Question

the base of triangle us (-8y square +2y+4) and the height is(-4y square -6y -2) then what is the area
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Answer 1

Base of triangle = -8y² + 2y + 4

Height of triangle = -4y²- 6y - 2

Area of triangle = 1/2 × height × base

$\sf = \dfrac{1}{2} \times ( - 4y {}^{2} - 6y - 2) \times ( - 8y {}^{2} + 2y + 4) \\ \\ \sf \: = \frac{1}{ \cancel{2} } \times \cancel2( - 2y {}^{2} - 3y - 1) \times (-8y {}^{2} + 2y + 4) \\ \\ \sf = ( - 2y {}^{2} )(-8y {}^{2} ) + ( - 2y {}^{2} )(2y) + ( - 2y {}^{2} )(4) \\ \sf + ( - 3y)(-8y {}^{2}) + ( - 3y)(2y) + ( - 3y)(4) \\ \sf+ ( - 1)(-8y {}^{2} ) + ( - 1)(2y) + ( - 1)(4) \\ \\ \sf = 16y {}^{4} - 4y {}^{3} - 8y {}^{2} +24y {}^{3} - 6y {}^{2} - 12y +8y {}^{2} - 2y - 4 \\ \\ \sf \: = 16{y}^{4} + 20{y}^{3} -6{y}^{2} -14y - 4$

∴ Area of traingle is 16y^4 + 20y^3 -6y^2 -14y -4.

Extra information:-

• Area of traingle = 1/2 × height × base, when height and breadth is given.

• Area of triangle by using Heron's Formula is given by calculating average of sum of traingle's sides called S, and formula created is equal to:

$\sf \implies \sqrt{s(s-a)(s-b)(s-c)}$

where, a,b and c are traingle's sides.

Answer 2

Please refer to the attachment !

Hope it helps uh ⭐