V K Raman and S N Sharma Solutions for Exercise 3: EXERCISE 24(C)

The side $\mathrm{AB}$ of $∆\mathrm{ABC}$ is produced on either side to $\mathrm{D}, \mathrm{E}$. Given that$\angle \mathrm{CBE}=148°$ and $\angle \mathrm{CAD}= 85°$, show that $\mathrm{BC}>\mathrm{AC}$ .

In$∆\mathrm{ABC}, \mathrm{AC} = 7 \mathrm{cm}$ and $\mathrm{BC}=8 \mathrm{cm}$. Show that $\mathrm{AB}<15 \mathrm{cm}$ and $\mathrm{AB}>1 \mathrm{cm}$ .

In $∆\mathrm{ABC}$, side $\mathrm{BA}$ is produced to $\mathrm{D}$and $\mathrm{CD}$ is joined. Given that $\mathrm{BA}= 5 \mathrm{cm}, \mathrm{AD}= 4 \mathrm{cm}$ and $\mathrm{BC}= 3.8 \mathrm{cm}$ show that$\mathrm{AC}<8.8 \mathrm{cm}$ and $\mathrm{CD}<12.8 \mathrm{cm}$.

In the quadrilateral $\mathrm{ABCD}$ show that

$\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>2 \mathrm{AC}$

In the quadrilateral $\mathrm{ABCD}$ show that

 $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>2\mathrm{BD}$

In the quadrilateral $\mathrm{ABCD}$, show that  $\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}>\mathrm{AC}+\mathrm{BD}$.

In $∆\mathrm{ABC}$, side $\mathrm{BA}$ is produced to any point $\mathrm{D}$. Given that $\angle \mathrm{ABC}=50°$ and $\angle \mathrm{DAC}=87°$. Show that $\mathrm{BC}>\mathrm{AC}$.