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Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

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

For a parallelogram ABCD, opposite angles are equal.

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

It is also known 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 ||gm. Find the angles A, B, C, and D.

Class 9 Maths Chapter 8 HOTS Questions - QuadrilateralsSol.

In ∆ACD, 4x + 5x + 6x = 180°
⇒ 15x = 180°
⇒ x = (180o/15) = 12°
∴ ∠D = 6 × 12° = 72°
⇒ B = 72°                 [∵ opposite angles of ||gm are equal]
∵ ∠ A + ∠ D = 180°                 [cointerior 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).

Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

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), hemce 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.

Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

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 ||gm. PS is produced to M so that SM = SR and MR produced meet PQ produced at N. Prove that QN = QR.

Class 9 Maths Chapter 8 HOTS Questions - QuadrilateralsSol.

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.

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.

Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

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

The document Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FAQs on Class 9 Maths Chapter 8 HOTS Questions - Quadrilaterals

1. What are the properties of a quadrilateral?
Ans. A quadrilateral is a polygon with four sides. Some of its properties include: the sum of its interior angles is equal to 360 degrees, opposite sides are parallel, opposite angles are equal, and the diagonals bisect each other.
2. Can a quadrilateral have all sides of different lengths?
Ans. Yes, a quadrilateral can have all sides of different lengths. It is called a general quadrilateral and does not have any specific properties or special characteristics.
3. How can we classify quadrilaterals based on their angles?
Ans. Quadrilaterals can be classified into different types based on their angles. Some of the common classifications include rectangles (opposite angles are equal and all angles are right angles), squares (all angles are right angles and all sides are equal), parallelograms (opposite angles are equal), and trapezoids (one pair of opposite sides is parallel).
4. Are all rectangles also squares?
Ans. No, not all rectangles are squares. While all squares are rectangles (because they have four right angles), rectangles can have different side lengths, whereas squares have all sides equal in length.
5. How can we calculate the area of a quadrilateral?
Ans. The formula to calculate the area of a quadrilateral depends on its type. For example, the area of a rectangle can be calculated by multiplying its length and width, while the area of a square can be found by squaring the length of one of its sides. Other quadrilaterals may require different formulas, such as the area of a trapezoid being calculated by multiplying the sum of its parallel sides with its height and dividing by 2.
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