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In the figure given below, AB = BC = 2√2 units and AC = 4

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by swerve » Thu Aug 30, 2018 9:57 am

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$$\text{In the figure given below, } AB = BC = 2\sqrt{2} \text{ units and } AC = 4 \text{ units.}$$ $$\text{If } BD \text{ bisects the side } AC\text{, find the length of }BD.$$

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$$\text{A. } 1$$
$$\text{B. } \sqrt{2}$$
$$\text{C. }2$$
$$\text{D. } 2\sqrt{2}$$
$$\text{E. } 3$$

The OA is C

Source: e-GMAT
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Source: — Problem Solving |

swerve wrote:$$\text{In the figure given below, } AB = BC = 2\sqrt{2} \text{ units and } AC = 4 \text{ units.}$$ $$\text{If } BD \text{ bisects the side } AC\text{, find the length of }BD.$$

Image

$$\text{A. } 1 \, \, \, \text{B. } \sqrt{2} \, \, \, \text{C. }2 \, \, \, \text{D. } 2\sqrt{2} \, \, \, \text{E. } 3$$
\[? = BD\]
\[{\left( {AC} \right)^2} = {\left( {AB} \right)^2} + {\left( {BC} \right)^2}\,\,\,\, \Rightarrow \,\,\,\,\Delta ABC\,\,right\,\,,\,\,\,\angle ABC = {90^ \circ }\]
\[\frac{{AC \cdot BD}}{2} = {S_{\,\Delta \,ABC\,}} = \frac{{AB \cdot BC}}{2}\,\,\,\,\, \Rightarrow \,\,\,\,\,\frac{{4 \cdot ?}}{{}} = \frac{{2\sqrt 2 \cdot 2\sqrt 2 }}{{}}\,\,\,\,\,\, \Rightarrow \,\,\,\,? = 2\]

Obs.: the reason that guarantees BD perpendicular to AC is symmetry or, if you prefer, the fact that AC is a base of an isosceles triangle, therefore relative to the base, the median (given) coincides with the height.

The above follows the notations and rationale taught in the GMATH method.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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