Question

The area of a triangle is 30cm2.find the base if the aptitude exceeds the base by 7cm.

Answer 1

Let base = x cm
Altitude = (x + 7) cm


We know,


$area \: = \frac{1}{2} \times x(x + 7) \\ \\ 30 \times 2 = x(x + 7)$

60 = x² + 7x

0 = x² + 7x - 60

0 = x² + 12x - 5x - 60

0 = x(x + 12) -5(x + 12)

0 = (x + 12)(x - 5)

x = 5 or -12

Taking positive value, x = 5


Base = x = 5 cm
Altitude = x + 7 = 5 + 7 = 12 cm




I hope this will help you



(-;

Answer 2

$Heya$‼


Given that, area of ∆ = 30 cm²

Base of ∆ = x cm
And altitude = (x + 7) cm

As we know,
Area of ∆ = $1/2*base*altitude$
$30\:cm^2=1/2*x*(x+7)$
$30*2=x^2+7x$
$60=x^2+7x$
$x^2+7x-60=0$
$x^2-5x+12x-60=0$
$x(x-5)+12(x-5)=0$
$(x+12)(x-5)=0$
$x=-12\:or\:x=5$


So, the base of ∆ isn't be -ve.
Thus, the base of ∆ is 5 cm i.e., +ve.
And the altitude is x+7 i.e., 5+7 = 12 cm