Pakistan
Find the area of the rhombus in which each side is ^@ 25 \space cm^@ long and one of whose diagonals is ^@14 \space cm^@.
Answer:
^@336 \space cm^2^@
- Let ^@ABCD^@ be the given rhombus with
^@AB = 25 \space cm ^@ and ^@ AC = 14 \space cm ^@.
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 } \angle AOB = 90^\circ @^ - Using Pythagous' theorem in right ^@\triangle AOB^@, we have @^ \begin{aligned} & AB^2 = AO^2 + BO^2 \\ \implies & (25)^2 = (7)^2 + BO^2 \\ \implies & 625 = 49 + BO^2 \\ \implies & 576 = BO^2 \\ \implies & 24 \space cm = BO \\ \end{aligned} @^ As, ^@ BO = \dfrac { 1 } { 2 } BD^@, we have @^ \begin{aligned} & BO = \dfrac { 1 } { 2 } BD \\ \implies & 24 = \dfrac { 1 } { 2 } BD \\ \implies & 2 \times 24 = BD \\ \implies & 48 \space cm = BD \\ \end{aligned} @^
- We know, @^ \begin{aligned} \text { Area of rhombus } & = \dfrac { 1 } { 2 } \times \text { Product of its diagonals } \\ & = \dfrac { 1 } { 2 } \times AC \times BD \\ & = \dfrac { 1 } { 2 } \times 14 \times 48 \space cm \\ & = 336 \space cm^2 \end{aligned} @^ Thus, the area of the rhombus is ^@336 \space cm^2^@.
