Understanding Quadrilaterals Class 8 Extra Questions with Answers

These Class 8 Maths Extra Questions Chapter 3 Understanding Quadrilaterals will help students prepare well for the exams.

Class 8 Maths Chapter 3 Extra Questions Understanding Quadrilaterals

Class 8 Maths Understanding Quadrilaterals Extra Questions

Question 1.
Draw a rough diagram of concave quadrilateral.
Solution:

Question 2.
Find measure of each exterior angle of regular pentagon.
Solution:
For a regular pentagon each exterior angle = \(\frac{360^{\circ}}{5}\) = 72°

Question 3.
Find the angle ‘x’ in the given figure

Solution:
In the given figure
∠A = x°, ∠B = 90° [∵ AB⊥BC]
∠C = 60° and ∠D = 70°
Sum of the angles in a quadrilateral = 360°
∴ ∠A + ∠B + ∠C + ∠D = 360°
x° + 90° + 60° + 70° = 360°
x° + 220° = 360°
∴ x° = 360° – 220° = 140°

Question 4.
The ratio of two adjacent angles of a parallelogram is 5 : 4. Find each of the angle of that parallelogram.
Solution:
The ratio of two adjacent angles of a parallelogram = 5 : 4
Two adjacent angles of a parallelogram = 180°
Total parts of Ratio = 5 + 4 = 9
∴ Each part of ratio = \(\frac{180}{9}\) = 20°
∴ One angle = 5 × 20° = 100°
Another angle = 4 × 20° = 80°

Question 5.
State whether true or false.
a) All rectangles are trapeziums.
b) All parallelograms are not trapeziums.
Solution:
a) False, All angles are right angles in rectangle but not in trapezium.
b) True, In a parallelogram the pair of opposite sides are equal and parallel but not in trapezium.

Understanding Quadrilaterals Extra Questions Class 8

Question 1.
The measure of two adjacent angles of a parallelogram are in the ratio of 2 : 3. Find the measure of each angle.
The ratio of two adjacent angles of a parallelogram = 2 : 3
Let the angles = 2x : 3x
The sum of the adjacent angles of a parallelogram = 180°
∴ 2x + 3x = 180°
5x = 180°
∴ x = \(\frac{180^{\circ}}{5}\) = 36°
One angle = 2x = 2 × 36° = 72°
Another angle = 3x = 3 × 36° = 108°
∴ The measures of each angle = 108°, 72°, 108°, 72°

Question 2.
ABCD is a rectangle. With diagonals AC and BD intersecting at point O. If the length of AO is 6 units find the length of BD.

Solution:
In a Rectangle
If OA = 6 units, then OC also 6 units. [Diagonals bisects each other]
So AC = AO + OC
= 6 + 6
AC = 12 units
Then DB =12 units [Diagonals are equal AC = BD]
BD = BO + OD = 12
∴ BD = 6u + 6u = 12 units

Question 3.
If each interior angle of a regular polygon is 150°, how many sides does this polygon have ?
Solution:
Each interior angle of regular polygon = 150°
Each exterior angle = 180° – 150° = 30° (By Linear pair property)
Sum of measures of exterior angles = 360°
Let number of sides = n
∴ n × 30 = 360°
n = \(\frac{360}{30}\)
n = 12
∴ Number of sides of polygon with each interior angle 150° is 12.

Question 4.
In the figure, ABCD is a parallelogram. Find the values of unknowns x, y, z.

Solution:
Given ABCD is a parallelogram.
Opposite angles are equal
⇒ x = 110°
AB || CD, \(\overline{\mathrm{BD}}\) is a transversal
z = 50° (Alternate Interior angles)
Δle BCD,
Sum of three angles of a Δle is 180°
So, z + y + 110° = 180°
50 + y + 110° = 180°
y + 160° = 180°
y = 180° – 60°
y = 20°

Question 5.
PQRS is a rectangle. Its diagonals meet at O. If OR = 2x + 4, OQ = 3x + 1 then find x.

Solution:
Given PQRS is a rectangle. Its diagonals are PR and QS meet at ‘O’.
OR = 2x + 4 and OQ = 3x + 1
\(\overline{\mathrm{OQ}}\) is half of the diagonal \(\overline{\mathrm{QS}}\)
\(\overline{\mathrm{OR}}\) is half of the diagonal \(\overline{\mathrm{PR}}\).
We know that the diagonals of a rectangle are of equal length.
So, their halves are also equal .
∴ 3x + 1 = 2x + 4
3x – 2x = 4 – 1
x = 3

Question 6.
The measures of two adjacent angles of a parallelogram are in the ratio 3 : 2. Find the measures of each of the angles of the parallelogram.
Solution:
Let the measure of two adjacent angles ∠Aand ∠B of a parallelogram ABCD are in the ratio = 3 : 2
Let ∠A = 3x
∠B = 2x

We know that the sum of the measures of adjacent angles is 180° for a parallelogram
∠A + ∠B = 180°
2x + 3x = 180°
5x = 180°
x = \(\frac{180^{\circ}}{5}\) ⇒ x = 36°
∴ ∠A = ∠C = 3x = 3 × 36° = 108°
∴ ∠B = ∠D = 2x = 2 × 36° = 72°
Thus, the four angles of a parallelogram are 108°, 72°, 108° and 72°.

Extra Questions of Understanding Quadrilaterals Class 8

Question 1.
Can a quadrilateral ABCD be a paralle-logram if
i) ∠D + ∠B = 180° ?
ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm ?
iii) ∠A = 70° and ∠C = 65° ?
Solution:
i) ∠D + ∠B = 180°
ABCD can be a parallelogram with ∠B + ∠D = 180° but need not be a parallelogram.

ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm
No, the quadrilateral ABCD cannot be a parallelogram with these measurements.
Reason : In a parallelogram opposite sides are equal, But AD ≠ BC

iii) ∠A = 70° and ∠C = 65° ?
No, the quadrilateral ABCD cannot be a parallelogram.
Reason : Opposite angles of a parallelogram are equal. But here ∠A ≠ ∠C

Understanding Quadrilaterals Class 8 Extra Questions

Question 1.
Find measure x in Fig.
Solution:
x + 90°+ 50° + 110° = 360°

⇒ x + 250° = 360° ∴ x = 110°

Question 2.
Find the number of sides of a regular polygon whose each exterior angle has a measure of 45°.
Solution:
Total measure of all exterior angles = 360°
Measure of each exterior angle = 45°
Therefore, the number of exterior angles = \(\frac{360^{\circ}}{45}\) = 8
The polygon has 8 sides.

Question 3.
Find the perimeter of the parallelogram PQRS.
Solution:
In a parallelogram, the opposite sides have same length.
Therefore, PQ = SR = 12 cm and
QR = PS = 7 cm

So, Perimeter
= PQ + QR + RS + SP
= 12 cm + 7 cm + 12 cm + 7 cm + 38 cm

Question 4.
In Fig., BEST is a parallelogram. Find the values x, y and z.

Solution:
S is opposite to B.
So.
x = 100° (opposite angles property)
y = 100° (measure of angle corresponding to ∠x)
z = 80° (since ∠y, ∠z is a linear pair)

Question 5.
In a parallelogram RING, if m∠R = 70°, find all the other angles.

Solution:
Given m∠R = 70°
Then m∠N = 70°
because ∠R and ∠I are opposite angles of a parallelogram.
Since ∠R and ∠I are supplementary,
m∠I = 180° – 70° = 110°
Also, m∠G = 110° since ∠G is opposite to ∠I
Thus, m∠R = m∠N = 70° and m∠I = m∠G = 110°

Question 6.
In Fig. HELP is a parallelogram. (Lengths are in cms). Given that OE = 4 and HL is 5 more than PE? Find OH.

Solution:
If OE = 4, then OP also is 4 So PE = 8,
Therefore HL = 8 + 5 = 13
Hence OH = \(\frac{1}{2}\) × 13 = 6.5 (cms)

Question 7.
RICE is a rhombus. Find x, y, z. Justify your findings.

Solution:
x = OE = OI (diagonals bisect) = 5
y = OR = OC (diagonals bisect) = 12
z = side of the rhombus = 13 (all sides are equal)
Rectangle : A parallelogram can be called rectangle if all its angles are equal and 90° = ∠A = ∠B = ∠C = ∠D.

So, rectangle is equiangular.
As the rectangle is parallelogram, then the opposite sides are equal.
AB = CD ; AB || CD;
AD = BC ; AD || BC
Properties:

1. Opposite sides are equal and parallel.
2. Each angle of a rectangle is right angle.
3. The diagonals of a rectangle are of equal length and also bisect each other.

Note : Every rectangle is parallelogram but not Vice – Versa.

Question 8.
RENT is a rectangle.
Its diagonals meet at O.
Find x, if OR = 2x + 4 and OT = 3x + 1.

Solution:
\(\overline{\mathrm{OT}}\) is half of the diagonal \(\overline{\mathrm{TE}}\), \(\overline{\mathrm{OR}}\) is
half of the diagonal \(\overline{\mathrm{RN}}\).
Diagonals are equal here.
So, their halves are also equal.
Therefore 3x + 1 = 2x + 4
or x = 3
Square:

SAHI is a Square.
A Square is a rectangle with equal sides.
As square is rectangle, all its angles are equal and 90°.
So, all sides and all angles are equal in a square.
In SAHI Square, \(\overline{\mathrm{SA}}\) = \(\overline{\mathrm{AH}}\) = \(\overline{\mathrm{HI}}\) = \(\overline{\mathrm{IS}}\)
and ∠S = ∠A = ∠H = ∠I = 90°

In a square

1. the diagonals bisect each other (like parallelogram)
2. the diagonals are of same length (like rectangle)
3. and they are perpendicular (like rhombus)

Properties:

1. All sides are equal (SA = AH = HI = IS)
2. All angles are right angles
   ∠S = ∠A = ∠H = ∠I = 90°
3. Diagonals perpendicular bisect each other
   SO = OH = 01 = OA (∵ Same length)