CBSE Class 8 Mathematics Understanding Quadrilaterals Worksheet Set 07

SECTION-A

Question 1. Define a concave quadrilateral & convex quadrilateral with the help of figure.
Answer: 1. Convex Quadrilateral: A quadrilateral in which all interior angles are strictly less than \(180^\circ\), and both of its diagonals lie entirely inside the figure.
2. Concave Quadrilateral: A quadrilateral in which at least one interior angle is greater than \(180^\circ\) (reflex angle), and at least one diagonal lies outside the figure.
In simple words: A convex polygon points outwards everywhere, while a concave polygon has a cave-like indented vertex.

Exam Tip: Remember the line segment test: if any line segment connecting two interior points lies outside, it is concave.

Question 2. Give any two properties of a parallelogram.
Answer: 1. Opposite sides are equal in length.
2. Opposite angles are equal in measure.
In simple words: Opposite sides and angles in a parallelogram match each other in size.

Exam Tip: Also remember that diagonals of a parallelogram bisect each other.

Question 3. Find the number of sides of a polygon which has 20 diagonals.
Answer: Let the number of sides of the polygon be \(n\).
The formula for the number of diagonals in an \(n\)-sided polygon is:
\(\text{Number of diagonals} = \frac{n(n - 3)}{2}\)
Given number of diagonals = 20
\(\frac{n(n - 3)}{2} = 20\)
\(n(n - 3) = 40\)
\(n^2 - 3n - 40 = 0\)
Factorizing the quadratic equation:
\((n - 8)(n + 5) = 0\)
Since the number of sides cannot be negative, \(n = 8\).
In simple words: By using the diagonal formula and solving the quadratic equation, we find the polygon has 8 sides (octagon).

Exam Tip: Memorize the diagonal formula \(\frac{n(n-3)}{2}\) as it is frequently tested.

Question 4. Define a regular & irregular polygon.
Answer: 1. Regular Polygon: A polygon is regular if all its sides are of equal length (equilateral) and all its interior angles are of equal measure (equiangular).
2. Irregular Polygon: A polygon whose sides are not all equal in length or whose interior angles are not all equal in measure.
In simple words: Regular polygons are symmetric with equal sides and angles, whereas irregular polygons have varying side lengths or angles.

Exam Tip: An equilateral triangle and a square are classic examples of regular polygons.

Question 5. What are the different types of a polygon.
Answer: Polygons are classified based on their number of sides and regularity:
1. Triangle (3 sides)
2. Quadrilateral (4 sides)
3. Pentagon (5 sides)
4. Hexagon (6 sides)
5. Heptagon (7 sides)
6. Octagon (8 sides)
They are further categorized as convex or concave, and regular or irregular.
In simple words: Polygons are named according to how many sides they have, starting from three sides.

Exam Tip: Be familiar with names up to 10 sides (decagon).

Question 6. Differentiate between a trapezium & an isosceles trapezium.
Answer: 1. Trapezium: A quadrilateral with at least one pair of parallel opposite sides.
2. Isosceles Trapezium: A trapezium in which the non-parallel sides are equal in length, and base angles are equal.
In simple words: An ordinary trapezium has one pair of parallel sides, while an isosceles trapezium is symmetric with equal non-parallel sides.

Question 7. Give two properties of a kite.
Answer: 1. Two pairs of adjacent sides are equal in length.
2. The diagonals intersect each other at right angles (\(90^\circ\), and the longer diagonal bisects the shorter diagonal.
In simple words: A kite has adjacent sides equal and perpendicular diagonals.

Exam Tip: The angle between unequal sides is bisected by the main diagonal.

Question 8. Find each angle of a regular polygon.
Answer: *(Note: Assuming the question refers to a general regular polygon with \(n\) sides)*
The sum of all interior angles of an \(n\)-sided polygon is given by:
\(\text{Sum} = (n - 2) \times 180^\circ\)
Each interior angle of a regular polygon of \(n\) sides is:
\(\text{Interior Angle} = \frac{(n - 2) \times 180^\circ}{n}\)
Each exterior angle is:
\(\text{Exterior Angle} = \frac{360^\circ}{n}\)
In simple words: Divide the total sum of interior angles by the number of sides to get each interior angle.

Exam Tip: Always use exterior angles when possible as they are often easier to calculate.

SECTION - B

Question 1. Construct a quadrilateral ABCD with AB=3 cm, BC=4 cm, CD=4.5 cm, AD=3.5 cm & Diagonal AC=5.2 cm.
Answer: Steps of construction:
1. Draw a line segment \(AB = 3\) cm.
2. With center A and radius \(5.2\) cm, draw an arc.
3. With center B and radius \(4\) cm, draw another arc to intersect the previous arc at C.
4. Join AC and BC.
5. With center A and radius \(3.5\) cm, draw an arc on the other side of AC.
6. With center C and radius \(4.5\) cm, draw an arc to intersect the previous arc at D.
7. Join AD and CD. ABCD is the required quadrilateral.
In simple words: Construct the quadrilateral by dividing it into two triangles (ABC and ADC) using the given diagonal.

Exam Tip: Always draw a rough sketch first to determine the relative positions of vertices.

Question 2. Construct a parallelogram MORE with OR=6 cm, RE=4.5 cm, EO=7.5 cm.
Answer: *(Note: In parallelogram MORE, sides are MO = RE = 4.5 cm, OR = ME = 6 cm, and diagonal EO = 7.5 cm)*
Steps of construction:
1. Draw a line segment \(OR = 6\) cm.
2. With center O and radius \(7.5\) cm (diagonal EO), draw an arc.
3. With center R and radius \(4.5\) cm (side RE), draw an arc intersecting at E.
4. Join OE and RE.
5. With center O and radius \(4.5\) cm (side MO), draw an arc.
6. With center E and radius \(6\) cm (side ME), draw an arc intersecting at M.
7. Join OM and EM. MORE is the required parallelogram.
In simple words: Construct triangle ORE first, then complete the parallelogram using opposite side lengths.

Exam Tip: Identify all side lengths using the properties of a parallelogram before starting construction.

Question 3. Construct a rhombus HOLY if HO=6 cm and HL=9 cm.
Answer: *(Note: In a rhombus, all sides are equal, so HO = OL = LY = HY = 6 cm, and diagonal HL = 9 cm)*
Steps of construction:
1. Draw a diagonal \(HL = 9\) cm.
2. With centers H and L as centers and radius \(6\) cm, draw arcs on both sides of HL to intersect at O and Y.
3. Join HO, OL, HY, and LY. HOLY is the required rhombus.
In simple words: Construct two triangles sharing the common diagonal HL with side lengths equal to 6 cm.

Question 4. Construct a quadrilateral ABCD in which AB=4 cm, BC=4.5 cm, Angle A=90 degrees, Angle B=75 degrees & Angle D=80 degrees.
Answer: Steps of construction:
1. Draw a line segment \(AB = 4\) cm.
2. At point A, construct an angle of \(90^\circ\) and draw ray AX.
3. At point B, construct an angle of \(75^\circ\) and draw ray BY.
4. Along ray BC, cut off \(BC = 4.5\) cm.
*(Note: To find Angle C, sum of angles = \(360^\circ\); find Angle C after determining Angle D position)*
In simple words: Draw base AB, construct angles at A and B, cut off BC = 4.5 cm, and complete using Angle D.

Exam Tip: Use a protractor when standard compass angles cannot be easily constructed.

Question 5. Construct a square POST with diagonal PS=6 cm.
Answer: Steps of construction:
1. Draw diagonal \(PS = 6\) cm.
2. Draw the perpendicular bisector of PS intersecting it at point O.
3. With center O and radius equal to half of diagonal (\(3\) cm), cut arcs on the perpendicular bisector above and below O to locate points O and T.
4. Join PO, OS, ST, and TP. POST is the required square.
In simple words: Use the property that diagonals of a square are equal and bisect each other perpendicularly.

Exam Tip: Diagonals of a square are perpendicular bisectors of each other.

SECTION - C

Question 1. The measure of angles of a hexagon are \(x^\circ\), \((x - 5)^\circ\), \((x - 5)^\circ\), \((2x - 5)^\circ\), \((2x - 5)^\circ\), and \((2x + 20)^\circ\). Find the value of x.
Answer: Sum of interior angles of a hexagon (\(n = 6\)) is:
\(\text{Sum} = (6 - 2) \times 180^\circ = 4 \times 180^\circ = 720^\circ\)
Summing the given angles:
\(x + (x - 5) + (x - 5) + (2x - 5) + (2x - 5) + (2x + 20) = 720\)
Combining like terms:
\((x + x + x + 2x + 2x + 2x) + (-5 - 5 - 5 - 5 + 20) = 720\)
\(9x + 0 = 720\)
\(9x = 720\)
\(x = \frac{720}{9} = 80\)
In simple words: Add all six angle expressions, equate them to the total sum of a hexagon (\(720^\circ\)), and solve for \(x\).

Exam Tip: Always double check the summation of constant terms before dividing.

Question 2. Three angles of a quadrilateral are in the ratio 3 : 5 : 8. The mean of these angles is \(80^\circ\). Find all the four angles of the quadrilateral.
Answer: Let the three angles in ratio be \(3x\), \(5x\), and \(8x\).
Since their mean is \(80^\circ\):
\(\frac{3x + 5x + 8x}{3} = 80\)
\(\frac{16x}{3} = 80\)
\(16x = 240\)
\(x = \frac{240}{16} = 15\)
Therefore, the three angles are:
First angle = \(3(15) = 45^\circ\)
Second angle = \(5(15) = 75^\circ\)
Third angle = \(8(15) = 120^\circ\)
Let the fourth angle be \(y\).
Sum of all angles of a quadrilateral = \(360^\circ\)
\(45^\circ + 75^\circ + 120^\circ + y = 360^\circ\)
\(240^\circ + y = 360^\circ\)
\(y = 360^\circ - 240^\circ = 120^\circ\)
In simple words: Use the mean to find \(x\) and the three angles, then subtract their sum from \(360^\circ\) to find the fourth angle.

Exam Tip: Mean multiplied by the number of items gives the sum of those items.

Question 3. Interior angles of a pentagon are five consecutive multiples of 3. Find all the angles of the pentagon.
Answer: Let the five consecutive multiples of 3 be \(3x\), \(3x + 3\), \(3x + 6\), \(3x + 9\), and \(3x + 12\).
Sum of interior angles of a pentagon (\(n = 5\)) is:
\(\text{Sum} = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ\)
Summing the angles:
\(3x + (3x + 3) + (3x + 6) + (3x + 9) + (3x + 12) = 540\)
\(15x + 30 = 540\)
\(15x = 510\)
\(x = \frac{510}{15} = 34\)
Substituting \(x = 34\):
First angle = \(3(34) = 102^\circ\)
Second angle = \(102^\circ + 3^\circ = 105^\circ\)
Third angle = \(105^\circ + 3^\circ = 108^\circ\)
Fourth angle = \(108^\circ + 3^\circ = 111^\circ\)
Fifth angle = \(111^\circ + 3^\circ = 114^\circ\)
In simple words: Set up consecutive multiples of 3, equate their sum to \(540^\circ\), and solve for each angle.

Question 4. Find the value of x in the adjacent figure if AO and BO are the bisectors of Angle A and Angle B respectively.
Answer: In quadrilateral ABCD, let the interior angles be \(\angle A\), \(\angle B\), \(\angle C = 130^\circ\), and \(\angle D = 120^\circ\).
Sum of interior angles of quadrilateral ABCD = \(360^\circ\)
\(\angle A + \angle B + 120^\circ + 130^\circ = 360^\circ\)
\(\angle A + \angle B + 250^\circ = 360^\circ\)
\(\angle A + \angle B = 110^\circ\)
Since AO and BO are angle bisectors of \(\angle A\) and \(\angle B\):
\(\angle OAB = \frac{1}{2}\angle A\)
\(\angle OBA = \frac{1}{2}\angle B\)
In triangle AOB:
\(\angle OAB + \angle OBA + x = 180^\circ\)
\(\frac{1}{2}\angle A + \frac{1}{2}\angle B + x = 180^\circ\)
\(\frac{1}{2}(\angle A + \angle B) + x = 180^\circ\)
\(\frac{1}{2}(110^\circ) + x = 180^\circ\)
\(55^\circ + x = 180^\circ\)
\(x = 180^\circ - 55^\circ = 125^\circ\)
In simple words: Find the sum of half of angles A and B using the angle sum property of quadrilateral ABCD, then apply the angle sum property in triangle AOB to find \(x\).

Exam Tip: A standard property states that the angle formed by angle bisectors of two adjacent interior angles equals \(180^\circ - \frac{\text{sum of other two angles}}{2}\).

Question 5. In the adjacent figure 'SMART' MART is a rhombus with each of its angle as \(90^\circ\). In SM = 5 cm, find the perimeter of 'SMART'
Answer: Given that SMART is a rhombus with each angle equal to \(90^\circ\), it is a square.
Side length \(SM = 5\) cm.
All four sides of a square are equal in length.
Perimeter of square SMART = \(4 \times \text{side} = 4 \times 5 = 20\) cm.
In simple words: A rhombus with \(90^\circ\) angles is a square, so multiply the side length by 4 to get the perimeter.

Exam Tip: A rhombus with even one \(90^\circ\) angle is a square.

Question 6. Perimeter of a square HOPE is 40 cm. find the value of \((HP)^2 + (OE)^2\).
Answer: Perimeter of square HOPE = 40 cm
Length of each side \(s = \frac{40}{4} = 10\) cm.
In square HOPE, HP and OE are the diagonals of the square.
The diagonals of a square are equal in length.
Length of diagonal \(d = s\sqrt{2} = 10\sqrt{2}\) cm.
Thus, \(HP = 10\sqrt{2}\) cm and \(OE = 10\sqrt{2}\) cm.
Calculating \((HP)^2 + (OE)^2\):
\((HP)^2 = (10\sqrt{2})^2 = 100 \times 2 = 200\)
\((OE)^2 = (10\sqrt{2})^2 = 100 \times 2 = 200\)
\((HP)^2 + (OE)^2 = 200 + 200 = 400\)
In simple words: Find the side length from the perimeter, calculate the diagonal length, square each diagonal, and add them together.

Exam Tip: For a square with side \(s\), each diagonal equals \(s\sqrt{2}\).