CBSE Class 8 Mathematics Understanding Quadrilaterals Assignments Set 01

Understanding Quadrilaterals

Question. In rectangle PQRS PQ = 8 cm and PS = 6 cm. What is the length of SQ?

Answer : 10 cm

Question. ABCD is a rectangle if AB = 15 cm and AD = 8 cm then find AC

Answer : 17 cm

Question. RENT is a trapezium in which TN||RE. What is the value of x.

Answer : 50°

Question. ABCD is a square with sides 2.4 cm. What is the perimeter of the square?

Answer : 9.6 cm

Question. The perimeter of a square is 196 cm. What is the side of the square?
Answer : 49 cm

Question. What is the total angle sum of a polygon having nine sides?
Answer : 1260°

Question. The side of the square is 16 cm. What is the length of the diagonal?
Answer : 16√2 cm

Question. The perimeter of a parallelogram is 52 cm. If the length of one side is 16 cm. What is the length of adjacent side?
Answer : 10cm

Question. Complete the following :
A quadrilateral can be constructed uniquely if 4 sides and one _________ is known.
Answer : diagonal or one angle

Question. PQRS is a quadrilateral. Find x.

Answer : x = 80°

Question. How many triangles are made while constructing a quadrilateral with four sides and a diagonal? 
Answer : Two

Question. ABCD is a square of side 5 cm. What is the length of its diagonal.
Answer : 5√2 cm

Question. ABCD is a quadrilateral. Find y.

Answer : 35°

Question. Complete the following :
With constructing a quadrilateral with three sides and two diagonals _______ triangles are made on the same base.
Answer : Two.

Question. CLUE is a quadrilateral find w.

Answer : 110°

Question. Name of polygon whose each exterior angle is 60°
Answer : Hexagon

Question. In trapezium ABCD, AB││CD. What is the sum of ∠B and ∠C ?

Answer : 180°

Question. BELT is a rhombus. Find the length of the side of the rhombus

Answer : 5

Question. What is the length of the rectangle if its breadth is 10 cm and the diagonal is 26 cm?
Answer : 24 cm

Question. NICE is a rectangle. What is the length of the diagonal EI.

Answer : 22

Question 1. Name the regular polygon with 8 sides.
Answer: Octagon
In simple words: A regular polygon with 8 sides is called an octagon.

Exam Tip: Memorize standard polygon names: Triangle (3), Quadrilateral (4), Pentagon (5), Hexagon (6), Heptagon (7), Octagon (8), Nonagon (9), Decagon (10).

Question 2. Find the number of diagonals in the figure given below.
Answer: Number of sides of the given pentagon (\(n\)) = 5
Formula for number of diagonals = \(\frac{n(n - 3)}{2}\)
Number of diagonals = \(\frac{5(5 - 3)}{2} = \frac{5 \times 2}{2} = 5\)
In simple words: A pentagon has 5 diagonals connecting non-adjacent vertices.

Exam Tip: Always use the formula \(\frac{n(n - 3)}{2}\) for any polygon with \(n\) sides to find the total number of diagonals quickly.

Question 3. Find x in the following figure.
Answer: The given figure is a quadrilateral with interior angles: \(x\), \(110^\circ\), \(60^\circ\), and a right angle (\(90^\circ\)).
Sum of interior angles of a quadrilateral = \(360^\circ\)
\(x + 110^\circ + 60^\circ + 90^\circ = 360^\circ\)
\(x + 260^\circ = 360^\circ\)
\(x = 360^\circ - 260^\circ = 100^\circ\)
In simple words: The sum of all four interior angles in any quadrilateral is always \(360^\circ\).

Exam Tip: Look closely at perpendicular markings or square symbols to identify right angles (\(90^\circ\)) in geometric figures.

Question 4. Find the measure of each exterior angle of a regular polygon of 9 sides.
Answer: Number of sides (\(n\)) = 9
Sum of all exterior angles of any convex polygon = \(360^\circ\)
Measure of each exterior angle of a regular polygon = \(\frac{360^\circ}{n}\)
Measure of each exterior angle = \(\frac{360^\circ}{9} = 40^\circ\)
In simple words: Divide \(360^\circ\) by the number of sides to get the measure of each exterior angle of a regular polygon.

Question 5. Name the quadrilateral whose diagonals are perpendicular bisector of each other.
Answer: Rhombus and Square
In simple words: Both rhombus and square have diagonals that bisect each other at right angles (\(90^\circ\)).

Exam Tip: Remember that a parallelogram and rectangle have diagonals that bisect each other, but only in a rhombus and square are they perpendicular.

Question 6. Name a quadrilateral with exactly one pair of parallel sides.
Answer: Trapezium
In simple words: A trapezium is defined as a quadrilateral having one pair of opposite sides parallel.

Exam Tip: Do not confuse a trapezium with a parallelogram, which has two pairs of parallel sides.

Question 7. Find x in the following figure.
Answer: The given figure is a quadrilateral with interior angles: \(60^\circ\), \(130^\circ\), \(120^\circ\), and \(x\).
Sum of interior angles of a quadrilateral = \(360^\circ\)
\(60^\circ + 130^\circ + 120^\circ + x = 360^\circ\)
\(310^\circ + x = 360^\circ\)
\(x = 360^\circ - 310^\circ = 50^\circ\)
In simple words: Subtract the sum of the three known angles from \(360^\circ\) to find the unknown angle \(x\).

Exam Tip: Always double-check your addition of the given angles before subtracting from \(360^\circ\).

Question 8. Use the figure given below to find x + y + z.
Answer: In any triangle, the sum of exterior angles taken in an order is \(360^\circ\). Alternatively, let us find each exterior angle using linear pairs.
Let the interior angles of the triangle be \(A, B, C\). From the triangle, one interior angle is \(40^\circ\), and the other two interior angles can be found using linear pairs with \(60^\circ\) and \(80^\circ\).
Interior angle adjacent to \(60^\circ\) is \(180^\circ - 60^\circ = 120^\circ\)
Interior angle adjacent to \(80^\circ\) is \(180^\circ - 80^\circ = 100^\circ\)
Let us check the third interior angle: \(180^\circ - (120^\circ + 100^\circ)\) which matches \(40^\circ\).
Now, calculating exterior angles \(x, y, z\):
\(x = 180^\circ - 120^\circ = 60^\circ\) (or using the property that the sum of exterior angles of any polygon is \(360^\circ\))
Since the figure shows a triangle with exterior angles \(x, y, z\):
\(x + y + z = 360^\circ\)
In simple words: The sum of the exterior angles of any convex polygon (including a triangle) is always \(360^\circ\).

Exam Tip: The sum of exterior angles of any polygon is independent of the number of sides and is always \(360^\circ\).

Question 9. Find the sum of the angles in the figure given below.
Answer: The given figure is a pentagon with 5 sides (\(n = 5\)).
Formula for the sum of interior angles of an \(n\)-sided polygon = \((n - 2) \times 180^\circ\)
Sum of angles = \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)
In simple words: Use the formula \((n - 2) \times 180^\circ\) where \(n\) is the number of sides.

Exam Tip: Remember that a pentagon can be split into 3 triangles, giving \(3 \times 180^\circ = 540^\circ\).

Question 10. Find the sum of the angles in the following figure.
Answer: The given figure is a hexagon with 6 sides (\(n = 6\)).
Formula for the sum of interior angles = \((n - 2) \times 180^\circ\)
Sum of angles = \((6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ\)
In simple words: A hexagon has 6 sides, so its interior angles add up to \(720^\circ\).

Exam Tip: Keep standard values in mind: Triangle = \(180^\circ\), Quadrilateral = \(360^\circ\), Pentagon = \(540^\circ\), Hexagon = \(720^\circ\).

Question 11. Find the number of sides of a regular polygon whose each exterior angle has a measure of \(60^\circ\).
Answer: Measure of each exterior angle = \(60^\circ\)
Total sum of exterior angles = \(360^\circ\)
Number of sides (\(n\)) = \(\frac{360^\circ}{\text{Measure of each exterior angle}}\)
\(n = \frac{360^\circ}{60^\circ} = 6\)
In simple words: Divide \(360^\circ\) by the exterior angle measure to find the number of sides, which is 6 (a hexagon).

Exam Tip: This formula works for any regular polygon: \(n = \frac{360^\circ}{\text{Exterior Angle}}\).

Question 12. ABCD is a trapezium in which \(AB \parallel DC\). If \(\angle A = \angle B = 40^\circ\), find the measures of other two angles.
Answer: Since \(AB \parallel DC\), the adjacent angles on the same side of the transversal are supplementary.
\(\angle A + \angle D = 180^\circ\)
\(40^\circ + \angle D = 180^\circ\)
\(\angle D = 180^\circ - 40^\circ = 140^\circ\)
Similarly, for the other side:
\(\angle B + \angle C = 180^\circ\)
\(40^\circ + \angle C = 180^\circ\)
\(\angle C = 180^\circ - 40^\circ = 140^\circ\)
Therefore, the other two angles are \(140^\circ\) and \(140^\circ\).
In simple words: Consecutive interior angles between parallel lines add up to \(180^\circ\).

Exam Tip: Identify parallel lines and use interior alternate or consecutive interior angle properties.

Question 13. The length of two adjacent sides of a parallelogram are 4 cm and 3 cm. Find its perimeter.
Answer: Let adjacent sides be \(a = 4\) cm and \(b = 3\) cm.
Perimeter of a parallelogram = \(2 \times (a + b)\)
Perimeter = \(2 \times (4 + 3) = 2 \times 7 = 14\) cm
In simple words: Add the two adjacent side lengths and multiply the sum by 2 to get the perimeter.

Exam Tip: Perimeter is the total boundary length. Opposite sides of a parallelogram are equal, so sides are 4 cm, 4 cm, 3 cm, and 3 cm.

Question 14. In the following figure, given a parallelogram ABCD. Find x and y.
Answer: In a parallelogram, opposite sides are equal in length.
\(AD = BC\)
\(3x = 18\)
\(x = \frac{18}{3} = 6\)
Also, \(AB = DC\)
\(3y - 1 = 26\)
\(3y = 26 + 1 = 27\)
\(y = \frac{27}{3} = 9\)
Therefore, \(x = 6\) and \(y = 9\).
In simple words: Set opposite sides equal to each other to solve for the variables \(x\) and \(y\).

Exam Tip: Always state the geometric property used (e.g., "Opposite sides of a parallelogram are equal") before setting up equations.

Question 15. The diagonals AC and BD of rectangle ABCD intersect each other at point O. If OA = 5 cm, find AC and BD.
Answer: In a rectangle, diagonals are equal in length and bisect each other.
Since \(OA = 5\) cm, and diagonals bisect each other at O:
\(AC = 2 \times OA = 2 \times 5 = 10\) cm
Since diagonals of a rectangle are equal: \(BD = AC\)
Therefore, \(BD = 10\) cm.
In simple words: Diagonals of a rectangle are equal and cut each other in half, so doubling half a diagonal gives the full length.

Exam Tip: Memorize rectangle properties: all angles are \(90^\circ\), opposite sides are equal, and diagonals are equal and bisect each other.

Question 16. In parallelogram PQRS, given that OQ = 4 cm, and PR is 5 more than SQ. Find OP.
Answer: In parallelogram PQRS, diagonals bisect each other at point O.
Since \(OQ = 4\) cm, the total length of diagonal \(SQ = 2 \times OQ = 2 \times 4 = 8\) cm.
Given that \(PR\) is 5 cm more than \(SQ\):
\(PR = SQ + 5 = 8 + 5 = 13\) cm.
Since diagonals bisect each other, \(OP\) is half of \(PR\):
\(OP = \frac{PR}{2} = \frac{13}{2} = 6.5\) cm.
In simple words: Find the length of the diagonal SQ, add 5 to get PR, and then take half of PR to find OP.

Question 17. Given ABCD is a trapezium. Find \(m\angle C\).
Answer: In the given trapezium with \(AB \parallel DC\), the adjacent angles on the same side of the transversal are supplementary.
\(\angle B + \angle C = 180^\circ\)
From the figure, \(\angle B = 120^\circ\)
\(120^\circ + \angle C = 180^\circ\)
\(\angle C = 180^\circ - 120^\circ = 60^\circ\)
In simple words: Consecutive interior angles sum to \(180^\circ\), so subtract \(120^\circ\) from \(180^\circ\) to find angle C.

Exam Tip: Verify which sides are parallel before applying the consecutive supplementary angle rule.

Question 18. In parallelogram HOPE, find x and y.
Answer: In parallelogram HOPE, opposite sides are equal.
\(HE = OP\)
\(3x = 18\)
\(x = \frac{18}{3} = 6\)
Also, \(HP = EO\)
\(3y - 1 = 26\)
\(3y = 26 + 1 = 27\)
\(y = \frac{27}{3} = 9\)
Therefore, \(x = 6\) and \(y = 9\).
In simple words: Opposite sides of a parallelogram are equal, allowing us to form simple linear equations for \(x\) and \(y\).

Exam Tip: Double check your arithmetic when solving linear equations.

Question 19. Find the measure of x in the following figure.
Answer: In the given figure, opposite angles of a parallelogram are equal.
One interior angle is adjacent to \(60^\circ\), so its measure is \(180^\circ - 60^\circ = 120^\circ\) (using linear pair).
Alternatively, using properties of parallel lines and alternate/corresponding angles: the angle corresponding to \(110^\circ\) is equal to the angle opposite to it.
Let us use consecutive interior angles: the two adjacent angles add up to \(180^\circ\). If one interior angle is \(60^\circ\), the adjacent interior angle is \(180^\circ - 60^\circ = 120^\circ\).
By vertically opposite angles, the exterior angle \(x\) equals the interior alternate angle or corresponding angle.
Directly, consecutive interior angles are supplementary. Here, the interior angle at the bottom-left is \(180^\circ - 110^\circ = 70^\circ\).
Using the properties of the parallelogram, angle \(x = 110^\circ\).
In simple words: Opposite angles of a parallelogram are equal.

Exam Tip: Identify parallel line segments to quickly find alternate interior or corresponding angles.

Question 20. The measure of two adjacent angles of a parallelogram are in the ratio of 2 : 3. Find the measure of each of the angles of the parallelogram.
Answer: Let the two adjacent angles be \(2x\) and \(3x\).
We know that adjacent angles of a parallelogram are supplementary (sum to \(180^\circ\)).
\(2x + 3x = 180^\circ\)
\(5x = 180^\circ\)
\(x = \frac{180^\circ}{5} = 36^\circ\)
Therefore, the angles are:
First angle = \(2 \times 36^\circ = 72^\circ\)
Second angle = \(3 \times 36^\circ = 108^\circ\)
Since opposite angles of a parallelogram are equal, all four angles of the parallelogram are \(72^\circ, 108^\circ, 72^\circ,\) and \(108^\circ\).
In simple words: Adjacent angles add up to \(180^\circ\). Use the given ratio to find each angle and then use opposite angle equality for the remaining two.

Exam Tip: Always verify that adjacent angles sum to \(180^\circ\) (\(72^\circ + 108^\circ = 180^\circ\)).

Question 21. Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Gspan style="color:#ff0000;">Answer: Let each of the two adjacent angles be \(y\).
Since adjacent angles of a parallelogram are supplementary:
\(y + y = 180^\circ\)
\(2y = 180^\circ\)
\(y = \frac{180^\circ}{2} = 90^\circ\)
Since all four angles are equal in pairs and adjacent angles are \(90^\circ\), all four angles are \(90^\circ\). (The parallelogram is a rectangle or square).
In simple words: If two neighboring angles in a parallelogram are equal and add up to \(180^\circ\), each must be \(90^\circ\).

Exam Tip: A parallelogram with one right angle (or equal adjacent angles) is always a rectangle.

Question 22. EFGH is a parallelogram. Find the angle measures x, y and z.
Answer: In parallelogram EFGH:
1. Angle \(z\): Since \(HE \parallel GF\) and diagonal \(EG\) acts as a transversal, alternate interior angles are equal. Therefore, \(z = 40^\circ\).
2. Angle \(x\): Opposite angles of a parallelogram are equal, or using triangle sum property in \(\Delta EGH\).
Angle \(\angle EGH = y\) (or similar). Let us use alternate interior angles: \(\angle HEG = \angle FGE = z = 40^\circ\).
Angle at F is given as \(70^\circ\) (exterior angle) or interior angle \(\angle GFE = 180^\circ - 70^\circ = 110^\circ\).
Since opposite angles are equal, \(\angle H = \angle F = 110^\circ\).
Thus, \(x = 110^\circ\).
3. Angle \(y\): In triangle \(EGH\), the sum of angles is \(180^\circ\).
\(\angle HEG + \angle EHG + \angle EGH = 180^\circ\)
\(z + x + y = 180^\circ\) (Wait, let us check proper labeling: \(\angle H = x\), so \(x = 110^\circ\)).
In triangle \(EGH\): \(\angle H + \angle HEG + \angle EGH = 180^\circ\)
\(110^\circ + 40^\circ + y = 180^\circ\)
\(150^\circ + y = 180^\circ \implies y = 30^\circ\).
Therefore, \(x = 110^\circ, y = 30^\circ, z = 40^\circ\).
In simple words: Use alternate interior angles for \(z\), opposite angles for \(x\), and the angle sum property of triangles for \(y\).

Exam Tip: Break complex parallelogram figures into triangles by drawing diagonals to easily apply the angle sum property.

Question 23. The diagonal AC of rhombus ABCD is equal to one of its sides BC. Find all the angles of rhombus.
Answer: In rhombus ABCD, all sides are equal: \(AB = BC = CD = DA\).
Given that diagonal \(AC = BC\), it follows that \(AB = BC = AC\).
Thus, triangle ABC is an equilateral triangle. Therefore, \(\angle BAC = \angle BCA = \angle ABC = 60^\circ\).
Similarly, triangle ADC is also an equilateral triangle with \(\angle DAC = \angle DCA = \angle ADC = 60^\circ\).
Adding the angles for the rhombus vertices:
\(\angle A = \angle BAC + \angle DAC = 60^\circ + 60^\circ = 120^\circ\)
\(\angle C = \angle BCA + \angle DCA = 60^\circ + 60^\circ = 120^\circ\)
\(\angle B = 60^\circ\)
\(\angle D = 60^\circ\)
The angles of the rhombus are \(120^\circ, 60^\circ, 120^\circ,\) and \(60^\circ\).
In simple words: The diagonal splits the rhombus into two equilateral triangles, making the angles \(60^\circ\) and \(120^\circ\).

Exam Tip: Whenever a diagonal equals a side in a rhombus, recognize the formation of equilateral triangles.

Question 24. Two adjacent angles of a parallelogram are \((3x - 4)^\circ\) and \((3x + 10)^\circ\). Find the angles of the parallelogram.
Answer: Since adjacent angles of a parallelogram are supplementary:
\((3x - 4) + (3x + 10) = 180\)
\(6x + 6 = 180\)
\(6x = 180 - 6 = 174\)
\(x = \frac{174}{6} = 29\)
Now, substitute \(x = 29\) into the angle expressions:
First angle = \(3(29) - 4 = 87 - 4 = 83^\circ\)
Second angle = \(3(29) + 10 = 87 + 10 = 97^\circ\)
Since opposite angles are equal, the four angles of the parallelogram are \(83^\circ, 97^\circ, 83^\circ,\) and \(97^\circ\).
In simple words: Set the sum of the two adjacent expressions equal to \(180^\circ\), solve for \(x\), and find the angles.

Exam Tip: Always verify that \(83^\circ + 97^\circ = 180^\circ\).

Question 25. ABCD is a rectangle with AB = 12 and BC = 5. Find AC.
Answer: In a rectangle ABCD, each interior angle is a right angle (\(90^\circ\)).
Triangle ABC is a right-angled triangle at vertex B (\(\angle B = 90^\circ\)).
Using Pythagoras theorem:
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 12^2 + 5^2\)
\(AC^2 = 144 + 25 = 169\)
\(AC = \sqrt{169} = 13\) cm
In simple words: Use the Pythagoras theorem in the right-angled triangle formed by the sides and diagonal of the rectangle.

Exam Tip: Memorize common Pythagorean triples like \((5, 12, 13)\) to solve such problems instantly.

Question 26. In the following figure, ABCD is a rectangle and its diagonals meet at O. Find x, if \(OA = 2x + 4\) and \(OD = 3x + 1\). Also find BD.
Answer: In a rectangle, diagonals are equal in length and bisect each other.
Since diagonals bisect each other and are equal, all segments from the intersection point O to the four vertices are equal (\(OA = OB = OC = OD\)).
Therefore, \(OA = OD\)
\(2x + 4 = 3x + 1\)
\(4 - 1 = 3x - 2x\)
\(x = 3\)
Now, find the length of \(OA\):
\(OA = 2(3) + 4 = 6 + 4 = 10\)
Since \(OD = OA = 10\), and diagonals are bisected at O, the full diagonal \(BD = 2 \times OD = 2 \times 10 = 20\).
In simple words: The distance from the center O to any of the four corners of a rectangle is equal. Set \(OA = OD\) to find \(x\).

Exam Tip: Remember that in rectangles, rhombuses, and squares, the center is equidistant from all vertices.

Question 27. Find the values of x, y and z in a parallelogram ABCD shown in the figure given below.
Answer: In parallelogram ABCD:
1. Angle \(\angle D = 114^\circ\). Since opposite angles of a parallelogram are equal, \(\angle B = 114^\circ\).
Given \(\angle B = 3y\), so:
\(3y = 114^\circ \implies y = \frac{114}{3} = 38^\circ\).
2. Consecutive angles are supplementary: \(\angle A + \angle D = 180^\circ\).
\((\angle A) + 114^\circ = 180^\circ \implies \angle A = 180^\circ - 114^\circ = 66^\circ\).
From the figure, \(\angle A = 42^\circ + z\).
\(42^\circ + z = 66^\circ \implies z = 66^\circ - 42^\circ = 24^\circ\).
3. Since \(AB \parallel DC\) and diagonal \(AC\) is a transversal, alternate interior angles are equal: \(\angle BAC = \angle ACD\).
Here, \(z = \angle BAC = 24^\circ\), and \(x = \angle ACD\). Thus, \(x = z = 24^\circ\).
In simple words: Use opposite angles, supplementary adjacent angles, and alternate interior angles to find \(x, y,\) and \(z\).

Exam Tip: Look for Z-shapes (alternate interior angles) formed by diagonals in parallelograms.

Question 28. Find \(x + y + z + w\) shown in the following figure:
Answer: The given figure is a quadrilateral with exterior angles \(x, y, z, w\).
We know that the sum of all exterior angles of any convex polygon is always \(360^\circ\).
Therefore, \(x + y + z + w = 360^\circ\).
In simple words: The sum of exterior angles of any polygon is always \(360^\circ\), regardless of the number of sides.

Exam Tip: Do not waste time calculating individual exterior angles if the question asks for their total sum.

Question 29. Find the measure of angle x in the following figure: (i)
Answer: In the given triangle, interior angles and exterior angles are given.
Using the exterior angle property or linear pairs: the interior angles are \(50^\circ\) and the angle forming a linear pair with \(130^\circ\) is \(180^\circ - 130^\circ = 50^\circ\).
Let us sum the interior angles of the triangle: \(50^\circ + 50^\circ + 2x = 180^\circ\)
\(100^\circ + 2x = 180^\circ\)
\(2x = 180^\circ - 100^\circ = 80^\circ\)
\(x = \frac{80^\circ}{2} = 40^\circ\)
In simple words: Find the interior angles using linear pairs and use the triangle angle sum property (\(180^\circ\)) to solve for \(x\).

Exam Tip: Always convert exterior angles to interior angles using linear pairs (\(180^\circ\) minus exterior angle) when applying the triangle sum property.

Question 29. Find the measure of angle x in the following figure: (ii)
Answer: The given figure is a pentagon with interior angles: \(30^\circ\), \(2x\), \(2x\), and two interior angles adjacent to \(70^\circ\) and \(60^\circ\) on a straight line.
Interior angle at A = \(180^\circ - 70^\circ = 110^\circ\)
Interior angle at E = \(180^\circ - 60^\circ = 120^\circ\)
Sum of interior angles of a pentagon = \(540^\circ\)
\(30^\circ + 2x + 2x + 110^\circ + 120^\circ = 540^\circ\)
\(4x + 260^\circ = 540^\circ\)
\(4x = 540^\circ - 260^\circ = 280^\circ\)
\(x = \frac{280^\circ}{4} = 70^\circ\)
In simple words: Find all interior angles of the pentagon, sum them to \(540^\circ\), and solve for \(x\).

Question 30. Find the measure of angle x in the following figure: (i)
Answer: In the given triangle, interior angles are \(50^\circ\) and \(x\), with exterior angles \(130^\circ\) and \(120^\circ\).
Using the exterior angle property of a triangle (an exterior angle equals the sum of the two opposite interior angles):
\(130^\circ = 50^\circ + \text{adjacent interior angle} \dots\)
Alternatively, using the triangle sum property: interior angles are \(50^\circ\), \(180^\circ - 130^\circ = 50^\circ\), and \(x\).
\(50^\circ + 50^\circ + x = 180^\circ\)
\(100^\circ + x = 180^\circ \implies x = 80^\circ\).
In simple words: Find the missing interior angle using linear pairs, then use the fact that triangle angles add up to \(180^\circ\).

Exam Tip: Linear pairs always sum to \(180^\circ\).

Question 30. Find the measure of angle x in the following figure: (ii)
Answer: The given figure is a pentagon with interior angles: \(30^\circ\), \(x\), \(x\), and interior angles at the base calculated via linear pairs:
\(180^\circ - 70^\circ = 110^\circ\)
\(180^\circ - 60^\circ = 120^\circ\)
Sum of interior angles of a pentagon = \(540^\circ\)
\(30^\circ + x + x + 110^\circ + 120^\circ = 540^\circ\)
\(2x + 260^\circ = 540^\circ\)
\(2x = 540^\circ - 260^\circ = 280^\circ\)
\(x = \frac{280^\circ}{2} = 140^\circ\)
In simple words: Set the sum of all five interior angles equal to \(540^\circ\) and solve for \(x\).

Exam Tip: Always double-check your subtraction when finding interior angles from exterior linear pairs.

Question 31. Find \(x + y + z\) shown in the following figure:
Answer: In the given figure, a right-angled triangle has interior angles \(90^\circ\) (indicated by the square symbol), \(50^\circ\), and the third angle is \(180^\circ - (90^\circ + 50^\circ) = 40^\circ\).
Now find exterior angles \(x, y, z\):
- Angle \(z\) forms a linear pair with the \(50^\circ\) angle: \(z = 180^\circ - 50^\circ = 130^\circ\).
- Angle \(x\) is the exterior angle at the right-angle vertex: \(x = 180^\circ - 90^\circ = 90^\circ\).
- Angle \(y\) forms a linear pair with the third angle (\(40^\circ\)): \(y = 180^\circ - 40^\circ = 140^\circ\).
Sum \(x + y + z = 90^\circ + 140^\circ + 130^\circ = 360^\circ\).
In simple words: The sum of the exterior angles of any polygon is always \(360^\circ\).

Exam Tip: Whenever a question asks for the sum of all exterior angles of a triangle or polygon, the answer is always \(360^\circ\).

Question 32. Find the values of x, y and z in a parallelogram ABCD shown in the figure given below.
Answer: In parallelogram ABCD:
1. Angle \(\angle D = 114^\circ\). Opposite angles are equal, so \(\angle B = 114^\circ\).
Since \(\angle B = y\), we have \(y = 114^\circ\).
2. Consecutive angles are supplementary: \(\angle A + \angle D = 180^\circ\)
\((\angle A) + 114^\circ = 180^\circ \implies \angle A = 66^\circ\).
From the figure, \(\angle A = 42^\circ + z\).
\(42^\circ + z = 66^\circ \implies z = 24^\circ\).
3. Using alternate interior angles along diagonal \(AC\): \(x = z = 24^\circ\).
Therefore, \(x = 24^\circ, y = 114^\circ, z = 24^\circ\).
In simple words: Use opposite angles for \(y\), supplementary angles for \(z\), and alternate interior angles for \(x\).

Exam Tip: State clearly whether you are using opposite, supplementary, or alternate angle properties.

Question 33. Find \(x + y + z + w\) shown in the following figure:
Answer: The given figure represents the exterior angles of a quadrilateral (\(x, y, z, w\)).
The sum of exterior angles of any convex polygon is always \(360^\circ\).
Therefore, \(x + y + z + w = 360^\circ\).
In simple words: The sum of the exterior angles of any polygon is always \(360^\circ\).

Question 34. (i) In the figure given below, ABCD is a parallelogram. Find the value of x, y and z.
Answer: In parallelogram ABCD:
1. Consecutive angles are supplementary: \(\angle A + \angle B = 180^\circ\)
Here, angle \(\angle B\) forms a linear pair with the exterior angle \(110^\circ\):
Interior angle \(\angle B = 180^\circ - 110^\circ = 70^\circ\).
Since opposite angles are equal, \(\angle D = \angle B = 70^\circ\), so \(x = 70^\circ\).
2. Angle \(y = \angle A\). Since adjacent angles are supplementary: \(\angle A = 180^\circ - \angle B = 180^\circ - 70^\circ = 110^\circ\), so \(y = 110^\circ\).
3. Angle \(z\): Using corresponding or alternate interior angles with the exterior angle \(110^\circ\), we find \(z = 70^\circ\).
In simple words: Use linear pairs and parallelogram properties (opposite angles equal, consecutive angles supplementary) to find \(x, y,\) and \(z\).

Exam Tip: Linear pairs along the side of a parallelogram quickly give interior angles.

Question 34. (ii) Figure HELP shown below is a parallelogram. It is given that OE = 3 cm and HL is 7 more than PE, find OH.
Answer: In parallelogram HELP, diagonals bisect each other at point O.
Since \(OE = 3\) cm, the full diagonal \(PE = 2 \times OE = 2 \times 3 = 6\) cm.
Given that \(HL\) is 7 cm more than \(PE\):
\(HL = PE + 7 = 6 + 7 = 13\) cm.
Since diagonals of a parallelogram bisect each other, \(OH\) is half of diagonal \(HL\):
\(OH = \frac{HL}{2} = \frac{13}{2} = 6.5\) cm.
In simple words: Find the full diagonal PE by doubling OE, add 7 to find HL, and then take half of HL to find OH.

Exam Tip: Be careful with half-diagonal and full-diagonal relationships in parallelograms.

Question 35. Find \(x + y + z\) shown in the following figure:
Answer: The sum of exterior angles of any triangle (or polygon) is always \(360^\circ\).
Therefore, \(x + y + z = 360^\circ\).
In simple words: The sum of exterior angles of any polygon is always \(360^\circ\).

Exam Tip: Do not calculate individual angles if the question asks for the sum of all exterior angles.

Question 36. (i) In the figure given below, ABCD is a parallelogram. Find the value of x, y and z.
Answer: In parallelogram ABCD:
1. The exterior angle is \(100^\circ\). The adjacent interior angle \(\angle B = 180^\circ - 100^\circ = 80^\circ\).
2. Opposite angles are equal, so \(x = \angle B = 80^\circ\).
3. Adjacent angles are supplementary, so \(y = 180^\circ - 80^\circ = 100^\circ\).
4. Angle \(z = \angle B = 80^\circ\) (corresponding or alternate interior angle).
In simple words: Use linear pairs to find one interior angle, then use parallelogram properties to find the rest.

Exam Tip: Linear pairs and parallel line rules are essential for solving angle-chasing problems.

Question 36. (ii) Figure HELP shown below is a parallelogram. It is given that OE = 4 cm and HL is 7 more than PE, find OH.
Answer: Diagonals of parallelogram HELP bisect each other at O.
Since \(OE = 4\) cm, the full diagonal \(PE = 2 \times 4 = 8\) cm.
Given \(HL = PE + 7 = 8 + 7 = 15\) cm.
Since \(OH\) is half of diagonal \(HL\):
\(OH = \frac{15}{2} = 7.5\) cm.
In simple words: Double OE to get PE, add 7 to get HL, and divide HL by 2 to find OH.

Exam Tip: Always write down the diagonal bisecting property before plugging in numbers.

Question 37. The lengths of the diagonals AC and BD of a rhombus are 6 cm and 8 cm respectively. Find the length of each side of the rhombus.
Answer: Let diagonals \(AC = 6\) cm and \(BD = 8\) cm intersect at point O.
Diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Therefore:
\(OA = \frac{AC}{2} = \frac{6}{2} = 3\) cm
\(OB = \frac{BD}{2} = \frac{8}{2} = 4\) cm
Triangle AOB is a right-angled triangle at O (\(\angle AOB = 90^\circ\)).
Using Pythagoras theorem to find side \(AB\):
\(AB^2 = OA^2 + OB^2\)
\(AB^2 = 3^2 + 4^2\)
\(AB^2 = 9 + 16 = 25\)
\(AB = \sqrt{25} = 5\) cm
Thus, the length of each side of the rhombus is 5 cm.
In simple words: Half the diagonals form the legs of a right-angled triangle, and the hypotenuse is the side of the rhombus.

Exam Tip: Remember that rhombus diagonals bisect at \(90^\circ\), creating four congruent right-angled triangles.

Most Important Questions

Question 1. What is the sum of all the angles of a 11-sided polygon?
Answer: Number of sides (\(n\)) = 11
Formula for the sum of interior angles = \((n - 2) \times 180^\circ\)
Sum = \((11 - 2) \times 180^\circ = 9 \times 180^\circ = 1620^\circ\)
In simple words: Substitute \(n = 11\) into the polygon angle sum formula to get \(1620^\circ\).

Exam Tip: Always subtract 2 from the number of sides before multiplying by \(180^\circ\).

Question 2. Five angles of a hexagon are \(150^\circ, 95^\circ, 80^\circ, 135^\circ\) and \(125^\circ\). Find the sixth angle.
Answer: A hexagon has 6 sides (\(n = 6\)).
Sum of interior angles of a hexagon = \((6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ\)
Sum of the five given angles = \(150^\circ + 95^\circ + 80^\circ + 135^\circ + 125^\circ = 585^\circ\)
Sixth angle = \(720^\circ - 585^\circ = 135^\circ\)
In simple words: Subtract the sum of the five given angles from the total sum of a hexagon's angles (\(720^\circ\)).

Exam Tip: Verify your addition twice to avoid calculation errors.

Question 3. Find the measure of each angle of a regular pentagon.
Answer: Number of sides of a pentagon (\(n\)) = 5
Sum of interior angles = \((5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)
Since it is a regular pentagon, all 5 interior angles are equal.
Measure of each interior angle = \(\frac{540^\circ}{5} = 108^\circ\)
In simple words: Divide the total sum of angles (\(540^\circ\)) by the number of sides (5).

Exam Tip: Alternatively, find the exterior angle (\(360^\circ / 5 = 72^\circ\)) and subtract from \(180^\circ\) to get \(108^\circ\).

Question 4. How many diagonals are there in a hexagon?
Answer: Number of sides (\(n\)) = 6
Formula for number of diagonals = \(\frac{n(n - 3)}{2}\)
Number of diagonals = \(\frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9\)
In simple words: A hexagon has 9 diagonals.

Exam Tip: Memorize or quickly derive the diagonal formula \(\frac{n(n - 3)}{2}\).

Question 5. How many diagonals are there in an octagon?
Answer: Number of sides (\(n\)) = 8
Number of diagonals = \(\frac{8(8 - 3)}{2} = \frac{8 \times 5}{2} = 20\)
In simple words: An octagon has 20 diagonals.

Exam Tip: Substitute \(n = 8\) carefully into the diagonal formula.

Question 6. If each interior angle of a regular polygon is \(144^\circ\). Find the number of sides in it.
Answer: Measure of each interior angle = \(144^\circ\)
Measure of each exterior angle = \(180^\circ - 144^\circ = 36^\circ\)
Number of sides (\(n\)) = \(\frac{360^\circ}{\text{Exterior Angle}} = \frac{360^\circ}{36^\circ} = 10\)
In simple words: Find the exterior angle by subtracting the interior angle from \(180^\circ\), then divide \(360^\circ\) by it.

Exam Tip: Using exterior angles is usually faster than solving equations with interior angles.

Question 7. An exterior angle and the interior angle of a regular polygon are in the ratio 2 : 7. Find the number of sides in the polygon?
Answer: Let the exterior angle be \(2x\) and the interior angle be \(7x\).
Since interior and exterior angles form a linear pair:
\(2x + 7x = 180^\circ\)
\(9x = 180^\circ \implies x = 20^\circ\)
Exterior angle = \(2 \times 20^\circ = 40^\circ\)
Number of sides (\(n\)) = \(\frac{360^\circ}{40^\circ} = 9\)
In simple words: Use the given ratio to find the exterior angle, then divide \(360^\circ\) by it to get the number of sides.

Exam Tip: Interior and exterior angles at any vertex are always supplementary (\(180^\circ\)).

Question 8. Find the sum of the interior angles of a polygon with 8 sides.
Answer: Number of sides (\(n\)) = 8
Sum of interior angles = \((n - 2) \times 180^\circ = (8 - 2) \times 180^\circ = 6 \times 180^\circ = 1080^\circ\)
In simple words: An octagon's interior angles sum up to \(1080^\circ\).

Exam Tip: Use the formula \((n - 2) \times 180^\circ\) for any polygon's interior angle sum.

Question 9. Complete the following:
A quadrilateral has 4 sides.
A quadrilateral has 4 angles.
A quadrilateral has 2 diagonals.
A quadrilateral has 4 vertices.
Answer: 4, 4, 2, 4
In simple words: A standard quadrilateral has 4 sides, 4 angles, 2 diagonals, and 4 vertices.

Exam Tip: Remember basic geometric definitions for quadrilaterals.

Question 10. Can a polygon have the sum of its interior angles as: (i) \(2160^\circ\) (ii) \(2400^\circ\)?
Answer: (i) Let the sum of interior angles be \(2160^\circ\).
\((n - 2) \times 180^\circ = 2160^\circ\)
\(n - 2 = \frac{2160}{180} = 12\)
\(n = 12 + 2 = 14\)
Since 14 is a whole number, a polygon with a sum of \(2160^\circ\) is possible (a tetradecagon).
(ii) Let the sum be \(2400^\circ\).
\((n - 2) \times 180^\circ = 2400^\circ\)
\(n - 2 = \frac{2400}{180} = \frac{40}{3} = 13.33\)
\(n = 13.33 + 2 = 15.33\)
Since the number of sides must be a positive integer, a polygon with an angle sum of \(2400^\circ\) is not possible.
In simple words: Divide the given sum by \(180^\circ\) and add 2. If the result is a whole number, it is possible; otherwise, it is not.

Question 11. The angle of a quadrilateral are in the ratio 3 : 4 : 5 : 6. Find all its angles.
Answer: Let the angles be \(3x, 4x, 5x,\) and \(6x\).
The sum of interior angles of a quadrilateral is \(360^\circ\).
\(3x + 4x + 5x + 6x = 360^\circ\)
\(18x = 360^\circ\)
\(x = \frac{360^\circ}{18} = 20^\circ\)
Now, find each angle:
First angle = \(3 \times 20^\circ = 60^\circ\)
Second angle = \(4 \times 20^\circ = 80^\circ\)
Third angle = \(5 \times 20^\circ = 100^\circ\)
Fourth angle = \(6 \times 20^\circ = 120^\circ\)
In simple words: Set the sum of ratio terms equal to \(360^\circ\), solve for \(x\), and multiply by each ratio coefficient.

Exam Tip: Always verify that your calculated angles sum to \(360^\circ\). Check: \(60^\circ + 80^\circ + 100^\circ + 120^\circ = 360^\circ\).

Question 12. Three angle of a quadrilateral are in the ratio 4 : 6 : 3. If the fourth angle is \(100^\circ\). Find the other three angles of the quadrilateral.
Answer: Let the three angles be \(4x, 6x,\) and \(3x\).
The sum of all four angles of a quadrilateral is \(360^\circ\).
\(4x + 6x + 3x + 100^\circ = 360^\circ\)
\(13x + 100^\circ = 360^\circ\)
\(13x = 360^\circ - 100^\circ = 260^\circ\)
\(x = \frac{260^\circ}{13} = 20^\circ\)
Now, find the three angles:
First angle = \(4 \times 20^\circ = 80^\circ\)
Second angle = \(6 \times 20^\circ = 120^\circ\)
Third angle = \(3 \times 20^\circ = 60^\circ\)
In simple words: Subtract the fourth angle from \(360^\circ\), divide the remainder among the ratio parts, and solve for \(x\).

Exam Tip: Double check addition: \(80^\circ + 120^\circ + 60^\circ + 100^\circ = 360^\circ\).

Question 13. In the given figure, ABCD is a rectangle. BM and DN are perpendicular to AC from B and D respectively.
(i) AB = DC ? Why ?
(ii) Is \(\angle BMA = \angle DNC\) ? Why ?
(iii) Is \(\angle BAM = \angle DCN\) ? Why ?
(iv) Is \(\Delta ABM \cong \Delta DNC\) ? By which congruence condition?
(v) Is BM = DN ? Why ?
Answer:
(i) Yes, \(AB = DC\) because opposite sides of a rectangle are equal.
(ii) Yes, \(\angle BMA = \angle DNC\) because both are \(90^\circ\) (since BM and DN are perpendiculars to AC).
(iii) Yes, \(\angle BAM = \angle DCN\) because AB is parallel to DC and diagonal AC acts as a transversal, making alternate interior angles equal.
(iv) Yes, \(\Delta ABM \cong \Delta DNC\) by the AAS (Angle-Angle-Side) or ASA congruence condition (since \(\angle BMA = \angle DNC = 90^\circ\), \(\angle BAM = \angle DCN\), and side \(AB = DC\)).
(v) Yes, \(BM = DN\) because corresponding parts of congruent triangles are equal (CPCTC).
In simple words: Prove the two small triangles congruent using rectangle properties and alternate angles, then conclude that BM = DN.

Exam Tip: Master triangle congruence criteria (SSS, SAS, ASA, AAS, RHS) as they frequently appear in quadrilateral proofs.

Question 14. In the given figure, diagonals AC and BD of a rectangle ABCD intersect each other at a point O. If OA = 4 cm, find AC and BD.
Answer: In a rectangle, diagonals are equal in length and bisect each other.
Since \(OA = 4\) cm, and diagonals bisect at O:
\(AC = 2 \times OA = 2 \times 4 = 8\) cm.
Since diagonals of a rectangle are equal, \(BD = AC = 8\) cm.
In simple words: Double the length of OA to get AC, and note that BD equals AC.

Question 15. In figure ABCD is parallelogram in which \(\angle DAB = 75^\circ\) and \(\angle DBC = 60^\circ\), calculate \(\angle CDB\) and \(\angle ADB\).
Answer: In parallelogram ABCD:
1. Opposite angles are equal: \(\angle C = \angle DAB = 75^\circ\).
2. In triangle BCD, the sum of angles is \(180^\circ\):
\(\angle BCD + \angle DBC + \angle CDB = 180^\circ\)
\(75^\circ + 60^\circ + \angle CDB = 180^\circ\)
\(135^\circ + \angle CDB = 180^\circ\)
\(\angle CDB = 180^\circ - 135^\circ = 45^\circ\)
3. Since \(AB \parallel DC\) and diagonal \(BD\) acts as a transversal, alternate interior angles are equal: \(\angle ADB = \angle DBC = 60^\circ\).
Therefore, \(\angle CDB = 45^\circ\) and \(\angle ADB = 60^\circ\).
In simple words: Use the triangle sum property in triangle BCD for \(\angle CDB\) and alternate interior angles for \(\angle ADB\).

Exam Tip: Look for alternate interior angles (Z-angles) formed by diagonals across parallel sides.

Question 16. The diagonal of a Rhombus is 6cm and 8cm find the length of a side of rhombus.
Answer: Let the diagonals be \(d_1 = 6\) cm and \(d_2 = 8\) cm.
Diagonals of a rhombus bisect each other at right angles (\(90^\circ\)).
Half of the diagonals are:
\(\frac{d_1}{2} = \frac{6}{2} = 3\) cm
\(\frac{d_2}{2} = \frac{8}{2} = 4\) cm
Using the Pythagoras theorem to find the side length (\(s\)):
\(s^2 = 3^2 + 4^2\)
\(s^2 = 9 + 16 = 25\)
\(s = \sqrt{25} = 5\) cm
In simple words: Halve both diagonals, then apply the Pythagoras theorem to find the side length of the rhombus.

Exam Tip: The classic \(3-4-5\) right triangle appears frequently in rhombus diagonal problems.

Question 17. ABCD is a parallelogram. What special name will you give it if the following additional facts are known?
(i) AB = AD
(ii) \(\angle DAB = 90^\circ\)
(iii) AB = AD and \(\angle DAB = 90^\circ\)
Answer:
(i) Rhombus (since adjacent sides become equal in a parallelogram).
(ii) Rectangle (since one interior angle being \(90^\circ\) makes all angles \(90^\circ\)).
(iii) Square (since it has equal adjacent sides and a right angle).
In simple words: Equal adjacent sides make a rhombus, a right angle makes a rectangle, and both together make a square.

Question 18. State, whether the given statement is true or not.
(i) A rectangle is a parallelogram.
(ii) A square is a rectangle.
(iii) A rectangle is a rhombus.
(iv) A square is a rhombus.
(v) A rectangle is a square.
Answer:
(i) True
(ii) True
(iii) False
(iv) True
(v) False
In simple words: Squares and rectangles inherit parallelogram properties, but rectangles do not have all equal sides (so not rhombuses/squares).

Exam Tip: Learn subset relationships between quadrilaterals to answer true/false questions correctly.

Question 19. Which of the following are true for a rhombus?
(i) It has two pairs of parallel sides.
(ii) It has two pairs of equal angles.
(iii) It has only two pairs of equal sides.
(iv) Two of its angles are right angle.
(v) Its diagonals bisect each other at right angle.
(vi) Its diagonals are equal and perpendicular to each other.
(vii) It has all its sides of equal lengths.
Answer:
True statements: (i), (ii), (v), (vii)
False statements: (iii) (it has 4 equal sides), (iv) (not necessarily right angles), (vi) (diagonals are not equal in a general rhombus).
In simple words: A rhombus has parallel sides, equal opposite angles, perpendicular bisecting diagonals, and all sides equal.

Exam Tip: Diagonals are equal only in rectangles and squares, not in general rhombuses.

Question 20. How does a trapezium differ from a parallelogram?
Answer: A trapezium has only one pair of parallel sides, whereas a parallelogram has two pairs of parallel sides.
In simple words: Trapeziums have one pair of parallel sides, while parallelograms have two.

Question 21. How does a rhombus differ from a square?
Answer: All interior angles of a square are right angles (\(90^\circ\)) and its diagonals are equal in length, whereas a rhombus has equal sides but its angles are not necessarily \(90^\circ\) and its diagonals are unequal.
In simple words: Squares have \(90^\circ\) angles and equal diagonals, while rhombuses do not.

Exam Tip: A square is a special type of rhombus with \(90^\circ\) corners.

Question 22. How does a kite differ from a parallelogram?
Answer: A kite has two pairs of equal adjacent sides and no parallel sides (generally), whereas a parallelogram has two pairs of equal and parallel opposite sides.
In simple words: Kites have adjacent equal sides, while parallelograms have opposite parallel and equal sides.

Exam Tip: Remember the distinct side-length patterns for kites versus parallelograms.

Question 23. Let ABCD be a parallelogram. What special name would you give it, when:
(a) AB = AD
(b) \(\angle ABC = 90^\circ\)
(c) AB = AD and \(\angle ABC = 90^\circ\)
Answer:
(a) Rhombus
(b) Rectangle
(c) Square
In simple words: Equal adjacent sides form a rhombus, a right angle forms a rectangle, and both together form a square.

Exam Tip: These three conditions are classic tests for identifying special types of parallelograms.