Question

ABCD is a right angle triangle with angle B = 90 degree . M is the midpoint of AC and BM is √117cm sum of the lenghts of sides AB and BC is 30 cm find the area of the triangle​

Answer 1

The area of the triangle is 108 cm²

Step-by-step explanation:

We know that

If ABC is a right angle triangle then the mid-point of the hypotenuse is the circumcentre of the triangle

As shown in the figure, the point M will be circumcentre of the $\Delta ABC$       (Note: the circumcircle is shown with dotted circle)

$\therefore MA=MC=MB$     (Radius of the circumcircle)

But given that $BM=\sqrt{117}$

$\therefore MA=MC=\sqrt{117}$

$\therefore AC=MA+MC=2\sqrt{117}$

Also given that

$AB+BC=30$

$\implies BC=30-AB$

Let $AB=x$

Now in $\Delta ABC$ by Pythagoras Theorem

$AB^2+BC^2=AC^2$

$\implies x^2+(30-x)^2=(2\sqrt{117})^2$

$\implies x^2+x^2-60x+900=468$

$\implies 2x^2-60x+432=0$

$\implies x^2-30x+216=0$

$\implies x^2-18x-12x+216=0$

$\implies x(x-18)-12(x-18)=0$

$\implies (x-18)(x-12)=0$

$\implies x=18,12$

or, $AB=18,12$

If we take $AB=18$ cm then side $BC$ will be $12$ cm

And if we take $AB=12$ cm then side $BC$ will be $18$ cm

So basically they will be the same triangle with sides interchanged

The area of this triangle is

$\Delta=\frac{1}{2}\times AB\times BC$

$\implies \Delta=\frac{1}{2}\times 18\times 12$

$\implies \Delta=108$ cm²

Hope this answer is helpful.

Know More:

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