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coldgirl [10]
4 years ago
15

Given EAB and DCB are two right triangles. The figurebhas BED=BDE. point b is the midpoint of segment AC. prove EAB= DCB.

Mathematics
1 answer:
Ronch [10]4 years ago
3 0

Answer:

EAB= DCB

Step-by-step explanation:

Since EAB and DCB are two right triangles then the figure ACDE is rectangular. since the total sum of rectangular 360° there is no options except to be rectangular. If the figure ACDE is rectangular then the two sides AE & CD are equal. Since Point b is midway point the two sides are also equal.

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Answer:

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Step-by-step explanation:

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step - 1 :  \: define

- 4 {t}^{2}  + 12t + 100 = 0

step - 2 :  \\ divide \: both \: sides \: by \:  - 4

{t}^{2}  - 3t - 25 = 0

step - 3 :  \\ solve \: the \: quadratic

formula :  \\ x =  \frac{ - b± \sqrt{ {b}^{2} - 4ac }  }{2a}

define \: a ,b \: and \: c \\ which \: are \: 1, - 3 \: and \:  - 25 \: respectively

t =  \frac{ - ( - 3)± \sqrt{ {( - 3)}^{2} - 4.1. - 25 }  }{2.1}

t =  \frac{  3± \sqrt{ 9    + 100}  }{2}

t = \frac{3 +  \sqrt{ 109 } }{2}

t =  \frac{3  -  \sqrt{ 109 } }{2}

t = 6.7 \\ t =  - 3.7

as \: we \: know \: time \: cannot \: be \: negative

\huge \therefore \: t = 6.7

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