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In ∆ABC, angle abc = 45°.prove …

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In ∆ABC, angle abc = 45°.prove that AC² = AB² + BC² - 4ar(∆abc).

Posted by Sandeep Krishnaa 6 years, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Gaurav Seth 6 years, 11 months ago

Angle ABC measures 45° (given). Hence angle OAB will also be 45°
As a result x = h

Now the area of triangle ABC = 1/2(BC)(h) = 1/2(BC)(x) ; since h = x,
Or we can write BC = 2(Area of Triangle ABC)/x ----------------(i)

We can also write AC² = h² + y²
But y is nothing but (BC - x)

So we can write AC² = h² + (BC - x)²

Simplifying this we get,
AC² = h² + BC² + x² - 2.x.BC
Now x² + h² = AB² (by Pythagoras theorem for triangle AOB)

Hence AC² = AB² + BC² -2(x)(BC)
Substituting the value of BC obtained from (i), we get
AC² = AB² + BC² - 2(x)(2Area of Triangle ABC)/x

Simplifying this we get:
AC² = AB² + BC² -4.Area of Triangle ABC

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