Sample Question on Qudrilaterals

Practice Question Sheet

Q:1: The diagonals AC and BD of parallelogram ABCD intersect at O. What is the ratio of the area of parallelogram ABCD and ΔAOB?

Ans:13

Q:2: If the length of the diagonal of a square is cm, then what is the perimeter of the square?

Q:3: The three angles of a quadrilateral are in the ratio 2:5:7. The sum of the greatest and the smallest among these angles is 180°. What is the measure of the fourth angle?

Q: 4: In a rhombus of side 5 cm, if the length of one of the diagonals is 8 cm, then what is the length of the other diagonal?

Ans: 5000

Q:5: In a quadrilateral PQRS, if the diagonals PR and QS are perpendicular to each other and PR bisects QS, but QS does not bisect PR, then the quadrilateral PQRS is a

Q:6: If the angles of a quadrilateral are in the ratio 3:3:7:7, then the quadrilateral is a/an

Q:7: In the given figure, ABCD is a square and PQRS is a rhombus.

The relation in which alternative is correct with respect to the given figure?

Ans:5

Q:8: The given figure shows a parallelogram ABCD where AD = BD and ∠ABD = 50°.

What are the measures of the adjacent angles of the parallelogram?

Q:9: The width and length of a rectangle are in the ratio 3:4. If its perimeter is 28 cm, then what is the area of the rectangle?

Q: 10: If one of the angles of an isosceles trapezium is 125°, then what are the measures of the other angles?

Ans:8400$

Q:11: The length and width of a rectangle ABCD are respectively a and 3a. The bisector of ∠A meets CD at P. What is the value of CP:DP?

Q:12: If in quadrilateral PQRS (vertices being taken in order), PS = QR and PQ||RS, then PQRS is necessarily a/an

Q:13: A quadrilateral in which opposite angles are equal and diagonals are bisecting each other is always a

Ans: 3, 24, 6, and 4

Q:14: The given figure shows a parallelogram ABCD. M is a point on CD such that AM = BM.

If ∠DAM = 25° and ∠ADC = 100º, then what is the measure of ∠AMB?

Ans:14

Q:15: The given figure shows a rhombus ABCD.

Which of the following relations is correct with respect to the given figure?

A) x < y B) x > y C) x = y D) y=180 - x

Solutions:

Answer:1:

Parallelogram ABCD can be drawn as:

It is known that diagonals of a parallelogram divide it into two triangles of equal area.
∴ Area of ΔABD = Area of ΔBCD
∴ Area of ΔABD =× Area of ABCD
In a parallelogram, diagonals bisect each other.
∴ AO is the median of ΔABD.
and Area of ΔAOB = Area of ΔAOD
⇒ Area of ΔAOB = × Area of ΔABD = ×× Area of ABCD

Thus, the required ratio is 4: 1.

Answer:2:

Let the length of each side of the square be a cm.
It is given that the length of its diagonal is cm.
Therefore, applying Pythagoras theorem in the triangle formed by a pair of adjacent sides and the diagonal:
a² + a² =
2a² =
a² = 16²
a = 16
Thus, perimeter of the square = 4a cm = 4 × 16 cm = 64 cm

Answer:3:

Let the three angles of the quadrilateral be 2x, 5x, and 7x.
Here, the greatest angle is 7x and smallest angle is 2x.
It is given that the sum of the greatest and the smallest angles is 180°.
∴ 7x + 2x = 180°
⇒ 9x = 180°
⇒ x = 20°
∴ Measure of fourth angle = 360° − (2x + 5x + 7x)
= 360° − 14x
= 360° − 14 × 20°                        [Putting  x = 20°]
= 360° − 280°
= 80°
Thus, the measure of the fourth angle is 80°.

Answer:4:

Let ABCD be the rhombus such that AB = BC = CD = AD = 5 cm and AC = 8 cm

In a rhombus, diagonals are perpendicular bisectors of each other.
∴AO = OC = cm = 4 cm
In ΔAOB, ∠AOB = 90°
∴ AB² = AO² + OB² [Pythagoras theorem]
⇒ (5 cm)² = (4 cm)² + OB²
⇒ OB² = (25 − 16) cm² = 9 cm²
⇒ OB = 3 cm
∴BD = 2OB = 2 × 3 cm = 6 cm
Thus, the length of the other diagonal is 6 cm.

Answer:5:

A quadrilateral whose diagonals are perpendicular to each other and one of the diagonals bisect the other is a kite.

Answer:6:

It can be observed that in the given quadrilateral, there are exactly two pairs of adjacent equal angles.
Thus, the given quadrilateral is an isosceles trapezium.

Answer:7:

In a square, all the angles are 90°.
s + q = 90° + 90° = 180°
In a rhombus (which is also a parallelogram), adjacent angles are supplementary.
∴a + b = 180°
Thus, s + q = a + b

Answer:8:

Since AD = BD and ∠ABD = 50°, ΔABD is isosceles with ∠DAB = ∠ABD = 50°.
In a parallelogram, adjacent angles are supplementary.
∴∠ABC = 180° − ∠A = 180° − 50° = 130°
Thus, the measures of adjacent angles of the parallelogram are 50° and 130°.

Answer:9:

Since the width and length of the rectangle are in the ratio 3:4, let the width be 3x cm and the length be 4x cm.
∴ 2(3x + 4x) = 28 [Perimeter = 28 cm]
⇒ 14x = 28
⇒ x = 2
∴Length = 3 × 2 cm = 6 cm and width = 4 × 2 cm = 8 cm
Thus, area of the rectangle = 6 cm × 8 cm = 48 cm²

Answer:10:

Let PQRS be an isosceles trapezium where ∠S = 125°

Since, SRPQ and ∠P = ∠Q.
So, ∠R =∠S= 125°

Also the sum of all angles of trapezium is 360°.
So,  ∠P + ∠Q + ∠R + ∠S = 360°
 ⇒  2∠P + 250° = 360°
 ⇒  2∠P = 110°
⇒ ∠P = 55°
Thus, the measures of the other angles are 55°, 55°, and 125°.

Answer:11:

The given rectangle ABCD can be drawn as:

Here, AD = a, DC = 3a
∠A = 90°
∴∠DAP = ∠A = × 90° = 45°
In ΔADP:
∠DAP + ∠ADP + ∠DPA = 180°
⇒ ∠DPA = 180° − 90° − 45° = 45°
⇒∠DAP = ∠DPA
∴ ΔADP is isosceles where ∠DAP = ∠DPA
Therefore, AD = DP = a
CP = DC − DP = 3a − a = 2a
Thus, CP: DP = 2a: a = 2:1

Answer:12:

The given quadrilateral PQRS can be drawn as:

It is known that a quadrilateral with exactly one pair of parallel sides where the non-parallel sides are of equal lengths is called an isosceles trapezium.
Thus, the given quadrilateral PQRS is an isosceles trapezium.

Answer:13:

A quadrilateral in which opposite angles are equal and diagonals are bisecting each other is always a parallelogram.

Answer:14:

In ΔAMD, ∠DMA + ∠MDA + ∠DAM = 180º
⇒ ∠DMA + 100º + 25º = 180º
⇒ ∠DMA + 125º = 180º
⇒ ∠DMA = 180º − 125º
∴∠DMA = 55º
Since ABCD is a parallelogram, AB||CD
∴∠MAB = ∠DMA = 55º [Alternate interior angles]
It is also given that AM = BM
∴∠MAB = ∠MBA = 55º
In ΔMAB:
∠MAB + ∠AMB + ∠MBA = 180º
⇒ ∠AMB + 55º + 55º = 180º
⇒ ∠AMB + 110º = 180º
⇒ ∠AMB = 180º − 110º = 70º
Thus, the measure of ∠AMB is 70°.

Answer:15:

Since ABCD is a rhombus, all the sides are equal.
∴ΔABD is isosceles.
⇒ ∠ADB = ∠ABD
∴ x = y