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What angle is formed at the intersection of the diagonals of a rhombus?

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Mensuration is a branch of mathematics that deals with the measurement of two-dimensional (2D) and three-dimensional (3D) geometric figures and their parameters like length, shape, surface area, lateral surface area, volume, etc. In simple words, it outlines the principles of calculation based on algebraic equations, mathematical formulas, and also the properties of various geometric shapes. Circles, triangles, quadrilaterals, and pentagons are some examples of 2D geometric shapes, while spheres, cubes, prisms, cylinders, etc. are examples of 3D geometric shapes. Archimedes is known as the father of mensuration. 

Rhombus

In Euclidean geometry, a rhombus is a quadrilateral whose opposite sides are parallel to each other and the lengths of all four sides are equal. A rhombus is a special type of parallelogram with diagonals bisecting and intersecting each other at right angles, i.e., 90°. The opposite interior angles of a rhombus are equal, and the sum of the four angles of a rhombus is equal to 360°. A rhombus is in the shape of a diamond, hence it is also called a diamond. A square is a special case of a rhombus as it also has four equal sides with diagonals bisecting each other at right angles. Apart from that, the four interior angles of a square are right angles.

 

Diagonals of a rhombus

A diagonal is a line segment that connects the opposite vertices of a polygon. A rhombus has two diagonals bisecting each other at right angles, i.e., 90°, and thus, four right-angled congruent triangles are formed. Let’s look at some of the properties of rhombus diagonals.

  1. The diagonals of a rhombus bisect each other at right angles.
  2. The diagonals of a rhombus bisect the opposite angles of a rhombus. 
  3. The length of the diagonals of a rhombus may or may not be equal.
  4. The length of the diagonals can be found using various formulae, i.e., the area of a rhombus, the Pythagorean theorem.
  5. Another rhombus can be formed in a rhombus by joining the midpoints of the half diagonals.

 

Area of a rhombus

The area covered by a rhombus in a 2D plane is called the area of a rhombus. The formula for the area of a rhombus is equal to the product of the lengths of the diagonals divided by 2.

Area of rhombus = (d1 × d2)/2 square units

Where d1 and d2 are the lengths of the diagonals.

Perimeter of a rhombus

The perimeter of a rhombus is the sum of all the four sides or the total length of its boundaries. The formula for the perimeter of a rhombus is equal to four times the length of its side.

The perimeter of a rhombus = 4a units

Where a is the length of the side of a rhombus. 

Angle formed at the intersection of the diagonals of a rhombus

Let’s consider a rhombus ABCD and O to be the point of intersection of the diagonals AC and BD.

 

Now, consider ΔAOB and ΔAOD

OB = OD [Since the diagonals of a rhombus bisect each other]

OA = OA [Common side of both the triangles]

AB = DA [Since sides of a rhombus are equal]

Hence, by SSS (side side side) theorem, ΔAOB and ΔAOD are congruent.

Since ΔAOB and ΔAOD are congruent, all the corresponding sides and angles must be equal.

Therefore, ∠AOB = ∠AOD.

As BD is a straight line, the sum of angles ∠AOB and ∠AOD should be equal to 180°.

So, ∠AOB + ∠AOD = 180°

⇒ 2 ∠AOD = 180°

⇒ ∠AOD = 180°/2 = 90°

⇒ ∠AOB = ∠AOD = 90°

Hence, the angle formed at the intersection of the diagonals of a rhombus is a right angle, i.e., 90°.

Sample Problems

Problem 1: Find the diagonal of a rhombus if its area is 150 cm2 and the length of its shortest diagonal is 10 cm. 

Solution: 

Given data, 

Area of a rhombus = 150 cm

The length of the shortest diagonal = 10 cm.

We know that,

Area of rhombus = ½ d1d2  square units 

⇒ 150 = (d1 × 10)/2

⇒ 300 = d1 ×10 ⇒ d1= 30 cm

Hence, the length of the longest diagonal of the rhombus is 30 cm.

Problem 2: Find the perimeter of the rhombus if the length of each side is 11 cm.

Solution:

Given data,

Length of the side of rhombus = 11 cm. 

We know that,

The perimeter of a rhombus = 4a units

= 4 × 11 = 11 units.

Hence, the perimeter of the rhombus is 44 cm.

Problem 3: Find the value of x if AO = 4x-7 and OC = 3x+11, where AC is the diagonal of a rhombus and O is its centre.

Solution: 

Given data,

AO = 4x-7 and OC = 3x + 11

We know that,

The diagonals of the rhombus bisect each other.

Hence, AO = OC

⇒ 4x – 7 = 3x + 11

⇒ 4x – 3x = 11 + 7

⇒ x = 18

Hence, the value of x = 18

Problem 4: Find the value of y in the figure given below.

 

Solution:

Given data,

∠OCD = 41° and ∠CDO = y 

We know that the angle formed at the intersection of the diagonals of a rhombus is a right angle, i.e., 90°.

Hence, ΔDOC is a right-angled triangle.

⇒ ∠DOC = 90° 

We know that the sum of angles in a triangle is 180°. 

⇒ ∠OCD + ∠DOC + ∠CDO = 180°

⇒ 41° + 90° + y = 180° 

⇒ y = 180° – 90° – 41° = 49°

Hence, the value of y = 49°. 

Problem 5: Find the area of a rhombus if its perimeter is 60 cm and the length of one of its diagonal is 18 cm.

Solution:

Given data,

 

Let “a” be each side of a rhombus

The perimeter of a rhombus = 60 cm

The length of one diagonal = 18 cm

We know that,

The perimeter of a rhombus = 4a

⇒ 4a = 60 ⇒ a = 60/4

⇒ a =15 cm

We know that the angle formed at the intersection of the diagonals of a rhombus is a right angle, i.e., 90°.

So, consider the right-angled triangle BOC

We know that, the diagonals of the rhombus bisect each other.

So, OB = BD/2 = 9 cm

By Pythagorean theorem,

BC2 = OB2 + OC2

⇒ OC2 = BC2 – OB2

= (15) – 9 = 225 – 81 = 144

⇒ OC = √144 ⇒ OC = 12 cm.

Now, AC = 2 × OC = 2 × 12 = 24 cm.

We know that,

Area of a rhombus (A) = (d1 × d2)/2 square units

⇒ A = (24 × 18)/2 = 432/2 = 216‬ sq. cm

Hence, the area of the given rhombus = 216‬ sq. cm

Problem 6: Find the value of y in the figure given below.

 

Solution:

Given data,

∠OAB = 36 and ∠OBC = x

We know that the angle formed at the intersection of the diagonals of a rhombus is a right angle, i.e., 90°.

Hence, ΔAOB is a right-angled triangle.

⇒ ∠AOB = 90°

We know that the sum of angles in a triangle is 180°.

⇒ ∠OAB + ∠AOB + ∠ABO = 180°

⇒ 36° + 90° + ∠ABO = 180° 

⇒ ∠ABO = 180° – 90° – 36° = 54°

We know that the diagonals of a rhombus bisect the opposite angles of a rhombus. 

So, ∠ABO = ∠OBC = 54°

Hence, the value of x is 54°.

Problem 7: Find the area of a rhombus if the length of each side is 10 cm and the length of one diagonal is 16 cm.

Solution:

Given data,

 

Length of each side of a rhombus = 10 cm

The length of one diagonal = 16 cm

We know that the angle formed at the intersection of the diagonals of a rhombus is a right angle, i.e., 90°.

Consider the right-angled triangle OAB

We know that, the diagonals of the rhombus bisect each other.

AB = 10 cm

OA = AC/2 = 16/2 = 8 cm

By Pythagorean theorem,

AB2 = OA2 + OB2 

⇒ OB2 = AB2 – OA2

= (10)2 – 82 = 100 – 64 = 36

⇒ OB = √36 ⇒ OB = 6 cm

Now, BD = 2 × OB = 2 × 6 = 12 cm

We know that,

Area of a rhombus (A) = (d1 × d2)/2

⇒ A = (16 × 12)/2 = 192/2 = 96 sq. cm

Hence, the area of the given rhombus = 96 sq. cm



Last Updated : 30 Jan, 2024
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