Question

In triangle ABC ab=bc=10cms,ac=12cm find its area

Answer 1

In ∆ABC AB=BC=10cm,AC=12cm
Area of ∆=√S(S-A)(S-B)(S-C). {Heron's formula}
Let AB=A,BC=B,AC=C
Area of the ∆ ABC=√S(S-A)(S-B)(S-C)
Semi-perimeter=A+B+C/2=10+10+12/2=32/2=16cm
Area of the ∆ ABC=√16(16-10)(16-10)(16-12)
=√16×6×6×4=√2304=48 sq.cm
Thus,area of the triangle ABC is 48sq.cm

Answer 2

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Drop a perpendicular from A to side BC and mark it as D

$= > AD= \sqrt{ {x}^{2} - {5}^{2} }$

$= > area= \frac{1}{2} ×AD×BC$

$= > 60 = \frac{1}{2} \times \sqrt{ {x}^{2} - 25 } \times 10$

$= > 144 = {x}^{2} - 25$

${x}^{2} = 169$

$x = 13cm$

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