Question

Length of the side of a rhombus is 13 cm and the sum of the lengths of diagonals is34cm . The area of the rhombus is:​

Answer 1

ANSWER:

The area of the triangle will be $120 cm^{2}$.

Step-by-step explanation:

Given: The length of the side is $13$$cm$.

           The sum of the length of the diagonals is $34cm^{2}$.

To find The area of the rhombus.

Solution:

• $(\frac{a}{2})^{2} + (\frac{b}{2})^{2} = side^{2}$
• $a^{2/4} +b^{2/4} =13^{2}$                                                                                
• $a^{2} +b^{2} = 676$
• Adding $2ab$ on both sides
• $a^{2} + b^{2} + 2ab = 676 + 2ab$
• $(a + b)^{2} = 676 + 2ab$
• Substituting the value of $a + b$ which are the diagonals
• $(34)^{2} = 676+2ab$
• $1156= 676+2ab$
• $480=2ab$
• $ab = 240$
• The area of a rhombus is $\frac{ab}{2} = \frac{240}2} = 120cm^{2}$
• Therefore, the area of the rhombus is $120cm^{2}$

PROJECT CODE: #SPJ1

1. For questions like this 'A square and rhombus have the same area. the square has a side of 9 CM. If one diagonal of the rhombus has a length of 12 CM, find the length of the other diagonal​' please refer to:

https://brainly.in/question/15002897

2. For sums like the above to find out the area of the rhombus please refer to:

https://brainly.in/question/22299679

Answer 2

Answer:

Ares of the rhombus  = 120cm²

Step-by-step explanation:

Given,

Length of a side of rhombus = 13cm

Sum of the length of the diagonals = 34cm

To find,

The area of the rhombus

Recall the concepts

The diagonals of a rhombus bisect each other at right angles

The area of the rhombus = $\frac{1}{2}$ ×d₁×d₂, where d₁, d₂ are diagonals

Pythagoras theorem

The square of the hypotenuse = sum of squares of the other two sides

Solution:

Let 'a' be the side of the rhombus and d₁ and d₂ be the diagonals.

Since the sum of the length of the diagonals = 34

d₁+d₂ = 34 -------------------(1)

Since the diagonals of the rhombus bisect each other at right angles, From the figure we have ΔAOB is a right-angled triangle.

AO + OB = $\frac{d_1 + d_2}{2}$

Then by Pythagoras theorem, we have,

AB² = AO² + OB²

13² = $(\frac{d_1}{2}) ^2 + (\frac{d_2}{2}) ^2$

169 = $\frac{d_1^2}{4} + \frac{d_2^2}{4}$

= $\frac{d_1^2+ d_2^2}{4}$

d₁² + d₂² = 169×4 = 676

d₁² + d₂² = 676 ----------------(2)

We know

(a+b)²  = a² + b² + 2ab

substituting a = d₁ and b = d₂ we get

(d₁+d₂)² =  d₁² + d₂² + 2×d₁×d₂

Substituting the values of d₁+d₂ and d₁×d₂ we get,

34² = 676 + 2×d₁×d₂

1156 = 676 + 2×d₁×d₂

2×d₁×d₂   = 1156 - 676

= 480

d₁×d₂  = 240

Ares of the rhombus =$\frac{1}{2}$×d₁×d₂  =$\frac{1}{2}$×240 = 120

Answer:

Ares of the rhombus  = 120cm²

#SPJ2