before writing the ratio.

Question 10. In a square ABCD, its diagonals bisect at O. Then the triangle AOB is
(a) an equilateral triangle.
(b) an isosceles but not right angled triangle.
(c) a right angled but not an isosceles triangle.
(d) an isosceles right angled triangle.
Answer: (d) an isosceles right angled triangle.
The diagonals of a square are equal in length and bisect each other perpendicularly. Thus, \( OA = OB \), making the triangle isosceles, and \( \angle AOB = 90^\circ \), making it right-angled. Therefore, triangle \( AOB \) is an isosceles right-angled triangle.
In simple words: The two diagonal halves are equal and meet at a right angle, forming an isosceles right-angled triangle.

Exam Tip: Remember that square diagonals possess both properties: equality (giving equal legs) and perpendicularity (giving a 90° angle).

Question 11. The angles of a quadrilateral are x°, x-10°, x + 52° and 3x°. Find the greatest angle.
(a) 159°
(b) 180°
(c) 68°
(d) None
Answer: (a) 159°
The sum of all four interior angles in any quadrilateral is \( 360^\circ \):
\( x + (x - 10) + (x + 52) + 3x = 360 \)
\( 6x + 42 = 360 \)
\( 6x = 318 \implies x = 53^\circ \)
Evaluating each angle:
First angle = \( 53^\circ \)
Second angle = \( 53^\circ - 10^\circ = 43^\circ \)
Third angle = \( 53^\circ + 52^\circ = 105^\circ \)
Fourth angle = \( 3(53^\circ) = 159^\circ \)
The maximum value among them is \( 159^\circ \).
In simple words: Add the expressions to 360 to find x, then calculate each angle to pick the largest one.

Exam Tip: Always evaluate every angle before concluding which one is the largest, rather than guessing from the variable expression alone.

Question 12. The sum of the angles of a quadrilateral is
(a) depends adjacent side.
(b) 360°.
(c) depends adjacent angle.
(d) depends on the quadrilateral.
Answer: (b) 360°.
A diagonal splits any four-sided polygon into two triangles. Because the angle sum of each triangle is \( 180^\circ \), the total angle sum for every quadrilateral is constantly \( 2 \times 180^\circ = 360^\circ \).
In simple words: Any quadrilateral can be cut into two triangles, making the total angle sum 360 degrees.

Exam Tip: This theorem applies universally to all convex quadrilaterals regardless of side lengths or regularity.

Question 13. The diameter of circumcircle of a rectangle is 13 cm and breadth of the rectangle is 5 cm. Its length is
(a) 6 cm
(b) 5 cm
(c) 12 cm
(d) none
Answer: (c) 12 cm
The circumcircle of a rectangle passes through all four vertices, meaning its diameter coincides with the diagonal of the rectangle. Therefore, diagonal \( d = 13\text{ cm} \) and breadth \( b = 5\text{ cm} \).
Using the Pythagorean relation:
\( \text{length}^2 + \text{breadth}^2 = \text{diagonal}^2 \)
\( l^2 + 5^2 = 13^2 \)
\( l^2 + 25 = 169 \)
\( l^2 = 144 \implies l = 12\text{ cm} \)
In simple words: The circle's diameter is the diagonal of the rectangle, so use Pythagoras' theorem with 13 and 5 to find 12.

Exam Tip: Recognize the common Pythagorean triplet (5, 12, 13) to save time during multiple-choice questions.

Question 14. Which of the following statement is/are false ?
(a) Each diagonal of a quadrilateral divides it into two triangles.
(b) Each side of a quadrilateral is less than the sum of the remaining three sides.
(c) A quadrilateral can at most have two obtuse angles.
(d) A quadrilateral has four diagonals.
Answer: (d) A quadrilateral has four diagonals.
A quadrilateral contains only two diagonals joining non-consecutive vertices. Statement (d) is clearly false. Statement (c) is also inaccurate because a quadrilateral may contain up to three obtuse angles (for instance, \( 100^\circ, 100^\circ, 100^\circ, 60^\circ \)), but (d) represents the fundamental geometric falsehood.
In simple words: Any four-sided polygon has only two diagonals, not four.

Exam Tip: Recall the formula for the number of diagonals in an n-sided polygon: \( \frac{n(n - 3)}{2} \). For \( n = 4 \), it gives 2.

Question 15. In a Rhombus ABCD, if AD = AC then \( \angle DCB \) ……..
(a) 60°
(b) 120°
(c) 72°
(d) 108°
Answer: (b) 120°
In rhombus \( ABCD \), all sides are equal: \( AD = CD \). Given that diagonal \( AC = AD \), we have \( AD = CD = AC \), meaning triangle \( ADC \) is equilateral. Thus:
\( \angle ADC = 60^\circ \)
Adjacent angles of a rhombus are supplementary:
\( \angle ADC + \angle DCB = 180^\circ \)
\( 60^\circ + \angle DCB = 180^\circ \)
\( \implies \angle DCB = 180^\circ - 60^\circ = 120^\circ \)
In simple words: Three equal sides make an equilateral triangle with a 60-degree angle, leaving 120 degrees for the neighboring angle.

Exam Tip: Keep in mind that an equilateral triangle formed by a diagonal immediately tells you that the corner angle is 60°.

Question 16. A quadrilateral is a rhombus but not a square if
(a) its diagonals do not bisect each other.
(b) its diagonals are not perpendicular.
(c) opposite angles are not equal.
(d) the length of diagonals are not equal.
Answer: (d) the length of diagonals are not equal.
Both a rhombus and a square have four equal sides and perpendicular diagonals that bisect each other. The key distinguishing feature is that a square has equal diagonals, whereas a rhombus that is not a square has diagonals of unequal length.
In simple words: If all sides are equal and diagonals cross at 90 degrees but differ in length, it is a rhombus and not a square.

Exam Tip: Remember: Diagonals equal + perpendicular = Square; Diagonals unequal + perpendicular = Rhombus.

Question 17. ABCD is a quadrilateral. If AC and BD bisect each other, then ABCD must be
(a) square.
(b) rectangle.
(c) parallelogram.
(d) rhombus.
Answer: (c) parallelogram.
By the fundamental quadrilateral theorem, if the diagonals of a four-sided figure mutually bisect each other, the figure is guaranteed to be a parallelogram. Additional conditions (such as perpendicularity or equal lengths) are needed to make it a rhombus, rectangle, or square.
In simple words: Mutual bisection of diagonals is the exact defining test for a parallelogram.

Exam Tip: Mutual bisection proves the general class (parallelogram); do not assume special sub-types unless more information is provided.

Question 18. The diagonals of a parallelogram ABCD intersect at O. If \( \angle BOC = 90^\circ \) and \( \angle BDC = 65^\circ \), then \( \angle OAB \) is
(a) 10°
(b) 25°
(c) 50°
(d) 90°
Answer: (b) 25°
Vertically opposite angles are equal, so \( \angle DOC = \angle BOC = 90^\circ \). In right-angled triangle \( DOC \):
\( \angle ODC + \angle OCD + \angle DOC = 180^\circ \)
\( 65^\circ + \angle OCD + 90^\circ = 180^\circ \)
\( \implies \angle OCD = 180^\circ - 155^\circ = 25^\circ \)
Since \( AB \parallel CD \) with diagonal \( AC \) acting as a transversal, alternate interior angles are equal:
\( \angle OAB = \angle OCD = 25^\circ \)
In simple words: Find the third angle in the right triangle, then use alternate interior angles to transfer it to angle OAB.

Exam Tip: Use parallel line properties (alternate interior angles) to connect angles at opposite vertices across a diagonal.

Question 19. If angles P, Q, R and S of the quadrilateral PQRS, taken in order, are in the ratio 3 : 7 : 6 : 4, then PQRS is a
(a) rhombus
(b) parallelogram
(c) kite
(d) none
Answer: (d) none
Let the angles be \( 3k, 7k, 6k, 4k \).
\( 3k + 7k + 6k + 4k = 360^\circ \implies 20k = 360^\circ \implies k = 18^\circ \)
Thus:
\( \angle P = 3(18^\circ) = 54^\circ \)
\( \angle Q = 7(18^\circ) = 126^\circ \)
\( \angle R = 6(18^\circ) = 108^\circ \)
\( \angle S = 4(18^\circ) = 72^\circ \)
Testing consecutive angle sums:
\( \angle P + \angle Q = 54^\circ + 126^\circ = 180^\circ \implies SP \parallel RQ \)
\( \angle Q + \angle R = 126^\circ + 108^\circ = 234^\circ \neq 180^\circ \implies PQ \nparallel SR \)
With exactly one pair of parallel opposite sides, \( PQRS \) is a trapezium. Because trapezium is not listed among choices (a), (b), or (c), the answer is none.
In simple words: The angles show that only one pair of opposite sides is parallel, making the shape a trapezium, which is not among the options.

Exam Tip: Check whether consecutive interior angle sums equal 180° to identify which specific pairs of sides are parallel.

Question 20. To construct a parallelogram, the minimum number of measurements required is
(a) 2
(b) 6
(c) 4
(d) None
Answer: (d) None
A general quadrilateral requires 5 independent measurements, but because a parallelogram has opposite sides equal and opposite angles equal, it requires a minimum of 3 independent measurements (such as two adjacent sides and an included angle, or two sides and one diagonal). Since 3 is not among the options, the correct option is None.
In simple words: Exactly three pieces of information are needed to draw a unique parallelogram, so none of the listed numbers is correct.

Exam Tip: Memorize the minimum measurements for standard figures: quadrilateral = 5, parallelogram = 3, rhombus = 2, square = 1.

Question 21. ABCD is a square. E, F, G, H are the mid-points of the four sides. Then the figure EFGH is
(a) square
(b) rectangle
(c) trapezium
(d) parallelogram
Answer: (a) square
Connecting the midpoints of the sides of a square yields four congruent right-angled isosceles triangles at the corners. By the Pythagorean theorem, each side of \( EFGH \) equals \( \frac{s}{\sqrt{2}} \) where \( s \) is the side of \( ABCD \). Furthermore, each corner angle of \( EFGH \) is \( 180^\circ - 45^\circ - 45^\circ = 90^\circ \). Hence, \( EFGH \) is a square.
In simple words: Joining the midpoints of a square's sides produces another smaller, rotated square.

Exam Tip: Joining midpoints of a rectangle gives a rhombus, while joining midpoints of a square produces a square.

Question 22. A quadrilateral having exactly one pair of parallel sides, is called?
Answer: A quadrilateral having exactly one pair of parallel sides is called a trapezium (or trapezoid).
In simple words: A four-sided shape with only one set of opposite lines running parallel is a trapezium.

Exam Tip: Explicitly state "trapezium" and note that the other pair of opposite sides is non-parallel.

Question 23. If each pair of opposite sides of a quadrilateral is equal, then it is a________
Answer: parallelogram
In simple words: A four-sided figure whose opposite sides match in length is always a parallelogram.

Exam Tip: This is a standard converse theorem: equal opposite side pairs guarantee parallel opposite sides.

Question 24. In a parallelogram sum of two consecutive angles is _____
Answer: 180° (or supplementary)
In simple words: Any two adjacent corner angles in a parallelogram add up to 180 degrees.

Exam Tip: Mention either "180°" or "supplementary" - both are accepted standard answers.

Question 25. ABCD is a parallelogram. The diagonals AC and BD intersect at a point O. If E, F, G and H are the mid points of AO, DO, CO and BO respectively, then the ratio of (EF+FG+GH+HE) to (AD+DC+CB+BA) is:
(a) 1:1
(b) 1:2
(c) 1:3
(d) 1:4
Answer: (b) 1:2
In triangle \( AOD \), \( E \) and \( F \) are midpoints of sides \( AO \) and \( DO \). By the midpoint theorem:
\( EF = \frac{1}{2}AD \)
Applying the midpoint theorem to triangles \( DOC \), \( COB \), and \( BOA \):
\( FG = \frac{1}{2}DC \)
\( GH = \frac{1}{2}CB \)
\( HE = \frac{1}{2}BA \)
Adding these four equations gives:
\( EF + FG + GH + HE = \frac{1}{2}(AD + DC + CB + BA) \)
Therefore, the ratio of the perimeters is \( 1 : 2 \).
In simple words: Each inner segment is half the length of its corresponding outer side by the midpoint theorem, making the total perimeter ratio 1 to 2.

Exam Tip: The midpoint theorem reduces linear boundary lengths by a factor of 2 in each triangle formed by the diagonals.

Question 26. ABCD is a parallelogram in which \( \angle A=70^\circ, \angle B=90^\circ \) and \( \angle C=100^\circ \). How many points in the plane P of the quadrilateral are there such that P is equidistant from its vertices?
(a) None
(b) 1
(c) 2
(d) 3
Answer: (a) None
A point in the plane equidistant from all four vertices of a quadrilateral is its circumcentre, which exists if and only if the quadrilateral is cyclic. In any cyclic quadrilateral, opposite angles must be supplementary. Here, \( \angle A + \angle C = 70^\circ + 100^\circ = 170^\circ \neq 180^\circ \). Because the vertices do not lie on a common circle, no point can be equidistant from all four vertices.
In simple words: A point can only be equally far from all four corners if the shape fits inside a circle, which requires opposite angles to sum to 180 degrees.

Exam Tip: A quadrilateral has a circumcentre if and only if opposite angles add up to 180°.

<2M>

Question 27. The angles of quadrilateral are respectively \( 100^\circ, 30^\circ, 92^\circ, x^\circ \) find the value of x
Answer: The sum of all four interior angles of a quadrilateral is always \( 360^\circ \):
\( 100^\circ + 30^\circ + 92^\circ + x^\circ = 360^\circ \)
\( 222^\circ + x^\circ = 360^\circ \)
\( \implies x = 360 - 222 \)
\( \implies x = 138^\circ \)
In simple words: Add the three known angles together and subtract their total from 360 to find x.

Exam Tip: State the angle sum property of quadrilaterals explicitly before carrying out the arithmetic.

<3M>

Question 28. Find the value of a and also find angles related to a as shown in figuer.
Answer: In triangle \( BCD \), the sum of all interior angles equals \( 180^\circ \):
\( \angle DBC + \angle BDC + \angle BCD = 180^\circ \)
From the figure, these angles are represented as \( 2a \), \( 3a \), and \( 5a \):
\( 2a + 3a + 5a = 180^\circ \)
\( 10a = 180^\circ \)
\( \implies a = 18^\circ \)
Now, calculating the measures of the angles related to \( a \):
\( 2a = 2 \times 18^\circ = 36^\circ \)
\( 3a = 3 \times 18^\circ = 54^\circ \)
\( 5a = 5 \times 18^\circ = 90^\circ \)
In simple words: The three angles in the triangle add up to 10a, which equals 180 degrees, giving a = 18 degrees.

Exam Tip: Be sure to list both the value of \( a \) and all required individual angle values to earn full credit.

Question 29. Prove that angle bisectors of a parallelogram form a rectangle.
Answer: Let \( ABCD \) be a parallelogram and let the bisectors of consecutive interior angles \( \angle A, \angle B, \angle C, \angle D \) intersect at points \( P, Q, R, S \).
Consecutive angles of a parallelogram are supplementary:
\( \angle A + \angle B = 180^\circ \)
Because lines \( AS \) and \( BS \) bisect \( \angle A \) and \( \angle B \):
\( \frac{1}{2}\angle A + \frac{1}{2}\angle B = \frac{180^\circ}{2} = 90^\circ \)
In triangle \( ASB \):
\( \angle ASB = 180^\circ - \left(\frac{1}{2}\angle A + \frac{1}{2}\angle B\right) = 180^\circ - 90^\circ = 90^\circ \)
Since vertically opposite angles are equal, the interior angle of quadrilateral \( PQRS \) at this vertex is \( 90^\circ \).
By following the exact same procedure for triangles \( BQC \), \( CRD \), and \( DPA \), every interior angle of quadrilateral \( PQRS \) is found to be \( 90^\circ \).
A quadrilateral whose interior angles are all right angles is a rectangle.
Hence, the angle bisectors of a parallelogram form a rectangle.
In simple words: Half of two adjacent angles equals 90 degrees, forcing each corner of the inner shape to be 90 degrees, which makes it a rectangle.

Exam Tip: Clearly show the step \( \frac{1}{2}\angle A + \frac{1}{2}\angle B = 90^\circ \) using consecutive interior angles before calculating the third angle of the triangle.

Question 30. ABC be an isosceles triangle with AB = AC and let D,F,E are the mid-points of BC,CA and AB respectively. Show that AD perpendicular to EF and AD bisector of EF.
Answer: In isosceles triangle \( ABC \) with \( AB = AC \), the median \( AD \) drawn to the non-equal base \( BC \) also serves as the altitude, so:
\( AD \perp BC \quad \text{--- (1)} \)
Since \( E \) and \( F \) are the midpoints of sides \( AB \) and \( AC \) respectively, the midpoint theorem states that:
\( EF \parallel BC \quad \text{--- (2)} \)
From (1) and (2), because a line perpendicular to one of two parallel lines is perpendicular to the other:
\( AD \perp EF \)
Let \( AD \) intersect \( EF \) at point \( M \). In triangle \( ABD \), \( E \) is the midpoint of \( AB \) and \( EM \parallel BD \). By the converse of the midpoint theorem, \( M \) is the midpoint of \( AD \), and:
\( EM = \frac{1}{2}BD \)
Similarly, in triangle \( ADC \):
\( MF = \frac{1}{2}DC \)
Since \( D \) is the midpoint of \( BC \), \( BD = DC \), which implies:
\( EM = MF \)
Hence, \( M \) is also the midpoint of \( EF \), showing that \( AD \) bisects \( EF \).
Therefore, \( AD \) is perpendicular to \( EF \) and \( AD \) bisects \( EF \).
In simple words: The central median line is perpendicular to the base and also cuts the parallel midpoint segment into two equal halves at right angles.

Exam Tip: Use both parts of the midpoint theorem: first the parallelism to prove perpendicularity, then the segment lengths to establish bisection.

Question 31. In a triangle ABC median AD is produced to X such that AD=DX. Prove that ABXC is a parallelogram.
Answer: In triangle \( ABC \), \( AD \) is a median, which means that \( D \) is the midpoint of side \( BC \):
\( BD = CD \quad \text{--- (1)} \)
It is given that segment \( AD \) is extended to \( X \) such that:
\( AD = DX \quad \text{--- (2)} \)
Equations (1) and (2) show that the line segments \( AX \) and \( BC \) bisect each other at their common point of intersection \( D \).
Since a quadrilateral whose diagonals mutually bisect each other is a parallelogram, it follows directly that \( ABXC \) is a parallelogram.
Hence proved.
In simple words: The diagonals AX and BC cross each other right at their midpoints, which proves that the figure is a parallelogram.

Exam Tip: Quoting the property "a quadrilateral whose diagonals bisect each other is a parallelogram" makes this proof concise and complete.

Question 32. ABCD is parallelogram. P is a point on AD such that AP = 1/3 AD and Q is a point on BC such that CQ = 1/3 BC. Prove that AQCP is a parallelogram.
Answer: Given that \( ABCD \) is a parallelogram, its opposite sides are both parallel and equal:
\( AD \parallel BC \quad \text{and} \quad AD = BC \)
Because point \( P \) lies on side \( AD \) and point \( Q \) lies on side \( BC \), the segments \( AP \) and \( QC \) remain parallel:
\( AP \parallel QC \quad \text{--- (1)} \)
Additionally, we are given:
\( AP = \frac{1}{3}AD \quad \text{and} \quad CQ = \frac{1}{3}BC \)
Since \( AD = BC \), their one-third fractions must be equal:
\( AP = CQ \quad \text{--- (2)} \)
In quadrilateral \( AQCP \), equations (1) and (2) demonstrate that one pair of opposite sides is both parallel and equal in length.
Therefore, \( AQCP \) is a parallelogram.
In simple words: Side AP and side CQ are parallel and equal in length, which is sufficient to make AQCP a parallelogram.

Exam Tip: Proving that a single pair of opposite sides is both equal and parallel is the most efficient way to prove a quadrilateral is a parallelogram.

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Question 33. ABCD is a Rhombus AD is produced to E so that DE = DC and EC produced meets AB produced in F. prove that BF=BC
Answer: In rhombus \( ABCD \), all four sides are equal:
\( AB = BC = CD = DA \quad \text{--- (1)} \)
It is given that \( AD \) is extended to \( E \) such that \( DE = DC \). Combining this with (1):
\( DE = DA = DC \)
In triangle \( CDE \), because \( DE = DC \), the base angles are equal:
\( \angle DEC = \angle DCE \)
Since line \( AB \) is parallel to side \( CD \), the extended line \( AF \) is parallel to \( CD \).
Considering transversal \( EF \), alternate interior angles are equal:
\( \angle BFC = \angle DCE \)
Considering transversal \( AE \), corresponding angles are equal:
\( \angle AFE = \angle DCE = \angle DEC \)
In triangle \( AEF \), since \( \angle AFE = \angle AEF \), the sides opposite to equal angles are equal:
\( AF = AE \)
From the construction of line \( AE \):
\( AE = AD + DE = AB + AB = 2AB \)
Therefore:
\( AF = 2AB \)
Since \( AF = AB + BF \):
\( AB + BF = 2AB \implies BF = AB \)
Because \( AB = BC \) from (1), we conclude:
\( BF = BC \)
Hence proved.
In simple words: Parallel lines and equal segments make triangle AEF isosceles, which shows that AF is twice AB, leaving BF equal to AB and therefore equal to BC.

Exam Tip: Clearly write out the chain of equal segments: \( DE = DC = AB \) and use the alternate interior angle property with \( CD \parallel AF \).

Question 34. In a quadrilateral ABCD, CO and DO are the bisector of angle C and angle D respectively . prove that \( \angle COD = \frac{1}{2}(\angle A + \angle B) \)
Answer: In triangle \( COD \), the sum of all interior angles is \( 180^\circ \):
\( \angle COD + \angle OCD + \angle ODC = 180^\circ \)
Since \( CO \) and \( DO \) bisect \( \angle C \) and \( \angle D \) respectively:
\( \angle OCD = \frac{1}{2}\angle C \quad \text{and} \quad \angle ODC = \frac{1}{2}\angle D \)
Substituting these into the triangle equation:
\( \angle COD + \frac{1}{2}\angle C + \frac{1}{2}\angle D = 180^\circ \)
\( \implies \angle COD = 180^\circ - \frac{1}{2}(\angle C + \angle D) \quad \text{--- (1)} \)
In quadrilateral \( ABCD \), the sum of all four interior angles is \( 360^\circ \):
\( \angle A + \angle B + \angle C + \angle D = 360^\circ \)
\( \implies \angle C + \angle D = 360^\circ - (\angle A + \angle B) \quad \text{--- (2)} \)
Substituting equation (2) into equation (1):
\( \angle COD = 180^\circ - \frac{1}{2}[360^\circ - (\angle A + \angle B)] \)
\( \angle COD = 180^\circ - 180^\circ + \frac{1}{2}(\angle A + \angle B) \)
\( \implies \angle COD = \frac{1}{2}(\angle A + \angle B) \)
Hence proved.
In simple words: Replace the sum of angles C and D in the small triangle using 360 minus the sum of angles A and B to get the formula.

Exam Tip: Be careful with the minus sign when expanding \( -\frac{1}{2}[360^\circ - (\angle A + \angle B)] \); the double negative becomes positive.

Question 35. In Triangle ABC, AD is the median through A and E is the mid-point of AD. BE produced meets AC in F proved that AF = 1/3 AC.
Answer: Through vertex \( D \), construct a line parallel to \( BF \) that meets \( AC \) at point \( G \):
\( DG \parallel BF \)
In triangle \( ADG \):
\( E \) is the midpoint of \( AD \), and \( EF \parallel DG \) (as \( EF \) lies on line \( BF \)).
By the converse of the midpoint theorem, \( F \) must be the midpoint of \( AG \):
\( AF = FG \quad \text{--- (1)} \)
In triangle \( BCF \):
\( D \) is the midpoint of \( BC \) (since \( AD \) is a median), and \( DG \parallel BF \).
By the converse of the midpoint theorem, \( G \) must be the midpoint of \( FC \):
\( FG = GC \quad \text{--- (2)} \)
Combining equations (1) and (2):
\( AF = FG = GC \)
Now, the full length of side \( AC \) is:
\( AC = AF + FG + GC = 3AF \)
\( \implies AF = \frac{1}{3}AC \)
Hence proved.
In simple words: Draw a parallel line from D to AC, which divides AC into three equal steps by using the converse midpoint theorem twice.

Exam Tip: Stating the auxiliary construction \( DG \parallel BF \) is essential to earning full marks in this classic proof.

Question 36. Show that the quadrilateral formed by joining the mid point of the consecutive sides of a rectangle is a rhombus.
Answer: Let \( ABCD \) be a rectangle and let \( P, Q, R, S \) be the midpoints of sides \( AB, BC, CD, DA \) respectively. Draw diagonals \( AC \) and \( BD \).
In a rectangle, diagonals are equal in length:
\( AC = BD \quad \text{--- (1)} \)
In triangle \( ABC \), \( P \) and \( Q \) are midpoints of \( AB \) and \( BC \). By the midpoint theorem:
\( PQ \parallel AC \quad \text{and} \quad PQ = \frac{1}{2}AC \quad \text{--- (2)} \)
In triangle \( ADC \), \( S \) and \( R \) are midpoints of \( AD \) and \( CD \). By the midpoint theorem:
\( SR \parallel AC \quad \text{and} \quad SR = \frac{1}{2}AC \quad \text{--- (3)} \)
From (2) and (3), \( PQ = SR = \frac{1}{2}AC \).
Similarly, applying the midpoint theorem to triangles \( ABD \) and \( CBD \):
\( SP = RQ = \frac{1}{2}BD \quad \text{--- (4)} \)
Since \( AC = BD \) from (1), we have:
\( PQ = QR = RS = SP = \frac{1}{2}AC \)
A quadrilateral whose four sides are all equal in length is a rhombus.
Thus, \( PQRS \) is a rhombus.
In simple words: The midpoint theorem makes each side of the inner shape half of a diagonal, and because rectangle diagonals are equal, all four inner sides are equal.

Exam Tip: Be sure to explicitly cite the equality of rectangle diagonals \( AC = BD \) as the reason all four sides become equal.

Question 37. P is the mid-point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R. prove that AR = 2BC
Answer: In parallelogram \( ABCD \):
\( AB \parallel CD \quad \text{and} \quad AB = CD \)
\( AD \parallel BC \quad \text{and} \quad AD = BC \quad \text{--- (1)} \)
Since \( P \) is the midpoint of \( AB \):
\( PB = \frac{1}{2}AB \)
It is given that line \( BR \) is parallel to \( PD \), so \( BQ \parallel PD \).
In quadrilateral \( PBQD \), \( PB \parallel DQ \) (as \( AB \parallel CD \)) and \( PD \parallel BQ \).
Therefore, \( PBQD \) is a parallelogram, which implies:
\( DQ = PB = \frac{1}{2}AB = \frac{1}{2}CD \)
Now consider triangle \( ARB \):
\( P \) is the midpoint of \( AB \), and \( PD \parallel BR \).
By the converse of the midpoint theorem, the line through the midpoint of one side parallel to another side bisects the third side.
Hence, \( D \) is the midpoint of \( AR \):
\( AR = 2AD \)
Substituting \( AD = BC \) from (1):
\( AR = 2BC \)
Hence proved.
In simple words: Quadrilateral PBQD is a parallelogram, which allows us to apply the converse midpoint theorem on triangle ARB to prove that AR is twice AD (and thus twice BC).

Exam Tip: Clearly show two stages: first prove \( PBQD \) is a parallelogram, then apply the converse midpoint theorem in \( \triangle ARB \).

Question 38. P,Q,R are, respectively, the mid points of sides BC,CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y. Prove that XY = ¼ BC.
Answer: In triangle \( ABC \), \( R \) and \( P \) are the midpoints of \( AB \) and \( BC \) respectively. By the midpoint theorem:
\( RP \parallel AC \quad \text{and} \quad RP = \frac{1}{2}AC \)
Since \( Q \) is the midpoint of \( AC \), \( QC = \frac{1}{2}AC \), so:
\( RP \parallel QC \quad \text{and} \quad RP = QC \)
Thus, quadrilateral \( RPCQ \) is a parallelogram.
Because the diagonals of a parallelogram bisect each other, their intersection point \( Y \) is the midpoint of diagonal \( PQ \).
Similarly, consider quadrilateral \( BPQR \):
\( QP \parallel AB \quad \text{and} \quad QP = \frac{1}{2}AB = RB \)
Hence, \( BPQR \) is a parallelogram.
Its diagonals \( PR \) and \( BQ \) bisect each other at \( X \), meaning \( X \) is the midpoint of \( PR \).
Now, in triangle \( PQR \):
\( X \) is the midpoint of \( PR \) and \( Y \) is the midpoint of \( PQ \).
Applying the midpoint theorem to triangle \( PQR \):
\( XY = \frac{1}{2}RQ \quad \text{--- (1)} \)
In triangle \( ABC \), since \( R \) and \( Q \) are midpoints of \( AB \) and \( AC \):
\( RQ = \frac{1}{2}BC \quad \text{--- (2)} \)
Substituting equation (2) into equation (1):
\( XY = \frac{1}{2}\left(\frac{1}{2}BC\right) = \frac{1}{4}BC \)
Hence proved.
In simple words: Diagonals of the formed parallelograms show that X and Y are midpoints of the sides of triangle PQR, making XY one-fourth of the base BC.

Exam Tip: State clearly that \( X \) and \( Y \) are midpoints because the diagonals of parallelograms \( BPQR \) and \( RPCQ \) bisect each other.