1. The problem asks to state one condition for two triangles to be similar. 2. One condition for two triangles to be similar is that their corresponding angles are equal (AAA criterion). 3. When two triangles are similar, the ratios of their corresponding sides are equal. 4. If the ratio of sides of two similar triangles is 3:5, then the ratio of their areas is the square of the ratio of their sides. $$\text{Area ratio} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$$ 5. To find the value of $x$ using similarity in triangles $MPN$ and $QOP$ (assuming $\triangle MPN \sim \triangle QOP$), use the ratio of corresponding sides: Given sides: $MN = 55$, $PQ = 66$, and $PN = 3x - 2$, $OQ = 48$ Set up the proportion: $$\frac{MN}{PQ} = \frac{PN}{OQ}$$ Substitute values: $$\frac{55}{66} = \frac{3x - 2}{48}$$ 6. Simplify the left side: $$\frac{\cancel{55}}{\cancel{66}} = \frac{5}{6}$$ So: $$\frac{5}{6} = \frac{3x - 2}{48}$$ 7. Cross multiply: $$5 \times 48 = 6 \times (3x - 2)$$ $$240 = 18x - 12$$ 8. Add 12 to both sides: $$240 + 12 = 18x$$ $$252 = 18x$$ 9. Divide both sides by 18: $$x = \frac{252}{18}$$ $$x = 14$$ 10. For the ratio of areas when the ratio of sides is 4:7: $$\text{Area ratio} = \left(\frac{4}{7}\right)^2 = \frac{16}{49}$$ 11. Given two similar triangular pyramids with surface areas 18 cm² and 450 cm², and volume of the larger pyramid 400 cm³, find the volume of the smaller pyramid. The ratio of surface areas is: $$\frac{18}{450} = \frac{1}{25}$$ Since surface area ratio = (scale factor)$^2$, the scale factor is: $$\sqrt{\frac{1}{25}} = \frac{1}{5}$$ Volume ratio = (scale factor)$^3$ = $\left(\frac{1}{5}\right)^3 = \frac{1}{125}$ Volume of smaller pyramid: $$\text{Volume}_{small} = \frac{1}{125} \times 400 = 3.2$$ cm³ 12. Water flows through a pipe of radius 3 cm at speed 5 cm/s. Find the volume of water flowing per second. Volume flow rate = cross-sectional area $\times$ speed Cross-sectional area: $$A = \pi r^2 = \pi \times 3^2 = 9\pi$$ cm² Volume flow rate: $$V = A \times v = 9\pi \times 5 = 45\pi$$ cm³/s 13. A pipe discharges 154 cm³/s of water. If the radius of the pipe is 7 cm, find the speed of water flow. Cross-sectional area: $$A = \pi r^2 = \pi \times 7^2 = 49\pi$$ cm² Speed: $$v = \frac{\text{Volume flow rate}}{A} = \frac{154}{49\pi} = \frac{154}{49\pi} = \frac{154}{49\pi}$$ cm/s Simplify numerator and denominator: $$\frac{154}{49} = \frac{154 \div 7}{49 \div 7} = \frac{22}{7}$$ So: $$v = \frac{22}{7\pi}$$ cm/s Numerically, since $\pi \approx 3.1416$: $$v \approx \frac{22}{7 \times 3.1416} \approx \frac{22}{21.9912} \approx 1.0$$ cm/s Final answers: - $x = 14$ - Ratio of areas for 3:5 sides = $9:25$ - Ratio of areas for 4:7 sides = $16:49$ - Volume of smaller pyramid = 3.2 cm³ - Volume flow rate for radius 3 cm and speed 5 cm/s = $45\pi$ cm³/s - Speed for discharge 154 cm³/s and radius 7 cm = approximately 1.0 cm/s