Embibe Experts Solutions for Chapter: Triangles, Exercise 1: Goa Board-2018

$\Delta ABC~\Delta PQR$ and $ar\left(\Delta ABC\right)=144{ \mathrm{cm}}^2,$ $ar\left(\Delta \mathrm{PQR}\right)=81{ \mathrm{cm}}^2$. If $\mathrm{QR}=27 \mathrm{cm}$ then $BC=$

With reference to the given figure and the given conditions write only the proof with reason of the following theorem.

Given: In $\Delta \mathrm{DEF},{\mathrm{DE}}^2+{\mathrm{EF}}^2={\mathrm{DF}}^2$

$\Delta XYZ$ is constructed such that : $\mathrm{XY}=\mathrm{DE},\mathrm{YZ}=\mathrm{EF}\text{ and }\angle \mathrm{Y}=90°$

Prove that: $∆\mathrm{DEF}$ is a right angled triangle.

Given : $D$ is any point on side $BC$ of $△ABC$. $\mathrm{DE}\Vert \mathrm{AB}\text{ and }\mathrm{DF}\Vert \mathrm{AC}\text{. }$$A-F-B$ and $A-E-C$. $EF$ and $CB$ meet at $H$ when produced as shown in the fig. Prove that: ${\mathrm{HD}}^2=\mathrm{HB}\times \mathrm{HC}$

$\Delta \mathrm{ABC}~\Delta \mathrm{DEF}$ and their areas are $64{ \mathrm{cm}}^2$ and $49{ \mathrm{cm}}^2$ respectively. If $\mathrm{AB}=5.6 \mathrm{cm}$ then $\mathrm{DE}=$

Given : In $∆\mathrm{ABC},\mathrm{DE}\Vert \mathrm{BC}$ where the points $D$ and $E$ lie on $AB$ and $AC$ respectively, $\mathrm{EM}\perp \mathrm{AB}$ and $\mathrm{DN}\perp \mathrm{AC}:$

Prove that : $\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

In the adjoining figure, smaller diagonal $BD$ of a parallelogram $ABCD$ is perpendicular to the sides $AB$ and $CD$. $P$ is the point of intersection of the two diagonals. Prove that : $3{\mathrm{AB}}^2={\mathrm{AC}}^2-{\mathrm{BC}}^2$