Question

The diagonal of a rhombus is in the ratio 5: 12. If the perimeter is 104cm, find the lengths of the side and the diagonals.
(a)Side – 26cm, diagonals- 20cm, 48cm
(b)Side – 26cm, diagonals- 40cm, 96cm
(c)Side- 25cm, diagonals – 20cm, 48cm
(d)None of those

Answer 1

Hint – In this question use the concept that the perimeter of a rhombus will be the sum of all its sides, that is AB+ BC+ CD + AD. Assume any one diagonal to be a variable and use the property that diagonals bisect each other and are perpendicular to each other along with the given ratio to get the answer.

Complete step by step answer: -
Let ABCD be a rhombus and AC and BD are the diagonals of the rhombus.
As we know that in the rhombus all sides are equal.
Therefore AB = BC = CD = DA............................ (1)
Now as we know that the perimeter (S) of any shape is the sum of all sides.
Therefore S = AB + BC + CD + DA
Now from equation (1)
$\Rightarrow S = AB + AB + AB + AB = 4AB$
Now it is given that the perimeter (S) of the rhombus is 104 cm.
$\Rightarrow S = 104 = 4AB$
$\Rightarrow AB = \dfrac{{104}}{4} = 26$ cm.
So the side of the rhombus is 26 cm.
Now it is given that the diagonals of the rhombus are in the ratio ( 5 : 12).
Let first diagonal = 5x
And second diagonal = 12x, (where x is any real value).
Therefore AC = 12x................... (2)
And BD = 5x........................... (3)
Now we know that the diagonal of a rhombus bisect each other and are perpendicular to each other as shown in figure.
Therefore OA = OC and OB = OD.
Therefore $OA = \dfrac{{AC}}{2}$ and $OB = \dfrac{{BD}}{2}$
Now from equation (2) and (3) we have,
$\Rightarrow OA = \dfrac{{12x}}{2} = 6x$ And $OB = \dfrac{{5x}}{2}$
Now in triangle AOB apply Pythagoras theorem we have,
$\Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
$\Rightarrow {\left( {AB} \right)^2} = {\left( {OA} \right)^2} + {\left( {OB} \right)^2}$
Now substitute the values in this equation we have,
$\Rightarrow {\left( {26} \right)^2} = {\left( {6x} \right)^2} + {\left( {\dfrac{{5x}}{2}} \right)^2}$
$\Rightarrow 676 = 36{x^2} + \dfrac{{25{x^2}}}{4} = \dfrac{{169{x^2}}}{4}$
$\Rightarrow {x^2} = \dfrac{{676 \times 4}}{{169}} = 4 \times 4 = {4^2}$
$\Rightarrow x = 4$
Now from (2) and (3)
AC = $\left( {12 \times 4} \right) = 48$ cm
And BD = $\left( {5 \times 4} \right) = 20$ cm.
So the side of the rhombus is 26 cm and the diagonals are 20 and 48 cm respectively.
Hence option (A) is correct.

Note – Tricky part here was the application of Pythagoras theorem, we were able to use it because triangle AOB is a right angled triangle, due to the property of diagonals being perpendicular. Pythagoras theorem is applicable to only right angled triangles. Some other properties of rhombus that are very useful while solving problems of this kind are, adjacent angles of rhombus are supplementary, all the opposite angles are congruent.

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