Question-100543

Question Number 100543 by mhmd last updated on 27/Jun/20

Commented by mr W last updated on 27/Jun/20

(1)

y^{2} = 4 + 4 - 2 y

\Rightarrow y^{2} + 2 y - 8 = 0

Δ = 2^{2} + 4 \times 8 = 36

A r e a = \frac{Δ \sqrt{Δ}}{6 a^{2}} = \frac{36 \sqrt{36}}{6 \times 1^{2}} = 36

(4)

2 y^{2} + 2 x = 0

\Rightarrow 2 y^{2} + y - 1 = 0

Δ = 1^{2} + 4 \times 1 \times 1 = 5

A r e a = \frac{Δ \sqrt{Δ}}{6 a^{2}} = \frac{5 \sqrt{5}}{6 \times 2^{2}} = \frac{5 \sqrt{5}}{24}

s e e Q 89781

Answered by Rio Michael last updated on 27/Jun/20

1 6

12 2 4 = 4

- 12 - 3 7 = 14

7 1 6 = - 18

\Rightarrow Area of triangle = \frac{1}{2} [4 + 14 - 18 - (12 - 12 + 7)] = \frac{7}{2} square units

Answered by Rio Michael last updated on 27/Jun/20

Curves intersect at the points (- 2, 5), (2.3, 0.7) and (1.25, - 1.50)

\Rightarrow - 2 5

16 2.3 0.7 - 1.4

0.851.25 - 1.5 3.45

3 - 2 5 6.25

A = \frac{1}{2} [- 1.4 + 3.45 + 6.25 - (16 + 0.85 + 3)] \approx 5.78 square units