Answer:
In \[\Delta ABD,\angle D=90{}^\circ \](Median of isosceles triangle) Hence, \[\Delta \]ABC is right angled triangle \[\therefore \] \[A{{B}^{2}}=B{{D}^{2}}+A{{D}^{2}}\] \[{{\left( 37 \right)}^{2}}={{\left( 12 \right)}^{2}}+{{\left( \frac{x}{2} \right)}^{2}}\] \[1369=144+{{\left( \frac{x}{2} \right)}^{2}}\] \[\frac{{{x}^{2}}}{4}\]=1369-144=1225 \[{{x}^{2}}=\text{ }4\times 1225\] \[x\text{ }=\sqrt{4\times l225}\] \[x=2\sqrt{5\times 5\times 7\times 7}\] \[=2\times 5\times 7=70\]
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