1. **State the problem:** We have a triangle and a rectangle with given dimensions and areas. The triangle's height is $h$ and base is $x$. The rectangle's height is $(h+2)$ and length is $(x+1)$. The triangle's area is 11 cm² and the rectangle's area is 39 cm². 2. **(a) Expression for the height of the rectangle:** The height of the rectangle is given as $h + 2$. 3. **(b) Show that $2x^2 - 15x + 22 = 0$:** - Area of triangle: $\frac{1}{2} \times x \times h = 11$ so $xh = 22$. - Area of rectangle: $(x + 1)(h + 2) = 39$. - Substitute $h = \frac{22}{x}$ into rectangle area: $$ (x + 1) \left( \frac{22}{x} + 2 \right) = 39 $$ - Simplify inside parentheses: $$ \frac{22}{x} + 2 = \frac{22 + 2x}{x} $$ - So: $$ (x + 1) \times \frac{22 + 2x}{x} = 39 $$ - Multiply both sides by $x$: $$ (x + 1)(22 + 2x) = 39x $$ - Expand left side: $$ (x)(22) + (x)(2x) + (1)(22) + (1)(2x) = 39x $$ $$ 22x + 2x^2 + 22 + 2x = 39x $$ - Combine like terms: $$ 2x^2 + 24x + 22 = 39x $$ - Bring all terms to one side: $$ 2x^2 + 24x + 22 - 39x = 0 $$ $$ 2x^2 - 15x + 22 = 0 $$ 4. **(c) Factorise and solve $2x^2 - 15x + 22 = 0$:** - Multiply $2 \times 22 = 44$. - Find two numbers that multiply to 44 and add to -15: -11 and -4. - Rewrite middle term: $$ 2x^2 - 11x - 4x + 22 = 0 $$ - Factor by grouping: $$ x(2x - 11) - 2(2x - 11) = 0 $$ - Factor out common binomial: $$ (2x - 11)(x - 2) = 0 $$ - Set each factor to zero: $$ 2x - 11 = 0 \Rightarrow x = \frac{11}{2} = 5.5 $$ $$ x - 2 = 0 \Rightarrow x = 2 $$ - Find corresponding heights $h$ using $h = \frac{22}{x}$: - For $x = 5.5$, $h = \frac{22}{5.5} = 4$ cm. - For $x = 2$, $h = \frac{22}{2} = 11$ cm. **Final answer:** The two possible heights of the triangle are 4 cm and 11 cm.