Understanding Shapes — Exercise 13.3

Identify all the quadrilaterals that have

(i) four sides of equal length

(ii) four right angles.

Answer

(i) Four sides of equal length

A square has all its four sides equal in length and a rhombus also has all its four sides equal in length.

Hence, a square and a rhombus have four sides of equal length.

(ii) Four right angles

A square has all its four angles equal to 90° and a rectangle also has all its four angles equal to 90°.

Hence, a square and a rectangle have four right angles.

Explain how a square is

(i) a quadrilateral

(ii) a parallelogram

(iii) a rhombus

(iv) a rectangle.

Answer

(i) A quadrilateral

A quadrilateral is a polygon bounded by four line segments.

A square is a polygon bounded by four line segments.

Hence, a square is a 4-sided polygon, so it is a quadrilateral.

(ii) A parallelogram

A parallelogram is a quadrilateral in which both the pairs of opposite sides are parallel.

In a square, both the pairs of opposite sides are parallel.

Hence, a square has its opposite sides parallel, so it is a parallelogram.

(iii) A rhombus

A rhombus is a parallelogram in which all the four sides are equal.

In a square, all the four sides are equal.

Hence, a square is a parallelogram with all the 4 sides equal, so it is a rhombus.

(iv) A rectangle

A rectangle is a parallelogram in which each angle is a right angle.

In a square, each angle is a right angle.

Hence, a square is a parallelogram with each angle a right angle, so it is a rectangle.

Name the quadrilaterals whose diagonals

(i) bisect each other

(ii) are perpendicular bisectors of each other

(iii) are equal.

Answer

(i) The diagonals of a parallelogram bisect each other. As a rhombus, a rectangle and a square are all parallelograms, their diagonals also bisect each other.

Hence, the quadrilaterals are parallelogram, rhombus, square and rectangle.

(ii) The diagonals of a rhombus bisect each other at right angles. As a square is also a rhombus, its diagonals are also perpendicular bisectors of each other.

Hence, the quadrilaterals are rhombus and square.

(iii) The diagonals of a rectangle are equal in length. As a square is also a rectangle, its diagonals are also equal in length.

Hence, the quadrilaterals are rectangle and square.

One of the diagonals of a rhombus and its sides are equal. Find the angles of the rhombus.

Answer

It is given that one of the diagonals of a rhombus and its sides are equal.

Let ABCD be the rhombus in which the diagonal AC is equal to each of its sides.

So, AB = BC = CD = DA = AC

In ΔABC,

AB = BC = AC

As all the three sides are equal, ΔABC is an equilateral triangle.

So, ∠BAC = ∠ABC = ∠BCA = 60°

Similarly, in ΔACD,

AC = CD = DA

As all the three sides are equal, ΔACD is an equilateral triangle.

So, ∠CAD = ∠ACD = ∠ADC = 60°

Thus, the angles of the rhombus are:

⇒ ∠A = ∠BAC + ∠CAD = 60° + 60° = 120°

⇒ ∠B = ∠ABC = 60°

⇒ ∠C = ∠BCA + ∠ACD = 60° + 60° = 120°

⇒ ∠D = ∠ADC = 60°

Hence, the angles of the rhombus are 60°, 120°, 60° and 120°.

In the adjoining figure ABCD is a rhombus, find the values of x, y and z.

Answer

It is given that ABCD is a rhombus in which the diagonals AC and BD intersect at O, with DO = 6, OC = 8, AO = x, OB = y and AB = z.

As the diagonals of a rhombus bisect each other,

AO = OC

⇒ x = 8

Also,

OB = DO

⇒ y = 6

As the diagonals of a rhombus intersect at right angles, ∠AOB = 90°.

So, ΔAOB is a right angled triangle.

By Pythagoras theorem,

⇒ AB² = AO² + OB²

⇒ z² = 8² + 6²

⇒ z² = 64 + 36

⇒ z² = 100

⇒ z = $\sqrt{100}$

⇒ z = 10

Hence, x = 8, y = 6 and z = 10.

In the adjoining figure, ABCD is a trapezium. If ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ∠C = (5x - 31)°, then find all the angles of the trapezium.

Answer

It is given that ABCD is a trapezium in which AB ∥ DC, ∠A : ∠D = 5 : 7, ∠B = (3x + 11)° and ∠C = (5x - 31)°.

As AB ∥ DC and AD is a transversal, the co-interior angles ∠A and ∠D are supplementary.

⇒ ∠A + ∠D = 180°

Let the common factor of ∠A and ∠D be a. So, ∠A = 5a and ∠D = 7a.

⇒ 5a + 7a = 180°

⇒ 12a = 180°

⇒ a = $\dfrac{180\degree}{12}$

⇒ a = 15°

Thus,

∠A = 5a = 5 x 15° = 75°

∠D = 7a = 7 x 15° = 105°

As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

⇒ (3x + 11)° + (5x - 31)° = 180°

⇒ 3x° + 11° + 5x° - 31° = 180°

⇒ 8x° - 20° = 180°

⇒ 8x° = 180° + 20°

⇒ 8x° = 200°

⇒ x° = $\dfrac{200\degree}{8}$

⇒ x = 25

Thus,

∠B = (3x + 11)° = (3 x 25 + 11)° = (75 + 11)° = 86°

∠C = (5x - 31)° = (5 x 25 - 31)° = (125 - 31)° = 94°

Hence, ∠A = 75°, ∠B = 86°, ∠C = 94° and ∠D = 105°.

In the adjoining figure, ABCD is a rectangle. If ∠CEB : ∠ECB = 3 : 2, find

(i) ∠CEB

(ii) ∠DCF

Answer

It is given that ABCD is a rectangle and ∠CEB : ∠ECB = 3 : 2.

(i) As ABCD is a rectangle, each of its angles is 90°.

So, ∠ABC = 90°, which means ∠EBC = 90°.

Let the common factor be a. So, ∠CEB = 3a and ∠ECB = 2a.

In ΔEBC, the sum of all the angles of a triangle is 180°.

⇒ ∠CEB + ∠ECB + ∠EBC = 180°

⇒ 3a + 2a + 90° = 180°

⇒ 5a = 180° - 90°

⇒ 5a = 90°

⇒ a = $\dfrac{90\degree}{5}$

⇒ a = 18°

Thus,

∠CEB = 3a = 3 x 18° = 54°

Hence, ∠CEB = 54°.

(ii) ∠ECB = 2a = 2 x 18° = 36°

As ABCD is a rectangle, ∠DCB = 90°.

From the figure,

⇒ ∠DCE = ∠DCB - ∠ECB

⇒ ∠DCE = 90° - 36°

⇒ ∠DCE = 54°

As EC is produced to F, ∠DCE and ∠DCF form a linear pair.

⇒ ∠DCF + ∠DCE = 180°

⇒ ∠DCF + 54° = 180°

⇒ ∠DCF = 180° - 54°

⇒ ∠DCF = 126°

Hence, ∠DCF = 126°.

In the adjoining figure, ABCD is a rectangle and diagonals intersect at O. If ∠AOB = 118°, find

(i) ∠ABO

(ii) ∠ADO

(iii) ∠OCB

Answer

It is given that ABCD is a rectangle whose diagonals intersect at O and ∠AOB = 118°.

As the diagonals of a rectangle are equal and bisect each other,

OA = OB = OC = OD

(i) In ΔAOB, OA = OB, so ΔAOB is an isosceles triangle.

So, ∠ABO = ∠BAO (angles opposite to equal sides are equal)

As the sum of all the angles of a triangle is 180°,

⇒ ∠ABO + ∠BAO + ∠AOB = 180°

⇒ 2∠ABO + 118° = 180°

⇒ 2∠ABO = 180° - 118°

⇒ 2∠ABO = 62°

⇒ ∠ABO = $\dfrac{62\degree}{2}$

⇒ ∠ABO = 31°

Hence, ∠ABO = 31°.

(ii) As ∠AOB and ∠AOD form a linear pair,

⇒ ∠AOD = 180° - 118° = 62°

In ΔAOD, OA = OD, so ΔAOD is an isosceles triangle.

So, ∠ADO = ∠DAO (angles opposite to equal sides are equal)

As the sum of all the angles of a triangle is 180°,

⇒ ∠ADO + ∠DAO + ∠AOD = 180°

⇒ 2∠ADO + 62° = 180°

⇒ 2∠ADO = 180° - 62°

⇒ 2∠ADO = 118°

⇒ ∠ADO = $\dfrac{118\degree}{2}$

⇒ ∠ADO = 59°

Hence, ∠ADO = 59°.

(iii) In ΔAOD, OA = OD, so ∠OAD = ∠ADO = 59°.

As AD ∥ BC and AC is a transversal,

∠OCB = ∠OAD (alternate angles are equal) In the adjoining isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.

⇒ ∠OCB = 59°

Hence, ∠OCB = 59°.

In the adjoining figure, ABCD is a rhombus and ∠ABD = 50°. Find

(i) ∠CAB

(ii) ∠BCD

(iii) ∠ADC

Answer

It is given that ABCD is a rhombus and ∠ABD = 50°.

Let the diagonals AC and BD intersect at O.

As the diagonals of a rhombus intersect at right angles,

∠AOB = 90°

(i) In ΔAOB, the sum of all the angles of a triangle is 180°.

⇒ ∠CAB + ∠ABO + ∠AOB = 180°

⇒ ∠CAB + 50° + 90° = 180°

⇒ ∠CAB + 140° = 180°

⇒ ∠CAB = 180° - 140°

⇒ ∠CAB = 40°

Hence, ∠CAB = 40°.

(ii) As the diagonals of a rhombus bisect the angles of the rhombus, the diagonal BD bisects ∠ABC.

⇒ ∠ABC = 2 x ∠ABD

⇒ ∠ABC = 2 x 50°

⇒ ∠ABC = 100°

As the adjacent angles of a rhombus are supplementary,

⇒ ∠BCD + ∠ABC = 180°

⇒ ∠BCD + 100° = 180°

⇒ ∠BCD = 180° - 100°

⇒ ∠BCD = 80°

Hence, ∠BCD = 80°.

(iii) As the opposite angles of a rhombus are of equal measure,

∠ADC = ∠ABC = 100°

Hence, ∠ADC = 100°.

In the adjoining isosceles trapezium ABCD, ∠C = 102°. Find all the remaining angles of the trapezium.

Answer

It is given that ABCD is an isosceles trapezium in which AB ∥ DC, AD = BC and ∠C = 102°.

As the angles on the same base of an isosceles trapezium are equal,

∠D = ∠C = 102°

As AB ∥ DC and BC is a transversal, the co-interior angles ∠B and ∠C are supplementary.

⇒ ∠B + ∠C = 180°

⇒ ∠B + 102° = 180°

⇒ ∠B = 180° - 102°

⇒ ∠B = 78°

As the angles on the same base of an isosceles trapezium are equal,

∠A = ∠B = 78°

Hence, ∠A = 78°, ∠B = 78° and ∠D = 102°.

In the adjoining figure PQRS is a kite. Find the values of x and y.

Answer

It is given that PQRS is a kite in which ∠Q = 120°, ∠R = 50°, ∠P = y and ∠S = x.

In kite PQRS, PQ = PS and RQ = RS.

So, the angles included between the pairs of unequal sides are equal.

⇒ ∠S = ∠Q

⇒ x = 120°

As we know, the sum of all the interior angles of a quadrilateral is 360°.

⇒ ∠P + ∠Q + ∠R + ∠S = 360°

⇒ y + 120° + 50° + 120° = 360°

⇒ y + 290° = 360°

⇒ y = 360° - 290°

⇒ y = 70°

Hence, x = 120° and y = 70°.