Understanding Quadrilaterals Class 8 Extra Questions

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Question 1.
The sum of the interior angles of a regular polygon is twice the sum of the exterior angles. Find the number of sides of the polygon Check your Answers

Let n be the sides,then
$(n-2) \times 180 = 2 \times 360$
$n=6$

Question 2.
Three angles of the quadrilaterals are in the ratio 2:3:4. The sum of the least and the greatest of these three angles is equal to 180°. Find all the angles of the quadrilateral Check your Answers

Three angles of the quadrilaterals are in the ratio 2:3:4, then it will be 2x,3x,4x.
Now it is given
$2x + 4x=180$
$x=30$
So three angles are 60°,90°,120°
Let y be the fourth angle,Now in a quadrilateral
Sum of all angles =360
$y + 60+90+120=360$
$y=90$
So All the four angles are 60°,90°,120°,90°

Question 3.
The exterior angle of a regular polygon is one-fifth of its interior angle. How many sides has the polygon? Check your Answers

let the interior angle be x,then exterior angle will be (180- x )
Now as per the question
$(180 - x ) = \frac {1}{5} x$
Solving this, we get
x= 150°
Therefore,Each interior angle is 150°
Now we know that
Sum of all interior angles is given by =$( n - 2 ) \times 180$
Therefore
$( n - 2 ) \times 180 = 150 n$
Simplifying this we get
6n - 5n = 12
n = 12
So, Number of sides= 12

Question 4.
One of the angle in the parallelogram is 80°. Find the other angles in the parallelogram Check your Answers

Adjacent angles are supplementary, So
$x + 80=180$
x=100°
So angles are 80°,100°,80°, 100°

Question 5.
The angles of the quadrilateral is in the ratio 1:2:3:4. Find all the angles Check your Answers

Let x be common ratio, then angles are 1x,2x,3x,4x
Sum of all angles =360
$1x+2x+3x+4x=360$
$x=36$
So angles are 36°, 72°, 108°,144°

Question 6.
One of the diagonals of a rhombus is equal to its sides. Find the measures of all the angles of the rhombus Check your Answers

Here AB=BC=CD=AD
Also AB=AC
Now in Triangle ABC,
AB=BC=AC
Hence equliateral triangle
$\angle B=60$, $\angle BAC=60$ and $\angle BCA=60$
Now in Triangle ACD
AC=CD=AD
Hence equliateral triangle
$\angle D=60$, $\angle DAC=60$ and $\angle DCA=60$
Now $\angle A= \angle BAC + \angle DAC=120$
$\angle C =\angle BCA + \angle DCA=120$
Hence angles are 60°, 120°, 60°,120°

Question 7.
In a parallelogram ABCD, the bisectors of ∠ A and ∠ B meet at O. Find ∠ AOB.

Question 8.
Find the value of x in the trapezium ABCD

Oppossite angles are equal
So 6y=120, y=20
Now adjacent angles are supplementary
$120 + 5x + 10=180$
$x=10$

Summary

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