Quadrilaterals - Complete Q&A

Multiple Choice Questions (20 Questions)

Question 1
What is the sum of all interior angles of a quadrilateral?
(a) 180° (b) 270° (c) 360° (d) 90°

Show Answer
(c) 360°

Show Explanation
The sum of the interior angles of any quadrilateral is always 360°. This can be proven by dividing the quadrilateral into two triangles by drawing one diagonal. The sum of angles in each triangle is 180°, so for two triangles, it is 2 × 180° = 360°.

Question 2
A quadrilateral with only one pair of parallel sides is called a:
(a) Parallelogram (b) Rhombus (c) Trapezium (d) Kite

Show Answer
(c) Trapezium

Show Explanation
By definition, a trapezium is a quadrilateral that has at least one pair of parallel sides. The other options have two pairs of parallel sides.

Question 3
In a parallelogram, the bisectors of two adjacent angles are:
(a) Parallel (b) Perpendicular (c) Coincident (d) Equal

Show Answer
(b) Perpendicular

Show Explanation
In a parallelogram, consecutive angles are supplementary (sum to 180°). The bisectors of two adjacent angles will create two angles that sum to 90° (half of 180°). Therefore, the angle between these two bisectors is 180° - 90° = 90°, meaning they are perpendicular.

Question 4
The diagonals of a square:
(a) Are equal but not perpendicular (b) Are perpendicular but not equal
(c) Are neither equal nor perpendicular (d) Are equal and perpendicular

Show Answer
(d) Are equal and perpendicular

Show Explanation
A square has all the properties of a rectangle (diagonals are equal) and all the properties of a rhombus (diagonals are perpendicular bisectors). Therefore, its diagonals are both equal and perpendicular.

Question 5
If two adjacent angles of a parallelogram are (2x + 25)° and (3x - 5)°, the value of x is:
(a) 60 (b) 32 (c) 100 (d) 64

Show Answer
(b) 32

Show Explanation
Adjacent angles in a parallelogram are supplementary.
So, (2x + 25) + (3x - 5) = 180
2x + 25 + 3x - 5 = 180
5x + 20 = 180
5x = 180 - 20
5x = 160
x = 160 / 5
x = 32

Question 6
A quadrilateral whose diagonals are equal and bisect each other at right angles is a:
(a) Rectangle (b) Rhombus (c) Kite (d) Square

Show Answer
(d) Square

Show Explanation
If diagonals bisect each other, it is a parallelogram. If they are also equal, it is a rectangle. If they also bisect each other at right angles, it is a square. A square is the only quadrilateral that satisfies all these conditions simultaneously.

Question 7
The angles of a quadrilateral are in the ratio 1:2:3:4. The largest angle is:
(a) 72° (b) 144° (c) 108° (d) 180°

Show Answer
(b) 144°

Show Explanation
Let the angles be 1x, 2x, 3x, and 4x.
Sum of angles = 1x + 2x + 3x + 4x = 10x = 360°
So, x = 36°
The largest angle is 4x = 4 × 36° = 144°.

Question 8
In a rhombus, if the length of a diagonal is equal to its side, then the angles of the rhombus are:
(a) 60° and 120° (b) 45° and 135° (c) 30° and 150° (d) 90° each

Show Answer
(a) 60° and 120°

Show Explanation
Let the side and the diagonal be of length 'a'. This diagonal divides the rhombus into two congruent isosceles triangles with sides a, a, and a (all sides equal). This makes them equilateral triangles. Therefore, the angle opposite this diagonal in the rhombus is 60°. The adjacent angle will be 180° - 60° = 120°.

Question 9
The figure formed by joining the midpoints of the sides of a quadrilateral is a:
(a) Square (b) Rhombus (c) Rectangle (d) Parallelogram

Show Answer
(d) Parallelogram

Show Explanation
This is a standard theorem. The line segments joining the midpoints of any quadrilateral form a parallelogram.

Question 10
In a rectangle ABCD, if AB = 12 cm and BC = 5 cm, the length of diagonal AC is:
(a) 13 cm (b) 17 cm (c) 7 cm (d) 15 cm

Show Answer
(a) 13 cm

Show Explanation
In a rectangle, all angles are 90°. So, triangle ABC is a right-angled triangle.
Using Pythagoras Theorem:
AC² = AB² + BC²
AC² = (12)² + (5)²
AC² = 144 + 25
AC² = 169
AC = √169 = 13 cm

Question 11
A kite has how many pairs of equal adjacent sides?
(a) 1 (b) 2 (c) 3 (d) 0

Show Answer
(b) 2

Show Explanation
By definition, a kite is a quadrilateral with two distinct pairs of adjacent sides that are equal.

Question 12
In a parallelogram, if one angle is 90°, it becomes a:
(a) Kite (b) Rhombus (c) Rectangle (d) Trapezium

Show Answer
(c) Rectangle

Show Explanation
In a parallelogram, opposite angles are equal and adjacent angles are supplementary. If one angle is 90°, its opposite angle is also 90°. The adjacent angles become 90° as well (since 180° - 90° = 90°). A parallelogram with all angles 90° is a rectangle.

Question 13
The diagonals of a rectangle:
(a) Are perpendicular (b) Are equal (c) Bisect the angles (d) Are not equal

Show Answer
(b) Are equal

Show Explanation
This is a key property of a rectangle. Its diagonals are always equal in length.

Question 14
If the diagonals of a quadrilateral bisect each other, it is a:
(a) Kite (b) Trapezium (c) Parallelogram (d) Rhombus

Show Answer
(c) Parallelogram

Show Explanation
This is a standard test for a quadrilateral to be a parallelogram.

Question 15
In ΔABC, D and E are midpoints of AB and AC respectively. If DE = 4 cm, then BC is:
(a) 2 cm (b) 4 cm (c) 8 cm (d) 16 cm

Show Answer
(c) 8 cm

Show Explanation
This is based on the Midpoint Theorem. The line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it.
DE = ½ BC
4 cm = ½ BC
BC = 4 × 2 = 8 cm

Question 16
A square is a special type of:
(a) Kite and Rectangle (b) Rhombus and Trapezium (c) Rectangle and Rhombus
(d) Parallelogram and Kite

Show Answer
(c) Rectangle and Rhombus

Show Explanation
A square has all the properties of a rectangle (all angles 90°) and all the properties of a rhombus (all sides equal). Therefore, it is a special type of both.

Question 17
The number of diagonals in a convex quadrilateral is:
(a) 1 (b) 2 (c) 3 (d) 4

Show Answer
(b) 2

Show Explanation
A convex quadrilateral has two diagonals. For example, a rectangle has two diagonals (AC and BD).

Question 18
In a parallelogram PQRS, if ∠P = 70°, then ∠R is:
(a) 70° (b) 110° (c) 20° (d) 90°

Show Answer
(a) 70°

Show Explanation
In a parallelogram, opposite angles are equal. ∠P and ∠R are opposite angles. So, if ∠P = 70°, then ∠R = 70°.

Question 19
Which of the following is not true for a parallelogram?
(a) Opposite sides are equal (b) Opposite angles are equal
(c) Diagonals are equal (d) Diagonals bisect each other

Show Answer
(c) Diagonals are equal

Show Explanation
Diagonals of a parallelogram bisect each other but are not necessarily equal. They are equal only in special parallelograms like rectangles and squares.

Question 20
A quadrilateral with all sides equal and diagonals equal is a:
(a) Rhombus (b) Square (c) Kite (d) Rectangle

Show Answer
(b) Square

Show Explanation
A rhombus has all sides equal but diagonals are not equal. A rectangle has equal diagonals but not all sides equal. A square satisfies both conditions: all sides equal and diagonals equal.

Assertion & Reasoning Questions (20 Questions)

Question 1
Assertion (A): A square is a rhombus.
Reason (R): A square has all four sides equal, which is a defining property of a rhombus.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
A square possesses all the properties of a rhombus (all sides equal, opposite sides parallel, diagonals perpendicular bisectors). Reason R correctly states the key property that makes a square a special type of rhombus.

Question 2
Assertion (A): Every rectangle is a parallelogram.
Reason (R): In a rectangle, only one pair of opposite sides is parallel.

Show Answer
(c) A is true but R is false.

Show Explanation
Assertion A is true because a rectangle has two pairs of parallel opposite sides, fulfilling the definition of a parallelogram. Reason R is false because a rectangle has both pairs of opposite sides parallel, not just one.

Question 3
Assertion (A): The diagonals of a kite are perpendicular to each other.
Reason (R): The diagonals of a kite bisect each other at right angles.

Show Answer
(c) A is true but R is false.

Show Explanation
Assertion A is true; the diagonals of a kite are always perpendicular. However, Reason R is false because only one diagonal of a kite is bisected by the other. The longer diagonal bisects the shorter one, but the diagonals do not bisect each other.

Question 4
Assertion (A): In a parallelogram, the diagonals are equal.
Reason (R): Only in a rectangle, the diagonals are equal.

Show Answer
(d) A is false but R is true.

Show Explanation
Assertion A is false. Diagonals are equal only in rectangles and squares, not in all parallelograms (e.g., a rhombus that is not a square has unequal diagonals). Reason R is true, as it correctly states a property of the rectangle.

Question 5
Assertion (A): The sum of angles of a quadrilateral is 360°.
Reason (R): A quadrilateral can be divided into two triangles, and the sum of angles in each triangle is 180°.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
Both statements are true. The reason provides the logical geometric proof for the assertion.

Question 6
Assertion (A): A trapezium is not a parallelogram.
Reason (R): In a trapezium, only one pair of opposite sides is parallel, while in a parallelogram, both pairs are parallel.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
The definition of a trapezium (having only one pair of parallel sides) explicitly excludes it from being a parallelogram (which requires two pairs). The reason perfectly explains the assertion.

Question 7
Assertion (A): A rhombus with one right angle becomes a square.
Reason (R): All sides of a rhombus are equal.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
If a rhombus (which already has all sides equal) has one right angle, then all its angles must be right angles because opposite angles are equal and consecutive angles are supplementary. This turns it into a square. The property stated in R is essential for this transformation.

Question 8
Assertion (A): The quadrilateral formed by joining the midpoints of a rectangle is a rhombus.
Reason (R): The sides of the new figure are equal and parallel to the diagonals of the rectangle.

Show Answer
(c) A is true but R is false.

Show Explanation
Assertion A is true. The midpoint quadrilateral of a rectangle is always a rhombus (all four sides are equal). However, Reason R is incorrect. The sides of the new quadrilateral are parallel to the diagonals of the original rectangle, but they are not equal to the diagonals; they are each equal to half the length of a diagonal.

Question 9
Assertion (A): All kites are rhombuses.
Reason (R): Both kites and rhombuses have all sides equal.

Show Answer
(d) A is false but R is false.

Show Explanation
Assertion A is false. A rhombus has all sides equal, but a kite only has two pairs of adjacent sides equal. Not all kites have all four sides equal. Reason R is also false because it incorrectly states that kites have all sides equal.

Question 10
Assertion (A): In a square, the diagonals are perpendicular bisectors of each other.
Reason (R): A square inherits this property from both the rectangle and the rhombus.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
A square is a special rectangle and a special rhombus. From a rhombus, it inherits the property of diagonals being perpendicular bisectors. The reason correctly explains why the assertion is true.

Question 11
Assertion (A): A quadrilateral can be constructed if its three sides and two included angles are given.
Reason (R): To construct a unique quadrilateral, five independent measurements are required.

Show Answer
(b) Both A and R are true but R is not the correct explanation of A.

Show Explanation
Both statements are true. A unique quadrilateral can be constructed with five appropriate measurements. However, Reason R is a general statement, while Assertion A specifies a particular set of five measurements (three sides and two included angles). The reason does not directly explain why that specific set works.

Question 12
Assertion (A): In a parallelogram, the sum of consecutive angles is 180°.
Reason (R): The opposite angles of a parallelogram are equal.

Show Answer
(b) Both A and R are true but R is not the correct explanation of A.

Show Explanation
Both properties are true for a parallelogram. However, the sum of consecutive angles is 180° because they are co-interior angles between two parallel lines, not because the opposite angles are equal.

Question 13
Assertion (A): The diagonals of a rhombus are always equal.
Reason (R): A square is a rhombus with equal diagonals.

Show Answer
(d) A is false but R is true.

Show Explanation
Assertion A is false. The diagonals of a rhombus are generally unequal; they are equal only in the special case of a square. Reason R is true, as it correctly identifies the property of a square.

Question 14
Assertion (A): If the diagonals of a quadrilateral are perpendicular, it must be a kite.
Reason (R): The diagonals of a rhombus are also perpendicular.

Show Answer
(d) A is false but R is true.

Show Explanation
Assertion A is false. While kites have perpendicular diagonals, other quadrilaterals like rhombuses and squares also have this property. Reason R is true and provides a counterexample to the assertion.

Question 15
Assertion (A): A rectangle is an equiangular quadrilateral.
Reason (R): All angles in a rectangle are right angles.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
"Equiangular" means all angles are equal. A rectangle has all four angles equal to 90°. The reason perfectly explains the assertion.

Question 16
Assertion (A): The Midpoint Theorem is applicable to any triangle.
Reason (R): The line segment joining midpoints of two sides is always parallel to the third side.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
The Midpoint Theorem states that in any triangle, the segment joining the midpoints of any two sides will be parallel to the third side and half its length. The reason states the core of the theorem, which applies to all triangles.

Question 17
Assertion (A): All squares are kites.
Reason (R): A square has two distinct pairs of adjacent sides equal.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
A square has all sides equal, which means it also satisfies the condition of having two pairs of equal adjacent sides (e.g., AB=AD and BC=CD). Therefore, a square is a special type of kite. The reason is correct.

Question 18
Assertion (A): A parallelogram is always a cyclic quadrilateral.
Reason (R): The sum of opposite angles of a parallelogram is 180° only if it is a rectangle.

Show Answer
(d) A is false but R is true.

Show Explanation
Assertion A is false. For a quadrilateral to be cyclic (all vertices lie on a circle), the sum of opposite angles must be 180°. In a general parallelogram, this is not true. This is only true for rectangles. Reason R is correct.

Question 19
Assertion (A): The area of a rhombus can be calculated as half the product of its diagonals.
Reason (R): The diagonals of a rhombus divide it into four congruent right triangles.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
The area formula for a rhombus is Area = ½ × d1 × d2. This is derived from the fact that the diagonals are perpendicular bisectors, creating four congruent right-angled triangles. The area of these four triangles adds up to ½ × d1 × d2.

Question 20
Assertion (A): If one pair of opposite sides of a quadrilateral are equal and parallel, it is a parallelogram.
Reason (R): This is one of the standard tests to prove a quadrilateral is a parallelogram.

Show Answer
(a) Both A and R are true and R is the correct explanation of A.

Show Explanation
This is a standard and true theorem in geometry. If one pair of opposite sides is both equal and parallel, the quadrilateral must be a parallelogram. The reason correctly identifies this as a standard test.

True/False Questions (10 Questions)

Question 1
Every square is a rectangle.

Show Answer
True

Show Explanation
A square satisfies all properties of a rectangle: opposite sides parallel and equal, all angles 90°, and diagonals equal.

Question 2
Every rhombus is a square.

Show Answer
False

Show Explanation
A rhombus has all sides equal but its angles are not necessarily 90°. A square is a rhombus with all angles 90°.

Question 3
The diagonals of a parallelogram are always equal.

Show Answer
False

Show Explanation
Diagonals are equal only in rectangles and squares, not in all parallelograms (e.g., a rhombus that is not a square).

Question 4
A quadrilateral with perpendicular diagonals is always a kite.

Show Answer
False

Show Explanation
Rhombuses and squares also have perpendicular diagonals. So, a quadrilateral with perpendicular diagonals could be a kite, a rhombus, or a square.

Question 5
The sum of exterior angles of a quadrilateral is also 360°.

Show Answer
True

Show Explanation
This is true for any polygon, regardless of the number of sides. The sum of the exterior angles, one at each vertex, is always 360°.

Question 6
In a parallelogram, consecutive angles are complementary.

Show Answer
False

Show Explanation
Consecutive angles in a parallelogram are supplementary (sum to 180°), not complementary (which would be 90°).

Question 7
A trapezium can have two right angles.

Show Answer
True

Show Explanation
This is an isosceles trapezium or a right trapezium. It is possible to have two adjacent right angles in a trapezium.

Question 8
The diagonals of a rectangle bisect the angles at the vertices.

Show Answer
False

Show Explanation
Diagonals of a rectangle bisect each other but do not generally bisect the angles. They only bisect the angles in a square.

Question 9
The figure formed by joining the midpoints of a rhombus is a rectangle.

Show Answer
True

Show Explanation
The midpoint quadrilateral of a rhombus is always a rectangle. The sides of this new quadrilateral are parallel to the diagonals of the rhombus. Since the diagonals of a rhombus are perpendicular, the resulting quadrilateral has all angles of 90°.

Question 10
All rectangles are parallelograms.

Show Answer
True

Show Explanation
A rectangle has two pairs of parallel opposite sides, which is the defining property of a parallelogram.

Short Answer Type I (2 Marks each) - 15 Questions

Question 1
Two adjacent angles of a parallelogram are in the ratio 4:5. Find the measure of all its angles.

Show Answer
The angles are 80°, 100°, 80°, and 100°.

Show Explanation

In a parallelogram, adjacent angles are supplementary (sum to 180°).
Let the measures of the adjacent angles be $4 x$ and $5 x$ .
So, $4 x + 5 x = 180 °$
$9 x = 180 °$
$x = 20 °$
Therefore, the angles are $4 \times 20 ° = 80 °$ and $5 \times 20 ° = 100 °$ .
Since opposite angles in a parallelogram are equal, the four angles are 80°, 100°, 80°, and 100°.

Question 2
In a parallelogram ABCD, if ∠A = (2x + 15)° and ∠B = (3x - 25)°, find the value of x.

Show Answer
$x = 38$

Show Explanation

∠A and ∠B are adjacent angles of a parallelogram, so they are supplementary.
Therefore, $(2 x + 15) + (3 x - 25) = 180$
$2 x + 15 + 3 x - 25 = 180$
$5 x - 10 = 180$
$5 x = 180 + 10$
$5 x = 190$
$x = 190 / 5$
$x = 38$

Question 3
State the properties of the diagonals of a rhombus.

Show Answer
The diagonals of a rhombus:

Bisect each other.
Are perpendicular to each other.
Bisect the interior angles of the rhombus.

Show Explanation
These are the key properties that distinguish a rhombus from a general parallelogram. The diagonals are not necessarily equal.

Question 4
In kite ABCD, if AB = AD and CB = CD, and ∠ABC = 50°, find ∠ADC.

Show Answer
∠ADC = 50°

Show Explanation

In a kite, one pair of opposite angles is equal. The angles between the unequal sides are equal.
Here, the angles at B and D are the angles between unequal sides (AB ≠ BC and AD ≠ CD).
Therefore, ∠ABC = ∠ADC.
So, if ∠ABC = 50°, then ∠ADC = 50°.

Question 5
Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side.

Show Answer
This is the statement of the Midpoint Theorem.

Show Explanation

Given: In ΔABC, D and E are the midpoints of AB and AC respectively.
To Prove: DE ∥ BC and DE = ½ BC.
Proof: Extend DE to F such that DE = EF. Join FC.
In ΔAED and ΔCEF:
- AE = CE (E is midpoint)
- ∠AED = ∠CEF (Vertically opposite angles)
- DE = EF (By construction)
- ∴ ΔAED ≅ ΔCEF (by SAS rule)
So, AD = CF (by CPCT) and ∠DAE = ∠FCE. But these are alternate interior angles, so AB ∥ CF.
Since AD = DB and AD = CF, we have DB = CF.
We have DB ∥ CF (since AB ∥ CF) and DB = CF. Thus, DBCF is a parallelogram (one pair of opposite sides is parallel and equal).
Therefore, DE ∥ BC (as DF is a part of the parallelogram DBCF and BC is its opposite side).
Also, DE = ½ DF = ½ BC (since DF = BC in parallelogram DBCF).
Hence, proved.

Question 6
The angles of a quadrilateral are 2x, 3x, 4x, and 6x. Find the value of the smallest angle.

Show Answer
The smallest angle is 48°.

Show Explanation

The sum of the interior angles of a quadrilateral is 360°.
So, $2 x + 3 x + 4 x + 6 x = 360 °$
$15 x = 360 °$
$x = 24 °$
The smallest angle is $2 x = 2 \times 24 ° = 48 °$ .

Question 7
Can all the angles of a quadrilateral be acute? Justify your answer.

Show Answer
No.

Show Explanation

An acute angle is less than 90°.
If all four angles were acute (each < 90°), their sum would be less than $4 \times 90 ° = 360 °$ .
However, the sum of the interior angles of any quadrilateral is always exactly 360°.
Therefore, it is impossible for all four angles to be acute.

Question 8
In a rectangle, one diagonal is 10 cm. What is the length of the other diagonal?

Show Answer
10 cm

Show Explanation

A key property of a rectangle is that its diagonals are equal in length.
Therefore, if one diagonal is 10 cm, the other diagonal must also be 10 cm.

Question 9
In an isosceles trapezium, if one of the base angles is 65°, what is the measure of the opposite angle?

Show Answer
65°

Show Explanation

In an isosceles trapezium, the base angles (angles on the same base) are equal.
The angles adjacent to each leg are supplementary, but the angles that are opposite are actually equal.
Specifically, the angles at each base are equal. So, if one base angle is 65°, the angle opposite to it (which is on the same base) is also 65°.

Question 10
Define a square and list any one of its unique properties.

Show Answer
A square is a quadrilateral that is both a rectangle (all angles are 90°) and a rhombus (all sides are equal).
Unique Property: Its diagonals are equal, perpendicular, and bisect each other. (Any one of these combinations is sufficient).

Show Explanation
A square inherits properties from both rectangles and rhombuses, making its set of properties unique.

Question 11
In a quadrilateral PQRS, the diagonals PR and QS bisect each other. What is the special name of this quadrilateral?

Show Answer
Parallelogram

Show Explanation

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
This is a standard test for a quadrilateral to be a parallelogram.

Question 12
If the diagonals of a quadrilateral are equal and bisect each other, prove that it is a rectangle.

Show Answer
Proof:

Show Explanation

Given: A quadrilateral ABCD where diagonals AC = BD and they bisect each other at O (so, OA=OC, OB=OD).
Since diagonals bisect each other, ABCD is a parallelogram.
Now, consider ΔABC and ΔDCB.
- AB = DC (Opposite sides of parallelogram)
- BC = BC (Common side)
- AC = DB (Given)
- ∴ ΔABC ≅ ΔDCB (by SSS rule)
Therefore, ∠ABC = ∠DCB (by CPCT).
But ∠ABC and ∠DCB are consecutive interior angles on the same side of transversal BC, and since AB ∥ DC, their sum is 180°.
So, ∠ABC + ∠DCB = 180°
=> 2∠ABC = 180° (since they are equal)
=> ∠ABC = 90°
If one angle of a parallelogram is 90°, then it is a rectangle.
Hence, proved.

Question 13
The perimeter of a parallelogram is 60 cm. If one side is 12 cm, find the length of the adjacent side.

Show Answer
18 cm

Show Explanation

Let the parallelogram be ABCD with AB = 12 cm.
Let the adjacent side BC = x cm.
Perimeter of parallelogram = 2(Sum of adjacent sides) = 2(AB + BC)
So, $2 (12 + x) = 60$
$12 + x = 30$
$x = 30 - 12$
$x = 18$ cm

Question 14
In ΔPQR, S and T are midpoints of sides PQ and PR respectively. If QR = 14 cm, find ST.

Show Answer
ST = 7 cm

Show Explanation

This is a direct application of the Midpoint Theorem.
The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half of it.
Here, S and T are midpoints of PQ and PR.
Therefore, ST ∥ QR and ST = ½ QR.
Given QR = 14 cm, ST = ½ × 14 = 7 cm.

Question 15
A field is in the shape of a rhombus whose perimeter is 400 m and one of its diagonals is 160 m. Find the area of the field.

Show Answer
The area is 9600 m².

Show Explanation

Perimeter of rhombus = 400 m. Since all sides are equal, each side = 400 / 4 = 100 m.
Let the diagonals be AC and BD, intersecting at O. Let AC = 160 m. So, AO = OC = 80 m.
Diagonals of a rhombus are perpendicular bisectors. So, ΔAOB is a right-angled triangle at O.
In right ΔAOB, AB (hypotenuse) = 100 m, AO (base) = 80 m.
Using Pythagoras theorem to find OB (height):
OB² = AB² - AO²
OB² = (100)² - (80)²
OB² = 10000 - 6400
OB² = 3600
OB = √3600 = 60 m
Therefore, the other diagonal BD = 2 × OB = 2 × 60 = 120 m.
Area of a rhombus = ½ × (Product of diagonals) = ½ × (AC × BD) = ½ × (160 × 120) = ½ × 19200 = 9600 m².

Short Answer Type II (3 Marks each) - 10 Questions

Question 1
Prove that the diagonals of a rectangle are equal.

Show Answer
Proof completed.

Show Explanation

Given: A rectangle ABCD with diagonals AC and BD.
To Prove: AC = BD.
Proof:
Consider triangles ΔABC and ΔDCB.
- AB = DC (Opposite sides of a rectangle are equal)
- BC = CB (Common side)
- ∠ABC = ∠DCB = 90° (All angles of a rectangle are 90°)
Therefore, ΔABC ≅ ΔDCB by the SAS (Side-Angle-Side) congruence rule.
Hence, AC = BD (Corresponding Parts of Congruent Triangles are equal - CPCT).
Thus, the diagonals of a rectangle are equal.

Question 2
ABCD is a rhombus. Show that the diagonal AC bisects ∠A as well as ∠C.

Show Answer
Proof completed.

Show Explanation

Given: A rhombus ABCD with diagonal AC.
To Prove: AC bisects ∠A and ∠C. i.e., ∠DAC = ∠BAC and ∠DCA = ∠BCA.
Proof:
Consider triangles ΔABC and ΔADC.
- AB = AD (All sides of a rhombus are equal)
- BC = DC (All sides of a rhombus are equal)
- AC = AC (Common side)
Therefore, ΔABC ≅ ΔADC by the SSS (Side-Side-Side) congruence rule.
Hence, by CPCT:
- ∠BAC = ∠DAC. So, AC bisects ∠A.
- ∠BCA = ∠DCA. So, AC bisects ∠C.
  Thus, diagonal AC bisects both ∠A and ∠C.

Question 3
In quadrilateral ABCD, E, F, G, H are the midpoints of sides AB, BC, CD, and DA respectively. Prove that EFGH is a parallelogram.

Show Answer
Proof completed.

Show Explanation

Given: A quadrilateral ABCD with E, F, G, H as midpoints of AB, BC, CD, and DA.
To Prove: EFGH is a parallelogram.
Construction: Join the diagonal AC.
Proof:
- In ΔABC, E and F are midpoints of AB and BC.
  Therefore, by the Midpoint Theorem, EF ∥ AC and EF = ½ AC. ...(1)
- In ΔADC, H and G are midpoints of AD and CD.
  Therefore, by the Midpoint Theorem, HG ∥ AC and HG = ½ AC. ...(2)
From equations (1) and (2):
- EF ∥ HG (Both are parallel to AC)
- EF = HG (Both are equal to half of AC)
A quadrilateral (EFGH) with one pair of opposite sides (EF and HG) that are both parallel and equal is a parallelogram.
Hence, proved.

Question 4
In a parallelogram, show that the angle bisectors of two adjacent angles intersect at a right angle.

Show Answer
Proof completed.

Show Explanation

Given: A parallelogram ABCD where AE is the bisector of ∠A and DF is the bisector of ∠D. AE and DF intersect at point O.
To Prove: ∠AOD = 90°.
Proof:
- In a parallelogram, adjacent angles are supplementary. So, ∠A + ∠D = 180°.
- Since AE and DF are bisectors, ∠OAD = ½ ∠A and ∠ODA = ½ ∠D.
- In ΔAOD, the sum of angles is 180°.
  So, ∠OAD + ∠ODA + ∠AOD = 180°
  => ½ ∠A + ½ ∠D + ∠AOD = 180°
  => ½ (∠A + ∠D) + ∠AOD = 180°
  => ½ (180°) + ∠AOD = 180° (Substituting ∠A + ∠D = 180°)
  => 90° + ∠AOD = 180°
  => ∠AOD = 180° - 90° = 90°.
  Hence, the angle bisectors of two adjacent angles intersect at a right angle.

Question 5
The ratio of two adjacent sides of a parallelogram is 5:4. Its perimeter is 90 cm. Find the lengths of all sides.

Show Answer
The sides are 25 cm, 20 cm, 25 cm, and 20 cm.

Show Explanation

Let the two adjacent sides be $5 x$ and $4 x$ .
The perimeter of a parallelogram is $2 \times (Sum of adjacent sides)$ .
So, $2 (5 x + 4 x) = 90$
$2 (9 x) = 90$
$18 x = 90$
$x = 90 / 18$
$x = 5$
Therefore, the sides are:
- $5 x = 5 \times 5 = 25$ cm
- $4 x = 4 \times 5 = 20$ cm
In a parallelogram, opposite sides are equal.
So, the lengths of all four sides are 25 cm, 20 cm, 25 cm, and 20 cm.

Question 6
Two parallel lines l and m are intersected by a transversal p. Show that the quadrilateral formed by the bisectors of the interior angles is a rectangle.

Show Answer
Proof completed.

Show Explanation

Given: Lines l ∥ m, transversal p. Let the bisectors of interior angles form quadrilateral ABCD.
To Prove: ABCD is a rectangle.
Proof:
- Since l ∥ m, the consecutive interior angles are supplementary. For example, ∠1 + ∠2 = 180°.
- The bisectors will create angles that are half of these. So, half of ∠1 + half of ∠2 = ½(∠1 + ∠2) = ½(180°) = 90°.
- Therefore, in the quadrilateral, one of the angles (e.g., ∠BAD) is 90°.
- Similarly, all other interior angles of the quadrilateral can be shown to be 90° by considering different pairs of consecutive interior angles.
Since all angles of the quadrilateral ABCD are 90°, it is a rectangle.

Question 7
In a trapezium ABCD, AB ∥ CD. If ∠A : ∠D = 3:2 and ∠B : ∠C = 4:5, find all the angles of the trapezium.

Show Answer
The angles are ∠A = 108°, ∠B = 96°, ∠C = 120°, ∠D = 72°.

Show Explanation

In a trapezium with AB ∥ CD, ∠A and ∠D are adjacent angles on the same side of the transversal AD, so they are supplementary.
So, ∠A + ∠D = 180°.
Let ∠A = 3x and ∠D = 2x.
Then, 3x + 2x = 180° => 5x = 180° => x = 36°.
Therefore, ∠A = 3 × 36° = 108° and ∠D = 2 × 36° = 72°.
Similarly, ∠B and ∠C are adjacent angles on the same side of the transversal BC, so they are also supplementary.
So, ∠B + ∠C = 180°.
Let ∠B = 4y and ∠C = 5y.
Then, 4y + 5y = 180° => 9y = 180° => y = 20°.
Therefore, ∠B = 4 × 20° = 80° and ∠C = 5 × 20° = 100°.
*Correction: The problem states ∠B : ∠C = 4:5, leading to 4y + 5y = 180°, y=20, so ∠B=80°, ∠C=100°. The initial answer was calculated with a different ratio. The final answer is ∠A=108°, ∠B=80°, ∠C=100°, ∠D=72°.*

Question 8
Prove that a diagonal of a parallelogram divides it into two congruent triangles.

Show Answer
Proof completed.

Show Explanation

Given: A parallelogram ABCD with diagonal AC.
To Prove: ΔABC ≅ ΔCDA.
Proof:
Consider triangles ΔABC and ΔCDA.
- AB = CD (Opposite sides of a parallelogram are equal)
- BC = DA (Opposite sides of a parallelogram are equal)
- AC = CA (Common side)
Therefore, ΔABC ≅ ΔCDA by the SSS (Side-Side-Side) congruence rule.
Hence, the diagonal AC divides the parallelogram into two congruent triangles.

Question 9
In kite ABCD, diagonals AC and BD intersect at O. If AB = 5 cm and BD = 8 cm, find the length of side AD and the area of the kite.

Show Answer
AD = 5 cm, Area = 24 cm².

Show Explanation

In a kite, two pairs of adjacent sides are equal. AB = AD and CB = CD.
Given AB = 5 cm, so AD = 5 cm.
The diagonals of a kite are perpendicular. The longer diagonal bisects the shorter one.
Given BD = 8 cm. Let's assume BD is the longer diagonal. So, OB = OD = 8/2 = 4 cm.
In right triangle ΔAOB, AB is the hypotenuse.
AB² = AO² + OB²
(5)² = AO² + (4)²
25 = AO² + 16
AO² = 25 - 16 = 9
AO = 3 cm
The area of a kite is given by ½ × (Product of diagonals) = ½ × (AC × BD).
We have AC = AO + OC. In a kite, the longer diagonal (BD) bisects the shorter one (AC), so OC = AO = 3 cm.
Thus, AC = 3 + 3 = 6 cm.
Area = ½ × (6 cm × 8 cm) = ½ × 48 = 24 cm².

Question 10
In ΔABC, D, E, and F are the midpoints of sides BC, CA, and AB respectively. Show that ΔABC is divided into four congruent triangles by segments DE, EF, and FD.

Show Answer
Proof completed.

Show Explanation

Given: D, E, F are midpoints of BC, CA, and AB in ΔABC. DE, EF, FD are joined.
To Prove: ΔAEF ≅ ΔFBD ≅ ΔEDC ≅ ΔFDE.
Proof:
- By the Midpoint Theorem:
  FE ∥ BC and FE = ½ BC = BD = DC.
  DF ∥ AC and DF = ½ AC = AE = EC.
  ED ∥ AB and ED = ½ AB = AF = FB.
Now, consider the triangles:
- In ΔAEF and ΔFBD:
  AF = FB (F is midpoint)
  ∠AFE = ∠FBD (Corresponding angles, FE ∥ BC)
  FE = BD (Proved above)
  ∴ ΔAEF ≅ ΔFBD (by SAS rule)
- Similarly, ΔAEF ≅ ΔEDC (by SAS rule).
- Now, for ΔFDE, we can see that its sides are FE, ED, DF.
  FE = BD = DC
  ED = AF = FB
  DF = AE = EC
  But from the congruences above, these sides are equal to the corresponding sides of ΔAEF, ΔFBD, and ΔEDC.
  Specifically, in ΔAEF, the sides are AF, FE, EA. We have DF = EA, FE = FE, ED = AF.
  So, by SSS rule, ΔFDE ≅ ΔAEF.
Therefore, all four small triangles are congruent to each other.
Long Answer Questions (5 Marks each) - 10 Questions
Question 1
Construct a quadrilateral ABCD where AB = 4.5 cm, BC = 5.5 cm, CD = 4 cm, AD = 6 cm, and diagonal AC = 7 cm. Measure the length of BD.
Show Answer
The constructed quadrilateral will have BD approximately 6.2 cm (Measurement may vary slightly).
Show Explanation
Steps of Construction:
1. Draw a line segment AB of length 4.5 cm.
2. With A as center and radius 7 cm, draw an arc.
3. With B as center and radius 5.5 cm, draw another arc to intersect the first arc at point C. Join BC.
4. With A as center and radius 6 cm, draw an arc.
5. With C as center and radius 4 cm, draw another arc to intersect the previous arc at point D.
6. Join AD and CD to complete quadrilateral ABCD.
7. Join BD and measure its length. It will be approximately 6.2 cm.
Question 2
Prove that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Show Answer
Proof completed.
Show Explanation
1. Given: A quadrilateral ABCD where diagonals AC = BD, they bisect each other (OA=OC, OB=OD), and they are perpendicular (AC ⊥ BD).
2. To Prove: ABCD is a square.
3. Proof:
  Since diagonals bisect each other, ABCD is a parallelogram.
  Now, consider ΔAOB and ΔAOD.
  OB = OD (Diagonals bisect each other)
  AO = AO (Common side)
  ∠AOB = ∠AOD = 90° (Diagonals are perpendicular)
  ∴ ΔAOB ≅ ΔAOD (by SAS rule)
  Therefore, AB = AD (by CPCT).
  In a parallelogram, if adjacent sides are equal, it is a rhombus. So, ABCD is a rhombus.
  Now, we have a rhombus ABCD with equal diagonals (AC = BD).
  In a rhombus, the diagonals are perpendicular bisectors. We are given that they are equal.
  Consider ΔABC and ΔDCB in rhombus ABCD.
  AB = DC (Opposite sides of rhombus)
  BC = BC (Common side)
  AC = DB (Given)
  ∴ ΔABC ≅ ΔDCB (by SSS rule)
  Therefore, ∠ABC = ∠DCB (by CPCT).
  But in a rhombus, consecutive angles are supplementary: ∠ABC + ∠DCB = 180°.
  So, 2∠ABC = 180° => ∠ABC = 90°.
  A rhombus with one right angle is a square.
  Hence, proved that ABCD is a square.
Question 3
ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square
(ii) Diagonal BD bisects ∠B as well as ∠D.
Show Answer
Proof completed.
Show Explanation
Part (i): Proving ABCD is a square
1. Given: ABCD is a rectangle, so all angles are 90°. AC bisects ∠A and ∠C.
2. Since AC bisects ∠A (which is 90°), ∠DAC = ∠BAC = 45°.
3. In ΔABC, ∠B = 90° and ∠BAC = 45°. Therefore, ∠BCA = 180° - (90°+45°) = 45°.
4. So, ∠BAC = ∠BCA = 45°. This means sides opposite to these angles are equal.
  So, BC = AB.
5. In a rectangle, adjacent sides AB and BC are equal. Therefore, ABCD is a square.
Part (ii): Proving BD bisects ∠B and ∠D
1. Since ABCD is a square, all properties of a square apply.
2. In a square, the diagonals bisect the interior angles.
3. Therefore, diagonal BD bisects ∠B (90° becomes 45° and 45°) and also bisects ∠D.
Question 4
A plot of land is in the shape of a quadrilateral. A farmer divides it equally among his two sons by building a straight boundary wall from one vertex to the midpoint of the opposite side. Which theorem did he use? Prove the theorem used.
Show Answer
The farmer used a corollary of the Midpoint Theorem. The theorem states: A line segment joining a vertex of a triangle to the midpoint of the opposite side (a median) divides the triangle into two regions of equal area.
Show Explanation
Proof of the Theorem:
1. Given: ΔABC, where D is the midpoint of BC. AD is the median.
2. To Prove: ar(ΔABD) = ar(ΔADC).
3. Proof:
  Draw AL ⊥ BC.
  Now, area of ΔABD = ½ × BD × AL.
  Area of ΔADC = ½ × DC × AL.
  Since D is the midpoint of BC, BD = DC.
  Therefore, ½ × BD × AL = ½ × DC × AL.
  Hence, ar(ΔABD) = ar(ΔADC).
4. The farmer applied this to the quadrilateral by dividing it into two triangles with a diagonal and then drawing a median in one of the triangles, thus creating two plots of equal area.
Question 5
In a parallelogram, prove that the bisectors of any two consecutive angles intersect at a right angle.
Show Answer
Proof completed.
Show Explanation
1. Given: A parallelogram ABCD. Let the bisectors of ∠A and ∠B meet at point O.
2. To Prove: ∠AOB = 90°.
3. Proof:
  In a parallelogram, consecutive angles are supplementary. So, ∠A + ∠B = 180°.
  Since AO and BO are bisectors, ∠OAB = ½ ∠A and ∠OBA = ½ ∠B.
  In ΔAOB, the sum of angles is 180°.
  So, ∠OAB + ∠OBA + ∠AOB = 180°
  => ½ ∠A + ½ ∠B + ∠AOB = 180°
  => ½ (∠A + ∠B) + ∠AOB = 180°
  => ½ (180°) + ∠AOB = 180° (Substituting ∠A + ∠B = 180°)
  => 90° + ∠AOB = 180°
  => ∠AOB = 180° - 90° = 90°.
  Hence, the bisectors of any two consecutive angles intersect at a right angle.
Question 6
The diagonal of a rhombus is 16 cm and 30 cm long. Find its perimeter and area.
Show Answer
Perimeter = 68 cm, Area = 240 cm².
Show Explanation
1. Let the rhombus be ABCD with diagonals AC = 16 cm and BD = 30 cm.
2. Diagonals of a rhombus bisect each other at right angles.
  So, OA = OC = 16/2 = 8 cm.
  OB = OD = 30/2 = 15 cm.
3. In right triangle ΔAOB, side AB (the side of the rhombus) is the hypotenuse.
  AB² = OA² + OB² (Pythagoras Theorem)
  AB² = (8)² + (15)²
  AB² = 64 + 225
  AB² = 289
  AB = √289 = 17 cm.
4. Perimeter of rhombus = 4 × side = 4 × 17 cm = 68 cm.
5. Area of rhombus = ½ × (Product of diagonals) = ½ × (16 × 30) = ½ × 480 = 240 cm².
Question 7
The given figure shows a pattern made from tiles shaped as parallelograms and rhombuses. Using the properties of these shapes, prove that the pattern can tessellate a plane without gaps.
Show Answer
The pattern can tessellate because the sum of angles around each common vertex is 360°.
Show Explanation
1. For a shape to tessellate, the arrangement of shapes around every point must cover a full 360° without gaps or overlaps.
2. Let's analyze the angles meeting at a common vertex in the pattern.
3. A parallelogram has opposite angles equal and consecutive angles supplementary. A rhombus has all sides equal and opposite angles equal.
4. In the given pattern, you will find that at each point where vertices meet, a combination of angles from the parallelograms and rhombuses adds up.
5. For example, two acute angles from two rhombuses and one obtuse angle from a parallelogram might meet, or vice-versa.
6. Due to the specific shapes and angles used, the sum of these angles at every meeting point is exactly 360°.
7. This property, derived from the angle measures of parallelograms and rhombuses, allows the pattern to cover the plane completely without any gaps, creating a tessellation.
Question 8
PQRS is a trapezium with PQ ∥ SR. A line parallel to the base PQ passes through the midpoint M of side PS and meets QR at N. Prove that N is the midpoint of QR.
Show Answer
Proof completed.
Show Explanation
1. Given: Trapezium PQRS with PQ ∥ SR. M is midpoint of PS. MN ∥ PQ (and hence MN ∥ SR, as PQ ∥ SR).
2. To Prove: N is the midpoint of QR.
3. Construction: Join the diagonal PR. Let MN intersect PR at O.
4. Proof:
  In ΔPSR:
  M is the midpoint of PS.
  MO ∥ SR (as MN ∥ SR and O lies on MN).
  Therefore, by the converse of the Midpoint Theorem, O is the midpoint of PR.
  Now, in ΔPQR:
  O is the midpoint of PR.
  ON ∥ PQ (Given MN ∥ PQ).
  Therefore, by the converse of the Midpoint Theorem, N is the midpoint of QR.
  Hence, proved.
Question 9
ABC is a triangle right-angled at C. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that:
(i) D is the midpoint of AC
(ii) MD ⊥ AC
(iii) CM = MA = ½ AB
Show Answer
Proof completed.
Show Explanation
Part (i): D is the midpoint of AC
1. In ΔABC, M is the midpoint of AB.
2. MD ∥ BC (Given).
3. Therefore, by the converse of the Midpoint Theorem, D is the midpoint of AC.
Part (ii): MD ⊥ AC
1. Given that ∠C = 90°.
2. MD ∥ BC (Given).
3. If a line is parallel to one line and perpendicular to another, then it is perpendicular to the first line as well. More directly:
  Since BC ⊥ AC and MD ∥ BC, therefore MD ⊥ AC.
Part (iii): CM = MA = ½ AB
1. Join MC.
2. In ΔAMD and ΔCMD:
  AD = CD (D is midpoint of AC)
  MD = MD (Common side)
  ∠ADM = ∠CDM = 90° (MD ⊥ AC)
  ∴ ΔAMD ≅ ΔCMD (by SAS rule)
3. Therefore, CM = AM (by CPCT).
4. But M is the midpoint of AB, so AM = ½ AB.
5. Thus, CM = MA = ½ AB.
Question 10
Construct a parallelogram whose diagonals are 6 cm and 8 cm long and the angle between them is 60°. Measure the sides of the parallelogram.
Show Answer
The constructed parallelogram will have sides approximately 5 cm and 3.6 cm (Measurements may vary slightly).
Show Explanation
Steps of Construction:
1. Draw a horizontal line segment AC of length 8 cm.
2. Find the midpoint O of AC (since diagonals of a parallelogram bisect each other).
3. At O, construct a line making an angle of 60° with AC.
4. On this line, on both sides of O, mark points B and D such that OB = OD = 3 cm (half of the other diagonal, 6 cm / 2 = 3 cm).
5. Join points A, B, C, and D to form parallelogram ABCD.
6. Measure the sides AB and BC. AB will be approximately 5 cm and BC will be approximately 3.6 cm.
  Case-Based Questions
  Case 1: The Playground
  A school playground is in the shape of a quadrilateral ABCD. The coach marks points E, F, G, H as the midpoints of sides AB, BC, CD, and DA respectively for a relay race.
  Question 1
  The quadrilateral EFGH formed is a:
  (a) Rectangle (b) Rhombus (c) Parallelogram (d) Square
  Show Answer
  (c) Parallelogram
  Show Explanation
  This is a standard theorem. The line segment joining the midpoints of any quadrilateral always forms a parallelogram.
  Question 2
  If the original quadrilateral ABCD is a rectangle, then EFGH will be a:
  (a) Rectangle (b) Rhombus (c) Parallelogram (d) Square
  Show Answer
  (b) Rhombus
  Show Explanation
  In a rectangle, the diagonals are equal (AC = BD). The sides of the midpoint quadrilateral EFGH are each equal to half of a diagonal (e.g., EF = ½ AC, FG = ½ BD). Since AC = BD, all sides of EFGH are equal. A parallelogram with all sides equal is a rhombus.
  Question 3
  If the original quadrilateral ABCD is a rhombus, then EFGH will be a:
  (a) Rectangle (b) Rhombus (c) Parallelogram (d) Square
  Show Answer
  (a) Rectangle
  Show Explanation
  In a rhombus, the diagonals are perpendicular (AC ⊥ BD). In the midpoint quadrilateral EFGH, the sides are parallel to the diagonals of the original rhombus (EF ∥ AC, FG ∥ BD). Since AC ⊥ BD, EF ⊥ FG. A parallelogram with one right angle is a rectangle.
  Question 4
  The theorem used to determine the nature of EFGH is the:
  (a) Pythagoras Theorem (b) Basic Proportionality Theorem
  (c) Midpoint Theorem (d) Angle Sum Property
  Show Answer
  (c) Midpoint Theorem
  Show Explanation
  The Midpoint Theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. This theorem is applied repeatedly to triangles formed within the quadrilateral to prove that EFGH is a parallelogram and to find its specific properties.
  Case 2: The Kite Festival
  During a kite festival, Rohan made a kite ABCD. He ensures that AB = AD and CB = CD. The diagonals AC and BD intersect at O. The length of AC is 24 cm and BD is 10 cm.
  Question 1
  The diagonals of a kite are:
  (a) Equal (b) Perpendicular (c) Parallel (d) None of these
  Show Answer
  (b) Perpendicular
  Show Explanation
  A key property of a kite is that its diagonals are perpendicular to each other.
  Question 2
  The area of Rohan's kite is:
  (a) 120 cm² (b) 240 cm² (c) 60 cm² (d) 34 cm²
  Show Answer
  (a) 120 cm²
  Show Explanation
  The area of a kite is given by half the product of its diagonals.
  Area = ½ × d₁ × d₂ = ½ × (AC) × (BD) = ½ × 24 cm × 10 cm = ½ × 240 = 120 cm².
  Question 3
  If ∠ABO = 35°, then what is the measure of ∠BAO?
  (a) 35° (b) 55° (c) 90° (d) 70°
  Show Answer
  (b) 55°
  Show Explanation
  In a kite, the longer diagonal (AC here) bisects the angles at the vertices it joins (A and C).
  Also, the diagonals are perpendicular, so in triangle AOB, ∠AOB = 90°.
  In right-angled ΔAOB, the sum of angles is 180°.
  So, ∠BAO + ∠ABO + ∠AOB = 180°
  => ∠BAO + 35° + 90° = 180°
  => ∠BAO + 125° = 180°
  => ∠BAO = 180° - 125° = 55°.
  Question 4
  The shorter diagonal of the kite is:
  (a) AC (b) BD (c) AB (d) AD
  Show Answer
  (b) BD
  Show Explanation
  The diagonals are given as AC = 24 cm and BD = 10 cm. Therefore, BD (10 cm) is the shorter diagonal.
  Case 3: The Picture Frame
  Mohan bought a rectangular picture frame. To check if it is perfectly rectangular, he measures the diagonals and finds they are 45 cm each.
  Question 1
  The property Mohan used to check the frame is:
  (a) Diagonals bisect each other (b) Diagonals are equal
  (c) Diagonals are perpendicular (d) All sides are equal
  Show Answer
  (b) Diagonals are equal
  Show Explanation
  A parallelogram is a rectangle if and only if its diagonals are equal. By verifying that the diagonals are equal, Mohan confirmed the frame is rectangular.
  Question 2
  If the length of the frame is 36 cm, what is its perimeter?
  (a) 126 cm (b) 108 cm (c) 90 cm (d) 81 cm
  Show Answer
  (a) 126 cm
  Show Explanation
  Let the rectangle be ABCD with AB = length = 36 cm, and diagonal AC = 45 cm.
  In right triangle ABC, by Pythagoras Theorem:
  BC² = AC² - AB²
  BC² = (45)² - (36)²
  BC² = 2025 - 1296
  BC² = 729
  BC = √729 = 27 cm. (This is the width/breadth).
  Perimeter of rectangle = 2 × (Length + Breadth) = 2 × (36 cm + 27 cm) = 2 × 63 cm = 126 cm.
  Question 3
  If the frame is deformed into a parallelogram without changing side lengths, what happens to the diagonal lengths?
  (a) They become equal (b) They become perpendicular (c) They become unequal (d) They bisect the angles
  Show Answer
  (c) They become unequal
  Show Explanation
  In a general parallelogram (that is not a rectangle), the diagonals are not equal. When a rectangle is deformed into a non-rectangular parallelogram, the diagonals change length and become unequal.
  Question 4
  A square is a special rectangle because it:
  (a) Has equal diagonals (b) Has all angles of 90° (c) Has all sides equal
  (d) Both (b) and (c)
  Show Answer
  (d) Both (b) and (c)
  Show Explanation
  A square satisfies all the properties of a rectangle (all angles 90°, opposite sides equal, diagonals equal) and has the additional property that all four sides are equal.
  Case 4: The Garden Plot
  A garden is in the shape of a parallelogram. The gardener wants to divide it into two equal parts by planting a hedge along one of its diagonals.
  Question 1
  The diagonal divides the parallelogram into:
  (a) Two similar triangles (b) Two congruent triangles
  (c) Two triangles of different areas (d) A triangle and a trapezium
  Show Answer
  (b) Two congruent triangles
  Show Explanation
  A diagonal of a parallelogram always divides it into two congruent triangles (which can be proven using the SSS congruence rule: two sides and the included diagonal are equal).
  Question 2
  If the adjacent sides of the plot are 15 m and 20 m, and one diagonal is 25 m, what is the shape of the triangular plot formed?
  (a) Right-angled triangle (b) Equilateral triangle
  (c) Isosceles triangle (d) Scalene triangle
  Show Answer
  (a) Right-angled triangle
  Show Explanation
  Check if the triangle formed by sides 15 m, 20 m, and the diagonal 25 m satisfies the Pythagoras Theorem.
  (15)² + (20)² = 225 + 400 = 625.
  (25)² = 625.
  Since (15)² + (20)² = (25)², the triangle is right-angled.
  Question 3
  If the gardener wants to divide it into four equal parts, he should plant hedges along:
  (a) The diagonals (b) The medians (c) The angle bisectors (d) The lines joining midpoints
  Show Answer
  (a) The diagonals
  Show Explanation
  The diagonals of a parallelogram bisect each other. When both diagonals are drawn, they divide the parallelogram into four triangles. These triangles are congruent in pairs (opposite triangles are congruent) and, in fact, all four are equal in area. Thus, planting hedges along both diagonals creates four plots of equal area.
  Question 4
  The diagonal of a parallelogram divides it into two congruent triangles. This property is based on which congruence rule?
  (a) SSS (b) SAS (c) ASA (d) RHS
  Show Answer
  (a) SSS
  Show Explanation
  Consider parallelogram ABCD and diagonal AC.
  In ΔABC and ΔCDA:
  AB = CD (Opposite sides are equal)
  BC = DA (Opposite sides are equal)
  AC = CA (Common side)
  Therefore, ΔABC ≅ ΔCDA by the SSS (Side-Side-Side) congruence rule.
  Case 5: The Tile Design
  An interior designer uses a combination of square and rhombus-shaped tiles. All tiles have the same side length of 10 cm.
  Question 1
  The main difference between the square and rhombus tiles is:
  (a) Number of sides (b) Length of diagonals
  (c) Measure of angles (d) Both (b) and (c)
  Show Answer
  (d) Both (b) and (c)
  Show Explanation
  Both squares and rhombuses have 4 sides. The differences are:
  Angles: A square has all angles 90°. A rhombus has opposite angles equal, but they are not necessarily 90°.
  Diagonals: A square's diagonals are equal. A rhombus's diagonals are generally unequal.
  Question 2
  If the diagonals of the rhombus tile are 16 cm and 12 cm, its area is:
  (a) 96 cm² (b) 160 cm² (c) 192 cm² (d) 100 cm²
  Show Answer
  (a) 96 cm²
  Show Explanation
  Area of a rhombus = ½ × (Product of diagonals) = ½ × (16 cm × 12 cm) = ½ × 192 = 96 cm².
  Question 3
  The designer can create a tessellation pattern without gaps because:
  (a) The sum of angles around a point is 360° (b) All sides are equal
  (c) Diagonals are perpendicular (d) Opposite sides are parallel
  Show Answer
  (a) The sum of angles around a point is 360°
  Show Explanation
  Tessellation works when the shapes can be arranged so that at every vertex where they meet, the sum of the angles from all the shapes is exactly 360°. The specific angles of the squares (90°) and rhombuses (e.g., 60° and 120°) are chosen to fit together perfectly to achieve this sum.
  Question 4
  If the rhombus tiles are replaced with rectangular tiles of the same area, the pattern will:
  (a) Have no gaps (b) Have gaps (c) Remain the same (d) Cannot be determined
  Show Answer
  (b) Have gaps
  Show Explanation
  Tessellation is highly dependent on the shapes and angles of the tiles. Simply having the same area does not guarantee that the rectangles will fit together with the squares in the same way the rhombuses did. The different angles of the rectangles would likely prevent them from fitting perfectly around a point with the squares, resulting in gaps.

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Saturday, November 15, 2025

Chapter 8 Quadrilaterals Question Bank Q & Answers

Quadrilaterals - Complete Q&A

Multiple Choice Questions (20 Questions)

Assertion & Reasoning Questions (20 Questions)

True/False Questions (10 Questions)

Short Answer Type I (2 Marks each) - 15 Questions

Short Answer Type II (3 Marks each) - 10 Questions

Long Answer Questions (5 Marks each) - 10 Questions

Case-Based Questions

Case 1: The Playground

Case 2: The Kite Festival

Case 3: The Picture Frame

Case 4: The Garden Plot

Case 5: The Tile Design

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Chapter 8 Quadrilaterals Figure it out Q & Answers

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Saturday, November 15, 2025

Chapter 8 Quadrilaterals Question Bank Q & Answers

Quadrilaterals - Complete Q&A

Multiple Choice Questions (20 Questions)

Assertion & Reasoning Questions (20 Questions)

True/False Questions (10 Questions)

Short Answer Type I (2 Marks each) - 15 Questions

Short Answer Type II (3 Marks each) - 10 Questions

Long Answer Questions (5 Marks each) - 10 Questions

Case-Based Questions

Case 1: The Playground

Case 2: The Kite Festival

Case 3: The Picture Frame

Case 4: The Garden Plot

Case 5: The Tile Design

No comments:

Post a Comment

Chapter 8 Quadrilaterals Figure it out Q & Answers

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