Question

Find the perimeter of a rhombus, if the length of its diagonals are 10 cm and 24 cm.​

Answer 1

Answer:

Diagonals meet at the centre and forms right-angled triangles.

So by using pythagoras theorem

Length of the base = 10/2 = 5cm

Length of the height = 24/2 = 12cm

Hypotenuse2 = side 2+ side2

Hypotenuse2= 52+ 122

Hypotenuse2 = 25 + 144

Hypotenuse2 = 169

On taking square root we get,

Hypotenuse = 13 { 13 X 13=169 => √169=13}

Hence the side of the rhombus is 13cm.

Perimeter of the rhombus = 4×side

= 4 × 13

= 52cm.

Answer 2

BY PYTHAGORAS THEOREM FIRST WE FIND THE SIDE

LET QUADRILATERAL ABCD IS A RHOMBUS THE DIAGONAL INTERSECT AT Y

AC = 10

BD = 24

THEREFORE

AY = 1/2 AB = 5CM

BY= 1/2 BD = 12CM      DIAGONALS OF RHOMBUS BISECTS EACH OTHER AND ARE PERPENDICULAR TO EACH OTHER

IN Δ AYB ,

ANGLE AYB = 90 DEGREE

BY PYTHAGORAS THEOREM

$AY^{2} + BY ^{2} = AB ^{2}$

$5^{2} + 12^{2} = AB^{2} \\25 + 144 = AB^{2} \\169 = AB^{2} \\\sqrt[2]{169} = AB\\AB = 13CM$

PERIMETER OF RHOMBUS = 4 x SIDE

= 4 x 13

= 52cm

HENCE THE PERIMETER IS 52 cm