Question

$\Delta ABC$ is a right angled triangle, where $\angle B={90}^{\circ }$. $CDandAE$ are medians. If $AE=xandCD=y$ then, correct statement is

• A. $x^2+y^2=A{C}^2$
• B. $x^2+y^2=2A{C}^2$
• C. $x^2+y^2=\frac{3}{2}A{C}^2$
• D. $x^2+y^2=\frac{5}{4}A{C}^2$

Answer 1

The correct option is D $x^2+y^2=\frac{5}{4}A{C}^2$

Consider $\Delta ABE,{AE}^2=A{B}^2+{BE}^2⟹{AB}^2=A{E}^2-{BE}^2$
Consider $\Delta DBC,{DC}^2=D{B}^2+{BC}^2⟹{BC}^2=D{C}^2-{DB}^2$
We know that BE = $\frac{1}{2}BC$ and DB = .$\frac{1}{2}AB$.
Consider $\Delta ABC,{AC}^2=A{B}^2+{BC}^2{AC}^2=A{E}^2-B{E}^2+{DC}^2-B{C}^2{AC}^2=A{E}^2-\frac{1}{4}B{C}^2+{DC}^2-\frac{1}{4}A{B}^2{AC}^2=A{E}^2+{DC}^2-\frac{1}{4}(B{C}^2+A{B}^2){AC}^2=A{E}^2+{DC}^2-\frac{1}{4}\left(A{C}^2\right)⟹\frac{5}{4}\left(A{C}^2\right)=A{E}^2+{DC}^2$
Substituting AE and DC with x and y respectively, we will arrive at option A