8th Grade > Mathematics
UNDERSTANDING QUADRILATERALS MCQs
Total Questions : 464
| Page 43 of 47 pages
Answer: Option A. ->
:
In a parallelogram, adjacent angles are supplementary.
∴120∘+(5x+10)∘=180∘⇒5x+10∘+120∘=180∘⇒5x=180∘−130∘⇒5x=50∘⇒x=10∘
Also, opposite angles are equal in a parallelogram.
Therefore, 6y=120∘⇒y=20∘
:
In a parallelogram, adjacent angles are supplementary.
∴120∘+(5x+10)∘=180∘⇒5x+10∘+120∘=180∘⇒5x=180∘−130∘⇒5x=50∘⇒x=10∘
Also, opposite angles are equal in a parallelogram.
Therefore, 6y=120∘⇒y=20∘
Answer: Option A. ->
:
The given figure is a kite.
In a kite, one pair of opposite angles are equal.
∴y=110∘
Now, by the angle sum property of a quadrilateral, we have
110∘+60∘+110∘+x=360∘⇒x=360∘−280∘⇒x=80∘
:
The given figure is a kite.
In a kite, one pair of opposite angles are equal.
∴y=110∘
Now, by the angle sum property of a quadrilateral, we have
110∘+60∘+110∘+x=360∘⇒x=360∘−280∘⇒x=80∘
Answer: Option A. ->
:
Given, a trapezium ABCD in which ∠A=(x−20)∘,∠D=(x+40)∘
Since, in a trapezium, the angles between the parallel lines are supplementary,
(x−20)+(x+40)=180∘⇒x−20+x+40=180∘⇒2x+20∘=180∘⇒2x=(180∘−20∘)=160∘⇒x=80∘
:
Given, a trapezium ABCD in which ∠A=(x−20)∘,∠D=(x+40)∘
Since, in a trapezium, the angles between the parallel lines are supplementary,
(x−20)+(x+40)=180∘⇒x−20+x+40=180∘⇒2x+20∘=180∘⇒2x=(180∘−20∘)=160∘⇒x=80∘
Answer: Option A. ->
:
Let ABCD be a quadrilateral,
where ∠A=∠C=75∘ and ∠B=∠D=x [say]
Then, by the angle sum property of a quadrilateral, we have
∠A+∠B+∠C+∠D=360∘⇒75∘+x+75∘+x=360∘⇒2x=360∘−150∘⇒2x=210∘⇒x=105∘
Thus, other two angles are of 105∘ each.
Since, opposite angles are equal, therefore the quadrilateral is a parallelogram.
:
Let ABCD be a quadrilateral,
where ∠A=∠C=75∘ and ∠B=∠D=x [say]
Then, by the angle sum property of a quadrilateral, we have
∠A+∠B+∠C+∠D=360∘⇒75∘+x+75∘+x=360∘⇒2x=360∘−150∘⇒2x=210∘⇒x=105∘
Thus, other two angles are of 105∘ each.
Since, opposite angles are equal, therefore the quadrilateral is a parallelogram.
Answer: Option A. ->
:
Given a quadrilateral PQRS, where
∠P=50∘,∠Q=50∘ and ∠R=60∘
Now, by the angle sum property of a quadrilateral, we have
∠P+∠Q+∠R+∠S=360∘⇒50∘+50∘+60∘+∠S=360∘⇒∠S=360∘−160∘⇒∠S=200∘
One interior angle of the given quadrilateral is greater than 180∘, therefore the quadrilateral is concave.
:
Given a quadrilateral PQRS, where
∠P=50∘,∠Q=50∘ and ∠R=60∘
Now, by the angle sum property of a quadrilateral, we have
∠P+∠Q+∠R+∠S=360∘⇒50∘+50∘+60∘+∠S=360∘⇒∠S=360∘−160∘⇒∠S=200∘
One interior angle of the given quadrilateral is greater than 180∘, therefore the quadrilateral is concave.
Answer: Option A. ->
:
Let ABCD be a quadrilateral, such that
∠A=∠C,∠B=∠D and also ∠A+∠C=180∘,∠B+∠D=180∘
Now, ∠A+∠A=180∘ [∵∠C=∠A]
⇒2∠A=180∘⇒∠A=90∘Similarly,∠B=90∘
Hence, each angle is a right angle.
:
Let ABCD be a quadrilateral, such that
∠A=∠C,∠B=∠D and also ∠A+∠C=180∘,∠B+∠D=180∘
Now, ∠A+∠A=180∘ [∵∠C=∠A]
⇒2∠A=180∘⇒∠A=90∘Similarly,∠B=90∘
Hence, each angle is a right angle.
Answer: Option A. ->
:
We know that, a pentagon has 5 sides and a decagon has 10.
Measure of each exterior angle of a regular pentagon
=360∘5=72∘
Measure of each exterior angle of a regular decagon
=360∘10=36∘
∴ Required ratio
=7236
=2:1
:
We know that, a pentagon has 5 sides and a decagon has 10.
Measure of each exterior angle of a regular pentagon
=360∘5=72∘
Measure of each exterior angle of a regular decagon
=360∘10=36∘
∴ Required ratio
=7236
=2:1
Answer: Option A. ->
:
Number of sides (n), in octagon = 8
Now, the sum of interior angles of a regular octagon =(n−2)×180∘
=(8−2)×180∘6×180∘=1080∘
Since, the octagon is regular, measure of each angle =1080∘8=135∘
:
Number of sides (n), in octagon = 8
Now, the sum of interior angles of a regular octagon =(n−2)×180∘
=(8−2)×180∘6×180∘=1080∘
Since, the octagon is regular, measure of each angle =1080∘8=135∘
Answer: Option A. ->
:
Let the measure of equal angles be x∘ each.
Then, by the angle sum property of a quadrilateral, we have
x∘+x∘+x∘+120∘=360∘⇒3x∘+120∘=360∘⇒3x∘=240∘⇒x∘=80∘
:
Let the measure of equal angles be x∘ each.
Then, by the angle sum property of a quadrilateral, we have
x∘+x∘+x∘+120∘=360∘⇒3x∘+120∘=360∘⇒3x∘=240∘⇒x∘=80∘