Question
In $\triangle\text{ABC},\ \angle\text{ABC}=135^\circ.$ Prove that $AC^2 = AB^2 + BC^2 + 4 \text{ar}(\triangle\text{ABC}).$
✓
Answer
We have the following figure.
Here $\triangle\text{ADB}$ is a right triangle right angled at $D.$ therefore by Pythagoras theorem we have
$AB^2 = AD^2 + DB^2$
Again $\triangle\text{ADC}$ is a right triangle right angled at $D.$
Therefore, by Pythagoras theorem, we have
$AC^2 = AD^2 + DC^2$
$AC^2 = AD^2 + (DB + BC)^2$
$AC^2 = AD^2 + DB^2 + BC^2 + 2 \times BC \times BD$
Since angle ABD is $45^\circ$ and therefore angle $BAD$ is also $45^\circ .$
Hence $AB = DB$
So,
$AC^2 = AD^2 + DB^2 + BC^2 + 2BC \times AD$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 2\times2\times\frac{1}{2}\text{BC}\times\text{AD}$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Since $AB^2 = AD^2 + DB^2$
So, $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Hence we have proved that $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Here $\triangle\text{ADB}$ is a right triangle right angled at $D.$ therefore by Pythagoras theorem we have
$AB^2 = AD^2 + DB^2$
Again $\triangle\text{ADC}$ is a right triangle right angled at $D.$
Therefore, by Pythagoras theorem, we have
$AC^2 = AD^2 + DC^2$
$AC^2 = AD^2 + (DB + BC)^2$
$AC^2 = AD^2 + DB^2 + BC^2 + 2 \times BC \times BD$
Since angle ABD is $45^\circ$ and therefore angle $BAD$ is also $45^\circ .$
Hence $AB = DB$
So,
$AC^2 = AD^2 + DB^2 + BC^2 + 2BC \times AD$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 2\times2\times\frac{1}{2}\text{BC}\times\text{AD}$
$= \text{AD}^2 + \text{DB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Since $AB^2 = AD^2 + DB^2$
So, $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
Hence we have proved that $\text{AC}^2 = \text{AB}^2 + \text{BC}^2 + 4\text{ar}(\triangle\text{ABC})$
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Marks
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$10-20$
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$20-30$
|
$30-40$
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$40-50$
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$50-60$
|
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Frequency
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$12$
|
$35$
|
$45$
|
$25$
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$13$
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