Ilango Arasiyal

 rhombus 

Question 4: Calculate the area of the rhombus having side equal to 17cm$17\mathrm{c}\mathrm{m}$ and one of its diagonals is 16cm.$16\mathrm{c}\mathrm{m}.$
Answer:
ABCD$ABCD$ is a rhombus,
So , AB=BC=CD=DA=17cm$AB=BC=CD=DA=17\text{cm}$
AC=16cm$AC=16\text{cm}$
Let O$O$ be the intersection point of diagonals.
So, AO=8cm$AO=8\text{cm}$
In ΔDOA$\mathrm{\Delta }DOA$
AD2=AO2+OD2$A{D}^2=A{O}^2+O{D}^2$
⇒172=82+OD2$\Rightarrow {17}^2=8^2+O{D}^2$
⇒289=64+OD2$\Rightarrow 289=64+O{D}^2$
⇒225=OD2$\Rightarrow 225=O{D}^2$
⇒OD=15$\Rightarrow OD=15$
Therefore, BD=2×OD$BD=2\times OD$
=2×15$=2\times 15$
=30cm$=30\mathrm{c}\mathrm{m}$
Now, area of rhombus =12×d1×d2$=\frac{1}{2}\times d_1\times d_2$
=12×16×30cm2$=\frac{1}{2}\times 16\times 30\mathrm{c}{\mathrm{m}}^2$
=240cm2$=240\mathrm{c}{\mathrm{m}}^2$

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