Question

in angle abc ab=ac=x bc=10cm and the area of angleabcis60cm2 find x​

Answer 1

Step-by-step explanation:

in triangle abc, if ab=ac then triangle is isosceles triangle.

draw a line perpendicular to the bc.

area of triangle = (1/2)×b×h

60 = 1/2×10×h

(60×2)/10 = h

h = 12 cm

in triangle adb, ab^2= ad^2+bd^2

x^2 = 12^2+5^2

x^2 = 144+25

x^2 = 169

x = 13cm

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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