Practice Questions with Solutions: Quadrilaterals

Q1: The value of x in the given diagram is ?

(a) 130^o
(b) 140^o
(c) 150^o
(d) 160^o
Ans:(b) 140^o
Sol: Use the angle-sum property of a quadrilateral: the four interior angles add to 360^o.
The given angles are 70^o, 60^o, 90^o and x.
So, 70^o + 60^o + 90^o + x = 360^o.
Therefore x = 360^o - (70^o + 60^o + 90^o) = 360^o - 220^o = 140^o.
Hence x = 140^o, so option (b) is correct.

Q2: Can all the four angles of a quadrilateral be obtuse angle?
(a) Yes
(b) No
(c) May be
(d) Cannot be determined
Ans: (b) No
Explanation: ⇒ An obtuse angle is greater than 90^o and by angle sum property, the sum of all four angles of a quadrilateral is 360^o.
⇒ If we take all four angles greater than 90^o then, there sum will be obviously greater than 360^o.
Hence, all the four angles of a quadrilateral cannot be obtuse.
Therefore, option B is correct.

Q3: Can a quadrilateral ABCD be a parallelogram for the following condition? (Enter 1 for True and 0 for False)AB = DC = 8 cm, AD = 4 cm and BC= 4.4 cm.
Ans: 0 (False)
Explanation:  The answer is false.
Its given that AB = DC = 8cm,
⇒ AD = 4cm and BC = 4.4cm
To form a parallelogram the opposite sides should be of same length here, AD=BC
Hence, the answer is False.

Q4: All rhombuses are parallelograms?
(a) True
(b) False
(c) Ambiguous
(d) Data Insufficient
Ans: (a) True
Explanation: A rhombus has all four sides equal. Opposite sides of a parallelogram are equal and parallel. Since a rhombus has opposite sides equal and parallel, every rhombus satisfies the definition of a parallelogram. Therefore every rhombus is a parallelogram, so option (a) is correct.

Q5: Find the unknown angle in the figure.

(a) 40º
(b) 50º
(c) 60º
(d) 70º
Ans: (c) 60º
Sol: By the angle-sum property of a quadrilateral, the four interior angles add to 360^o.
The given angles are 50^o, 130^o, 120^o and x.
So, 50^o + 130^o + 120^o + x = 360^o.
Therefore x = 360^o - (50^o + 130^o + 120^o) = 360^o - 300^o = 60^o.
Hence x = 60^o, so option (c) is correct.

Q6: If two adjacent angles of a parallelogram are in the ratio of 2:3 find all the angles of the parallelogram.
Ans:  Let the adjacent angles be 2x and 3x.
Adjacent angles of a parallelogram are supplementary, so 2x + 3x = 180^o.
Thus 5x = 180^o ⇒ x = 36^o.
Hence the angles are 2x = 72^o and 3x = 108^o.
Opposite angles are equal, so the four angles of the parallelogram are 72^o, 108^o, 72^o and 108^o.

Q7: The angles of a quadrilateral are of the measures 120^o , 90^o, 72^o and x^o , then find x.
Ans:  We know, by angle sum property,
Sum of angles = 360^o.
So 120^o + 90^o + 72^o + x = 360^o.
Therefore x = 360^o - (120^o + 90^o + 72^o) = 360^o - 282^o = 78^o.
Thus x = 78^o.

Q8: Angles of a quadrilateral are in the ratio 3 : 6 : 8 : 13. The largest angle is :
(a) 178º
(b) 90º
(c) 156º
(d) 36º
Ans: (c) 156º
Sol: Let the common multiple be x. Then the angles are 3x, 6x, 8x and 13x.
Sum: 3x + 6x + 8x + 13x = 30x = 360^o ⇒ x = 12^o.
The largest angle is 13x = 13 × 12^o = 156^o.
Hence option (c) is correct.

Q9: Three angles of a quadrilateral are of the measure 130º, 82º , 40º . Find the measure of the fourth angle.
Ans: Sum of angles of a quadrilateral = 360^o.
Let the fourth angle be x.
Then 130^o + 82^o + 40^o + x = 360^o ⇒ x = 360^o - 252^o = 108^o.
Therefore the fourth angle is 108^o.

Q10: The angles of a quadrilateral are in the ratio 2:3:5:8. Find the smallest angle of the quadrilateral.
Ans: Given ratio 2 : 3 : 5 : 8.Let the angles be 2x, 3x, 5x, 8x.
Sum = 2x + 3x + 5x + 8x = 18x = 360^o ⇒ x = 20^o.
The smallest angle is 2x = 2 × 20^o = 40^o.

Q11: One angle of a quadrilateral is of 108º  and the remaining three angles are equal. Find each of the three equal angles.
(a) 64^o
(b) 74^o
(c) 84^o
(d) 94^o
Ans: (c) 84^o
Sol: Sum of angles = 360^o. One angle is 108^o, remaining three are equal, each = x.
So 108^o + 3x = 360^o ⇒ 3x = 252^o ⇒ x = 84^o.
Thus each equal angle measures 84^o, so option (c) is correct.

Therefore, each of the three remaining angle is x^o = 84^o.
Hence, option C is correct.

Q12: D and E are the mid-points of the sides AB and AC of ABC and O is any point on side BC.O is joined to A. If P and Q are the mid-points of OB and OC respectively, then DEQP is:
(a) a square
(b) a rectangle
(c) parallelogram
(d) quadrilateral whose opposite angles are supplementary
Ans: (c) parallelogram
Explanation:  In ΔABC, D and E are mid-points of AB and AC, so by the midpoint theorem DE ∥ BC and DE = 1/2 BC. In ΔOBC, P and Q are mid-points of OB and OC, so PQ ∥ BC and PQ = 1/2 BC. Hence DE ∥ PQ and DE = PQ.
Similarly, in triangles ABO and ACO, DP ∥ AO and EQ ∥ AO, so DP ∥ EQ and DP = EQ.Since both pairs of opposite sides DE ∥ PQ and DP ∥ EQ, DEQP is a parallelogram. Therefore option (c) is correct.

Q13: Three angles of a quadrilateral are 75º ,90º and 75º. The measure of fourth angle is?
(a) 90º
(b) 95º
(c) 105º
(d) 120º
Ans: (d) 120º
Sol: Sum of angles = 360^o.
Given angles: 75^o, 90^o, 75^o. Let the fourth angle be x.
So 75^o + 90^o + 75^o + x = 360^o ⇒ x = 360^o - 240^o = 120^o.
Hence the fourth angle is 120^o, so option (d) is correct.

Q14: Can all the angles of a quadrilateral be acute angles?
(a) Yes
(b) No
(c) May be
(d) Cannot be determined
Ans:(b) No
Explanation: Let us take a qudrilateral ABCD with BD as diagonal.
In ΔABD we have,
∠ADB +∠ABD + ∠DAB = 180º .......(i)
And in ΔDBC we have
∠DCB+∠CBD+∠BDC=180 O ......(ii)
Adding (i) & (ii), we get,
∠ADB+∠ABD+∠DAB+∠DCB+∠CBD+∠BDC = 360º
⟹∠A+∠B+∠C+∠D = 360º
So, the sum of the angles of a quadrilateral is 360 O .....(iii).
Now, if the angles are acute then ∠A < 90º ,∠B<90º ,∠C<90º ,∠D<90º
 ⟹∠A+∠B+∠C+∠D<90º +90º +90º +90º
⟹∠A+∠B+∠C+∠D<360º.
But this does not comply with (iii) . Therefore, all the angles of a quadrilateral cannot be acute angles. Hence, option B is correct.

Q15: Can the angles 110º , 80^o , 70 ^º and 95º be the angles of a quadrilateral?
(a) Yes
(b) No
(c) May be
(d) Cannot be determined
Ans: (b) No
Explanation: Sum the given angles: 110^o + 80^o + 70^o + 95^o = 355^o, which is not equal to 360^o. Since the interior angles of a quadrilateral must sum to 360^o, these four numbers cannot be the angles of a quadrilateral. Therefore option (b) is correct.

Q16: The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,
(a) ABCD is a rhombus
(b) diagonals of ABCD are equal
(c) diagonals of ABCD are equal and perpendicular
(d) diagonals of ABCD are perpendicular.
Ans:(c) diagonals of ABCD are equal and perpendicular
Explanation: In given figure,
ABCD is a quadrilateral and P,Q,R & S are mid-pints of sides AB,BC,CD and DA respectively. Then, PQRS is a square.
∴PQ=QR=RS=PS    ---------  (1)
and PR=SQ
But PR=BC and SQ=AB
∴AB=BC
Thus, all sides of quadrilateral ABCD are equal.
Hence, quadrilateral ABCD is either a square or a rhombus.
Now, in △ADB,
By using Mid-point theorem,
SP∣∣DB;SP= 1/2 DB   ------ (2)
Similarly in △ABC,
PQ∣∣AC;PQ= 1/2 AC ----- (3)
From equation (1),
PS = PQ
From (2) and (3),

Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square. So, diagonals of quadrilateral also perpendicular.

Q17: The quadrilateral formed by joining the mid-points of the sides AB,BC,CD,DA of a quadrilateral ABCD is
(a) a trapezium but not a parallelogram
(b) a quadrilateral but not a trapezium
(c) a parallelogram only
(d) a rhombus
Ans: (c) a parallelogram only
Explanation: Given quadrilateral formed By joining mid points of the sides AB,BC,CD,DA
Let mid point of AB,BC,CD,DA are P,Q,R,S respectively.
To prove:- PQRS is a parallelogram.
Draw diagonal BD and PS is mid segment of ΔABD therefore PS∣∣BD also QR is mid segment ΔBCD therefore QR∣∣BD
∵PS∣∣BD and QR∣∣BD
∴PS∣∣QR
Draw diagonal AC,SR is the mid segment of ΔACD therefore SR∣∣AC also PQ is the mid segment of ΔABC therefore PQ∣∣AC
∵SR∣∣AC and PQ∣∣AC
∴SR∣∣PQ
∵PS∣∣QR and SR∣∣PQ
∴ quadrilateral PQRS is a parallelogram as. A parallelogram is a simple quadritiral with two pairs of parallel sides.

Q18: State the angle sum property of a quadrilateral.
Ans: Ans: The sum of the interior angles of any quadrilateral is 360^o. This is the angle-sum property of quadrilaterals.

Q19: ABCD is a parallelogram. The circle through A,B and C intersects CD produced at E. If AB=10 cm,BC=8 cm,CE=14 cm . Find AE
Ans: Given ABCD is a parallelogram, so AD = BC = 8 cm and AB = CD = 10 cm.
A circle through A, B and C meets the produced line CD at E, so A, B, C, E are concyclic.
In cyclic quadrilateral A, B, C, E, we have ∠AED = ∠ABC (angles subtending the same arc) and in parallelogram ∠ABC = ∠ADC (opposite angles).
From these relations we get ∠AED = ∠EDA, so triangle AED has two equal base angles and is therefore isosceles with AE = AD.
Since AD = BC and BC = 8 cm, we obtain AE = 8 cm.
Hence AE = 8 cm.

Q20: If the tangents  PA  and  PB  from a point  P  to the circle with cetre  O  inclined to each other at the angle of  110 ,  then find  ∠POA.
Ans: 70^o
Explanation: Let A and B be the points of contact. The angles between a tangent and the radius at the point of contact are right angles, so ∠OAP = ∠OBP = 90^o. In quadrilateral PAOB the sum of interior angles is 360^o.
Given ∠APB = 110^o, so ∠AOB = 360^o - (110^o + 90^o + 90^o) = 70^o.
Thus ∠POA (which is the same as ∠AOB) = 70^o.