Question

In $\Delta ABC,\angle B=45$ and $\overline{AM}\perp \overline{BC}\cdot$ If B C = 7 , A M = 4 then A C =

• A. 4
• B. 2.5
• C. 5
• D. 5.31

Answer 1

The correct option is D 5.31

$\Rightarrow$ In $△ABC,\angle Beta=45$

$AM\perp BC$

$BC=7,AM=4$

$AM\perp BC$

$\Rightarrow BM=\frac{1}{2}\times BC$

$\Rightarrow BM=\frac{1}{2}\times 7$

$\Rightarrow BM=3.5$

In $△AMB,$ by using Pythagoras thrm

$\Rightarrow A{B}^2=A{M}^2+B{M}^2$

$\Rightarrow A{B}^2=16+12.25$

$\Rightarrow A{B}^2=28.25$

$AB=5.31$

By using perpendicular bisector theorem

$\overline{AM}\perp \overline{BC}$ then $BM=MC$ & $AB=AC$

$\therefore AC=5.31$