1. **Problem 1: Find the median length of the leaves given the frequency distribution.** We have the length intervals and the number of leaves in each interval: | Length (mm) | Number of Leaves | |-------------|-----------------| | 118 - 126 | 3 | | 127 - 135 | 5 | | 136 - 144 | 9 | | 145 - 153 | 12 | | 154 - 162 | 5 | | 163 - 171 | 4 | | 172 - 180 | 2 | Total leaves $= 3 + 5 + 9 + 12 + 5 + 4 + 2 = 40$ 2. **Formula for median in grouped data:** $$\text{Median} = l + \left(\frac{\frac{n}{2} - F}{f}\right) \times h$$ where: - $l$ = lower boundary of median class - $n$ = total frequency - $F$ = cumulative frequency before median class - $f$ = frequency of median class - $h$ = class width 3. **Find the median class:** Half of total frequency $= \frac{40}{2} = 20$ Cumulative frequencies: - Up to 126: 3 - Up to 135: 3 + 5 = 8 - Up to 144: 8 + 9 = 17 - Up to 153: 17 + 12 = 29 Since 20 lies between 17 and 29, the median class is 145 - 153. 4. **Identify values:** - $l = 145$ - $F = 17$ - $f = 12$ - $h = 153 - 145 = 8$ 5. **Calculate median:** $$\text{Median} = 145 + \left(\frac{20 - 17}{12}\right) \times 8 = 145 + \left(\frac{3}{12}\right) \times 8 = 145 + 2 = 147$$ --- 6. **Problem 2 (a): Prove that $DF \times EF = FB \times FA$ in parallelogram $ABCD$ where diagonal $BD$ intersects segment $AE$ at $F$.** 7. **Given:** - $ABCD$ is a parallelogram. - $BD$ intersects $AE$ at $F$. - $E$ is any point on $AB$. 8. **To prove:** $$DF \times EF = FB \times FA$$ 9. **Proof:** - Since $ABCD$ is a parallelogram, $AB \parallel DC$ and $AD \parallel BC$. - Triangles $\triangle AFB$ and $\triangle DFE$ are similar by AA similarity: - $\angle AFB = \angle DFE$ (vertically opposite angles) - $\angle FAB = \angle EDF$ (alternate interior angles because $AB \parallel DC$) - From similarity, corresponding sides are proportional: $$\frac{AF}{DF} = \frac{FB}{EF}$$ - Cross-multiplying: $$AF \times EF = FB \times DF$$ - Rearranged: $$DF \times EF = FB \times FA$$ This completes the proof. --- 7. **Problem 2 (b): In $\triangle ABC$, if $AD \perp BC$ and $AD^2 = BD \times DC$, prove $\angle BAC = 90^\circ$.** 8. **Given:** - $AD$ is perpendicular to $BC$. - $AD^2 = BD \times DC$. 9. **To prove:** $$\angle BAC = 90^\circ$$ 10. **Proof:** - By the property of right triangles and the altitude drawn to the hypotenuse: $$AD^2 = BD \times DC$$ implies $AD$ is the altitude from $A$ to $BC$. - This relation holds only if $\triangle ABC$ is right angled at $A$. - Therefore, $\angle BAC = 90^\circ$. **Final answers:** - Median length of leaves $= 147$ mm. - Proof of $DF \times EF = FB \times FA$ is established. - Proof that $\angle BAC = 90^\circ$ under given conditions is established.