Maharashtra Board Solutions for Chapter: Pythagoras Theorem, Exercise 3: Problem set 2

In $\mathrm{\Delta ABC}, \angle \mathrm{BAC}=90°$ $\mathrm{seg} \mathrm{BL}$ and $\mathrm{seg} \mathrm{CM}$ are medians of $△\mathrm{ABC}$. Then prove that $4\left({\mathrm{BL}}^2+{\mathrm{CM}}^2\right)=5{\mathrm{BC}}^2$

Sum of the squares of adjacent sides of a parallelogram is $130 \mathrm{sq}.\mathrm{cm}$ and length of one of its diagonals is $14\mathrm{cm}.$ Find the length of the other diagonal.

In $\mathrm{\Delta ABC},\mathrm{AD}\perp \mathrm{BC}$ and $\mathrm{DB}=3\mathrm{CD}.$ Prove that $2{\mathrm{AB}}^2=2{\mathrm{AC}}^2+{\mathrm{BC}}^2$.

In an isosceles triangle, length of the congruent sides is $13 \mathrm{cm}$ and its base is $10 \mathrm{cm}.$ Find the distance between the vertex opposite the base and the centroid in centimetre.

In a trapezium $\mathrm{ABCD}$, $seg\mathrm{AB}\Vert seg\mathrm{DC}$ $seg\mathrm{BD}\perp seg\mathrm{AD}$ $seg\mathrm{AC}\perp seg\mathrm{BC}$. If $\mathrm{AD}=15,\mathrm{BC}=15$ and $\mathrm{AB}=25.$ Find $\mathrm{A}(\square \mathrm{ABCD})$

In the figure, $\mathrm{\Delta }$ $\mathrm{PQR}$ is an equilateral triangle. Point $\mathrm{S}$ is on $seg\mathrm{QR}$ such that $\mathrm{QS}=\frac{1}{3}\mathrm{QR}$. Prove that $: 9{\mathrm{PS}}^2=7{\mathrm{PQ}}^2$. 

Seg $\mathrm{PM}$ is a median of $\mathrm{\Delta PQR}$. If $\mathrm{PQ}=40, \mathrm{PR}=42$ and $\mathrm{PM}=29,$ find $\mathrm{QR}$.

$\mathrm{SegAM}$ is a median of $△\mathrm{ABC}$. If $\mathrm{AB}=22, \mathrm{AC}=34, \mathrm{BC}=24.$ Find $\mathrm{AM}$.