1. **State the problem:** We have trapezium $ABCD$ with $AD \parallel BC$ and $\angle ADC = 90^\circ$. $M$ is midpoint of $AB$, $CM = \frac{13}{2}$ cm, and $BC + CD + DA = 17$ cm. Find the area of trapezium $ABCD$. 2. **Set a coordinate system:** Place $D$ at origin $(0,0)$. Since $\angle ADC = 90^\circ$ and $AD \parallel BC$, assume $AD$ along x-axis and $DC$ along y-axis. - Let $AD = a$ along x-axis, so $A = (a,0)$. - Let $DC = d$ along y-axis, so $C=(0,d)$. 3. **Coordinates of points:** - $D = (0,0)$ - $A = (a,0)$ - $C = (0,d)$ - Since $AD \parallel BC$, $BC$ is parallel to $x$-axis, so $B$ has same y-coordinate as $C$: $B = (x_B,d)$. 4. **Midpoint $M$ of $AB$: ** $M = \left( \frac{a + x_B}{2}, \frac{0 + d}{2} \right) = \left( \frac{a + x_B}{2}, \frac{d}{2} \right)$. 5. **Length $CM = \frac{13}{2}$:** Calculate distance $CM$: $$CM = \sqrt{\left(0 - \frac{a + x_B}{2}\right)^2 + \left(d - \frac{d}{2}\right)^2} = \sqrt{\left(\frac{a + x_B}{2}\right)^2 + \left(\frac{d}{2}\right)^2} = \frac{13}{2}.$$ Square both sides: $$\left(\frac{a + x_B}{2}\right)^2 + \left(\frac{d}{2}\right)^2 = \left(\frac{13}{2}\right)^2 = \frac{169}{4}.$$ Multiply both sides by 4: $$(a + x_B)^2 + d^2 = 169. \quad (1)$$ 6. **Sum of sides $BC + CD + DA = 17$:** - $BC = |x_B - 0| = x_B$ - $CD = d$ - $DA = a$ So, $$x_B + d + a = 17. \quad (2)$$ 7. **$ABCD$ is trapezium with $AD \parallel BC$:** Since $AD$ is along the x-axis, $AD = a$. The line $BC$ is parallel to $AD$, so $BC = x_B$. 8. **Express in variables:** From (2), $$x_B = 17 - d - a.$$ Substitute in (1): $$(a + (17 - d - a))^2 + d^2 = 169,$$ which simplifies to $$(17 - d)^2 + d^2 = 169.$$ 9. **Simplify equation:** $$(17 - d)^2 + d^2 = 169$$ $$289 - 34 d + d^2 + d^2 = 169$$ $$2 d^2 - 34 d + 289 = 169$$ $$2 d^2 - 34 d + 120 = 0$$ Divide entire equation by 2: $$d^2 - 17 d + 60 = 0.$$ 10. **Solve quadratic:** $$d = \frac{17 \pm \sqrt{17^2 - 4 \times 60}}{2} = \frac{17 \pm \sqrt{289 - 240}}{2} = \frac{17 \pm \sqrt{49}}{2} = \frac{17 \pm 7}{2}.$$ So, $$d = 12 \text{ or } d = 5.$$ 11. **Calculate $x_B + a$: ** From (2), $$x_B + a = 17 - d.$$ If $d=12$, $$x_B + a = 5.$$ If $d=5$, $$x_B + a = 12.$$ 12. **Find area of trapezium:** The area formula is $$ \text{Area} = \frac{(AD + BC)}{2} \times CD= \frac{(a + x_B)}{2} \times d.$$ Test $d=12$: Area $= \frac{5}{2} \times 12 = 30$. Test $d=5$: Area $= \frac{12}{2} \times 5 = 30$. 13. **Conclusion:** Area of trapezium $ABCD$ is $30$ cm$^2$.