Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

Q1:  Find all the angles of a parallelogram if one angle is 80°.

Sol: 
For a parallelogram ABCD, opposite angles are equal.

So, the angles opposite to the given 80° angle will also be 80°.

We also know that the sum of angles of any quadrilateral = 360°.

So, if ∠A = ∠C = 80° then,

∠A + ∠B + ∠C + ∠D = 360°

Also, ∠B = ∠D

Thus,

80° + ∠B + 80° + ∠D = 360°

Or, ∠B +∠ D = 200°

Hence, ∠B = ∠D = 100°

Now, all angles of the quadrilateral are found which are:

∠A = 80°

∠B = 100°

∠C = 80°

∠D = 100°

Q2: In the adjoining figure, ABCD is a parallelogram. Find the angles A, B, C, and D.

Sol:

In ∆ACD, 4x + 5x + 6x = 180°
⇒ 15x = 180°
⇒ x = (180^o/15) = 12°
∴ ∠D = 6 × 12° = 72°
⇒ B = 72°                 [∵ opposite angles of ||gm are equal]
∵ ∠ A + ∠ D = 180°                 [co-interior angles]
∴ ∠A = 180° – ∠D = 180° – 72° = 180°
⇒ Therefore, ∠ C = 108°

Q3: The sides AD and BC of a quadrilateral are produced as shown in the given figure. Prove that x = ((a+b)/2).

Sol: 

We have ∠a + ∠ADC = 180°            [linear pair]
Similarly, ∠b + ∠BCD = 180°
Adding (a + b) + ∠ADC + ∠BCD = 360°               ...(1)
But     x + x + ∠ADC + ∠BCD = 360°                   ...(2)
From (1) and (2)
x + x + ∠ADC + ∠BCD = a + b + ∠ADC + ∠BCD
⇒ x + x = a + b
⇒ 2x = a + b
⇒ x = ((a+b)/2). Hence proved.

Q4: L, M, N, K are midpoints of sides BC, CD, DA and AB respectively of a square ABCD.
Prove that DL, DK, BM and BN enclose a rhombus.

Sol: 

BK = DM [halves of equal sides]
∴ BM || DK. Similarly, BN || DL
Also, ∆ ABN ≌ ∆ ADK   [SAS congruency]
⇒ ∠ 1 = ∠ 2
Also, ∆PND ≌ ∆PKB     [ASA congruency]
⇒ PB = PD
⇒ Therefore, DQBP is a rhombus.

Q5: PQRS is a parallelogram. PS is produced to M so that SM = SR and MR produced meet PQ produced at N. Prove that QN = QR.

Sol: 

In ∆SMR, SM = SR
⇒ ∠ 1 = ∠ 2                 [Angles opposite to equal sides are equal]
∠1 = ∠3                
[∵QR || PM, corresponding angles are equal]
Similarly, ∠2 = ∠4      [corresponding angles]
⇒ ∠ 3 = ∠ 4
⇒ Hence in ∆ QRN, QN = QR

Q6: Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.

Sol:

As the angles are in the ratio 2:5:4:1, they can be written as: 2x, 5x, 4x, and x.

Now, as the sum of the angles of a quadrilateral is 360°,

2x + 5x + 4x + x = 360°

Or, x = 30°

Now, all the angles will be,

2x =2 × 30° = 60°

5x = 5 × 30° = 150°

4x = 4 × 30° = 120°, and

x = 30°

Q7: Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.

Sol: 
Let the angle of the parallelogram given in the question statement be “x”.

Now, its adjacent angle will be 2x.

It is known that the opposite angles of a parallelogram are equal.

So, all the angles of a parallelogram will be x, 2x, x, and 2x

As the sum of interior angles of a parallelogram = 360°,

x + 2x + x + 2x = 360°

Or, x = 60°

Thus, all the angles will be 60°, 120°, 60°, and 120°.

Q8:  In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°.

Sol: 
In a trapezium ABCD, ∠A + ∠D = 180° and ∠B + ∠C = 180°

So, 55° + ∠D = 180°

Or, ∠D = 125°

Similarly,

70° + ∠C = 180°

Or, ∠C = 110°

Q9: In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.

Sol: 
Let ABCD be a rectangle where AC and BD are the two diagonals which are intersecting at point O.

Now, assume ∠BDC = 25° (given)

Now, ∠BDA = 90° – 25° = 65°

Also, ∠DAC = ∠BDA, (as diagonals of a rectangle divide the rectangle into two congruent right triangles)

So, ∠BOA = the acute angle between the two diagonals = 180° – 65° – 65° = 50°

Q10: If the bisectors of the angles of a quadrilateral enclose a rectangle, then show that it is a parallelogram.

Sol: 
Angle bisectors of the quadrilateral ABCD enclose a rectangle PQRS.
∴ ∠P = 90^o
⇒ In Δ PCD, ∠1+∠2 = 90^o
But, ∠1 and ∠2 are 1/2∠D and 1/2 ∠C respectively.
⇒ ∠D +∠C = 180^o [∵ 2∠1 + 2∠2 = 180^o]
⇒ ∠ D and ∠ C form a pair of co-interior supplementary angles AD || BC
Similarly, AB || DC ⇒ ABCD is a parallelogram.