Question

If ABC is an isosceles triangle with AB=AC = a and BC = b, then its area is

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Answer 1

Answer:

if we draw a perpendicular from A on the base BC then it bisects BC on the point D

then BD=b/2

ABD is a right angle triangle.

height AD^2=AB^2-BD^2

=a^2-b^2/4

=(4a^2-b^2)/4

then AD=1/2√(4a^2-b^2)

area of the triangle is

1/2 .b.1/2.√(4a^2-b^2)

=b/4√(4a^2-b^2) sq unit

Answer 2

Given,

$\[\begin{array}{l}AB = AC = a\\BC = b\end{array}\]$

Solution,

See the figure 1.

Calculate the length of AD, is perpendicular bisector from vertex A.

$\[AD = \sqrt {A{B^2} - {{\left( {\frac{{BD}}{2}} \right)}^2}} \]$

$\[ \Rightarrow AD = \sqrt {{a^2} - {{\left( {\frac{b}{2}} \right)}^2}} \]$

$\[ \Rightarrow AD = \frac{1}{2}\sqrt {4{a^2} - {b^2}} \]\left$

Calculate the area of the triangle,

$\[ = \frac{1}{2} \times BC \times AD\]$

$\[ = \frac{1}{2} \times b \times \frac{1}{2}\sqrt {4{a^2} - {b^2}} \]$

$\[ = \frac{1}{4}\left( {b\sqrt {4{a^2} - {b^2}} } \right)\]$

Hence the area of triangle is $\[\frac{1}{4}\left( {b\sqrt {4{a^2} - {b^2}} } \right)\]$.