Question

Area de triangulo 4x+6 de base y 4x+3 de altura , .

Answer 1

Area es (base × altura)/2
$\frac{(4x + 6)(4x + 3)}{2} \\ \frac{(16 {x}^{2} + 12x + 24x + 18)}{2} \\ \frac{16 {x}^{2} + 36x + 18 }{2} \\ \frac{2(8 {x}^{2} + 18x + 9)}{2} \\ 8 {x}^{2} + 18x + 9$

Answer 2

AREA DE UN TRIÁNGULO = (BASE x ALTURA) / 2

AREA = (4x+6)(4x+3)/ 2 = (16x²+12x+24x+18)/2 = 1/2 (16x²+12x+24x+18) =

= 1/2 (16 x² + 36 x +18) = 1/2 (2) (8 x² + 18 x + 9) = 8 x² + 18 x + 9 


A = 8 x² + 18 x + 9