If the area of the rhombus is ^@336 \space cm^2^@ and one of its diagonals is ^@14\space cm,^@ find its perimeter.


  1. Home
  2. Math
  3. All Questions
Answer:

^@100 \space cm ^@

Step by Step Explanation:
  1. Let ^@ABCD^@ be the given rhombus with ^@ AC = 14 \space cm ^@ and BD = ^@ x \space cm^@.



    We know, @^ \begin{aligned} \text { Area of rhombus } & = \dfrac { 1 } { 2 } \times \text { Product of its diagonals } \\ \implies & 336 = \dfrac { 1 } { 2 } \times AC \times BD \\ \implies & 336 = \dfrac { 1 } { 2 } \times 14 \times x \\ \implies & \dfrac { 2 \times 336 } { 14 } = x \\ \implies & 48 = x \\ \end{aligned} @^ Thus, the length of diagonal ^@ BD = 48 \space cm. ^@
  2. Let the diagonals ^@AC^@ and ^@BD^@ bisect at a point ^@O^@.
    We know that the diagonals of a rhombus bisect each other at right angles.
    So, ^@ AO = \dfrac { 1 } { 2 } AC ^@ and ^@ BO = \dfrac { 1 } { 2 } BD. ^@ @^ \therefore AO = \dfrac { 1 } { 2 } \times 14 = 7 \space cm \text { and } BO = \dfrac { 1 } { 2 } \times 48 = 24 \space cm.@^ Also, ^@\angle AOB = 90^\circ. ^@
  3. Using Pythagous' theorem in right ^@\triangle AOB^@, we have @^ \begin{aligned} & AB^2 = AO^2 + BO^2 \\ \implies & AB^2 = (7)^2 + (24)^2 \\ \implies & AB^2 = 49 + 576 \\ \implies & AB^2 = 625 \\ \implies & AB = 25 \space cm \\ \end{aligned} @^
  4. Now, @^ \begin{aligned} \text{ Perimeter of rhombus } & = 4 \times \text { Length of side } \\ & = 4 \times AB \\ & = 4 \times 25 \\ & = 100 \space cm \end{aligned} @^ Thus, the perimeter of the rhombus is ^@100 \space cm. ^@

You can reuse this answer
Creative Commons License