1. Let the length be $x$ meters, so the width is $x - 6$ meters. The area is $x(x - 6) = 112$. Solving $x^2 - 6x - 112 = 0$ gives $x = 14$ (discarding the negative root). So the dimensions are 14 m by 8 m.
2. Let the width be $x$ feet, so the length is $2x + 1$ feet. The area is $x(2x + 1) = 300$. Solving $2x^2 + x - 300 = 0$ gives $x = 12$ (discarding the negative root). So the dimensions are 25 ft by 12 ft.
3. The dimensions are $x + 6$ and $x - 2$, so the area is $(x + 6)(x - 2) = 65$. Solving $x^2 + 4x - 77 = 0$ gives $x = 7$ (discarding the negative root). So the dimensions are 13 in by 5 in.
4. Let the height be $x$ cm, so the base is $2x - 7$ cm. The area is $\frac{1}{2}x(2x - 7) = 102$. Solving $x^2 - \frac{7}{2}x - 102 = 0$ or $2x^2 - 7x - 204 = 0$ gives $x = 12$ (discarding the negative root). So the base is $2(12) - 7 = 17$ cm.
5. Let the shorter leg be $x$ feet, so the hypotenuse is $2x + 1$ feet and the longer leg is $x + 7$ feet. By the Pythagorean theorem: $x^2 + (x + 7)^2 = (2x + 1)^2$. Solving $x^2 + x^2 + 14x + 49 = 4x^2 + 4x + 1$ simplifies to $2x^2 - 10x - 48 = 0$ or $x^2 - 5x - 24 = 0$. This gives $x = 8$ (discarding the negative root). So the triangle has legs 8 ft and 15 ft, and hypotenuse 17 ft.
6. The volume is height × length × width = $3(x - 1)(x + 4) = 198$. So $(x - 1)(x + 4) = 66$. Solving $x^2 + 3x - 4 = 66$ or $x^2 + 3x - 70 = 0$ gives $x = 7$ (discarding the negative root).
Parent Tip: Review the logic above to help your child master the concept of quadratic applications worksheet.
