R. D. Sharma Solutions for Chapter: Triangles, Exercise 7: EXERCISE 7.7

In Figure, $\mathrm{\angle }\mathrm{B}<90°$ and segment $\mathrm{A}\mathrm{D}\perp \mathrm{B}\mathrm{C},$ show that ${\mathrm{b}}^2={\mathrm{a}}^2+{\mathrm{c}}^2-2\mathrm{ax}$

In figure, $D$ is the mid-point of side $BC$ and $\mathrm{A}\mathrm{E}\perp \mathrm{B}\mathrm{C}.$ If $\mathrm{B}\mathrm{C}=\mathrm{a}, \mathrm{A}\mathrm{C}=\mathrm{b}, \mathrm{A}\mathrm{B}=\mathrm{c}, \mathrm{E}\mathrm{D}=\mathrm{x}, \mathrm{A}\mathrm{D}=\mathrm{p}$ and $\mathrm{A}\mathrm{E}=\mathrm{h},$ Prove that ${\mathrm{c}}^2={\mathrm{p}}^2-\mathrm{ax}+\frac{{\mathrm{a}}^2}{4}$.

In the figure below, $D$ is the mid-point of side $BC$ and $\mathrm{A}\mathrm{E}\perp \mathrm{B}\mathrm{C}.$ If $\mathrm{B}\mathrm{C}=\mathrm{a}, \mathrm{A}\mathrm{C}=\mathrm{b}, \mathrm{A}\mathrm{B}=\mathrm{c}, \mathrm{E}\mathrm{D}=\mathrm{x}, \mathrm{A}\mathrm{D}=\mathrm{p}$ and $\mathrm{A}\mathrm{E}=\mathrm{h},$ prove that ${\mathrm{b}}^2+{\mathrm{c}}^2=2{\mathrm{p}}^2+\frac{{\mathrm{a}}^2}{2}$.

In $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}, \mathrm{\angle }\mathrm{A}$ is obtuse, $\mathrm{P}\mathrm{B}\perp \mathrm{A}\mathrm{C}$ and $\mathrm{Q}\mathrm{C}\perp \mathrm{A}\mathrm{B}.$ Prove that: $\mathrm{A}\mathrm{B}\times \mathrm{A}\mathrm{Q}=\mathrm{A}\mathrm{C}\times \mathrm{A}\mathrm{P}$.

In $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C},\mathrm{\angle }\mathrm{A}$ is obtuse, $\mathrm{P}\mathrm{B}\perp \mathrm{A}\mathrm{C}$ and $\mathrm{Q}\mathrm{C}\perp \mathrm{A}\mathrm{B}.$ Prove that: $\mathrm{B}{\mathrm{C}}^2=(\mathrm{A}\mathrm{C}\times \mathrm{C}\mathrm{P}+\mathrm{A}\mathrm{B}\times \mathrm{B}\mathrm{Q})$

In a right $\mathrm{\Delta }\mathrm{A}\mathrm{B}\mathrm{C}$ right-angled at $\mathrm{C}$, if $\mathrm{D}$ is the mid-point of $\mathrm{BC}$, prove that $\mathrm{B}{\mathrm{C}}^2=4(\mathrm{A}{\mathrm{D}}^2-\mathrm{A}{\mathrm{C}}^2).$

In a quadrilateral $\mathrm{ABCD}$, $\mathrm{\angle }\mathrm{B}=90°, \mathrm{A}{\mathrm{D}}^2=\mathrm{A}{\mathrm{B}}^2+\mathrm{B}{\mathrm{C}}^2+\mathrm{C}{\mathrm{D}}^2$. Prove that $\mathrm{\angle }\mathrm{A}\mathrm{C}\mathrm{D}=90°$.

An aeroplane leaves an airport and flies due North at a speed of $1000 \mathrm{km}/\mathrm{hr}$. At the same time, another aeroplane leaves the same airport and flies due West at a speed of $1200 \mathrm{km}/\mathrm{hr}$. How far apart will be the two planes after $1\frac{1}{2} \mathrm{hours}$?