9th Class Math Chapter 9 Notes

Q1 — Are the cuboids similar? Edges $3,4,5$ and $4.5,6,7.5$

Check whether the two solids are similar.

1. Compare ratios: $\frac{4.5}{3}=1.5$, $\frac{6}{4}=1.5$, $\frac{7.5}{5}=1.5$ — all equal.

Answer: Yes, similar (scale factor $1.5$)

Q2 — $△ABC\sim △DEF$: find $DF,EF$

$AB=12,AC=20,BC=16$ cm; $DE=6$ cm. Find $DF$ and $EF$.

1. Scale factor $=\frac{DE}{AB}=\frac{6}{12}=\frac{1}{2}$.
2. $DF=AC\times \frac{1}{2}=10$; $EF=BC\times \frac{1}{2}=8$.

Answer: $DF=10$ cm; $EF=8$ cm

Q3 — Find $x$: right angles at $A,C$; $AE=7.5,EC=3,DC=1.2$

Find $x=AB$.

1. $△ABE\sim △DCE$ (AA: right angles at $A,C$, vertical angles at $E$).
2. $\frac{AB}{DC}=\frac{AE}{EC}\Rightarrow \frac{x}{1.2}=\frac{7.5}{3}$.

Answer: $x=3$ cm

Q4 — Find $x$: $△BCD$, $EF\parallel CD$, $BE=6,BF=8,FD=3$

Find $x=EC$.

1. $△BEF\sim △BCD$: $\frac{BE}{BC}=\frac{BF}{BD}$.
2. $BD=BF+FD=11$; $\frac{6}{6+x}=\frac{8}{11}\Rightarrow 66=48+8x$.

Answer: $x=2.25$ cm

Q5 — Find $x$: equal angles at $C,F$; $CE=2.5,EF=2.4,FG=2.1$

Find $x=CD$.

1. $△CDE\sim △FGE$ (AA: equal angles at $C,F$, vertical angles at $E$).
2. $\frac{CD}{FG}=\frac{CE}{FE}\Rightarrow \frac{x}{2.1}=\frac{2.5}{2.4}$.

Answer: $x=2.1875$ cm

WP1 — Box width under a plank

Plank forms a right triangle (base $20$ cm, height $16$ cm); a box of height $8$ cm touches it. Find the box's width $x$.

1. Set the right angle at the origin, plank from $(0,16)$ to $(20,0)$: slope $=-\frac{16}{20}=-\frac{4}{5}$.
2. At height $y=8$: $8=16-\frac{4}{5}x\Rightarrow \frac{4}{5}x=8$.

Answer: $x=10$ cm

WP2 — Telephone pole height from a shadow

A $1.8$ m man casts a $0.76$ m shadow; a pole casts a $3$ m shadow. Find the pole's height.

1. Similar triangles (same sun angle): $\frac{h}{3}=\frac{1.8}{0.76}$.
2. $h=\frac{1.8\times 3}{0.76}=\frac{5.4}{0.76}=\frac{135}{19}$.

Answer: $h\approx 7.11$ m

Q6 — Find $x,y,z$: leg $=10$, hypotenuse segments $6$ and $x$

The altitude from the right angle divides the hypotenuse into $6$ (near the leg-10 side) and $x$; $y=$ altitude, $z=$ other leg.

1. Leg${}^2=$ (adjacent segment)$\times$(whole hypotenuse): $10^2=6(6+x)\Rightarrow 100=36+6x\Rightarrow x=\frac{64}{6}=\frac{32}{3}$.
2. $y^2=$ product of segments $=6x=6\times \frac{32}{3}=64\Rightarrow y=8$.
3. $z^2=x(6+x)=\frac{32}{3}\times \frac{50}{3}=\frac{1600}{9}\Rightarrow z=\frac{40}{3}$.
4. Check: $10^2+{\left(\frac{40}{3}\right)}^2=100+\frac{1600}{9}=\frac{2500}{9}={\left(\frac{50}{3}\right)}^2$ ✓

Answer: $x=\frac{32}{3}$ cm, $y=8$ cm, $z=\frac{40}{3}$ cm

Q7 — Dodecagon scaled by $\frac{1}{\sqrt{2}}$, original perimeter $72$

Find the new side length.

1. Original side $=\frac{72}{12}=6$ cm.
2. New side $=6\times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2}$.

Answer: $3\sqrt{2}\approx 4.24$ cm

Q8(i) — Area ratio for length ratio $1:3$

Find the ratio of areas.

1. Area ratio $=1^2:3^2$.

Answer: $1:9$

Q8(ii) — Area ratio for length ratio $3:4$

Find the ratio of areas.

1. Area ratio $=3^2:4^2$.

Answer: $9:16$

Q8(iii) — Area ratio for length ratio $2:7$

Find the ratio of areas.

1. Area ratio $=2^2:7^2$.

Answer: $4:49$

Q8(iv) — Area ratio for length ratio $8:9$

Find the ratio of areas.

1. Area ratio $=8^2:9^2$.

Answer: $64:81$

Q8(v) — Area ratio for length ratio $6:5$

Find the ratio of areas.

1. Area ratio $=6^2:5^2$.

Answer: $36:25$

Q9(i) — Rectangles, sides $10,6$; $A_1=240$

Find $A_2$.

1. ${\left(\frac{10}{6}\right)}^2=\frac{A_1}{A_2}\Rightarrow \frac{100}{36}=\frac{240}{A_2}$.

Answer: $A_2=86.4$ cm${}^2$

Q9(ii) — Triangles, sides $15,20$; $A_1=60$

Find $A_2$.

1. ${\left(\frac{15}{20}\right)}^2=\frac{9}{16}=\frac{60}{A_2}$.

Answer: $A_2=\frac{960}{9}\approx 106.67$ cm${}^2$

Q9(iii) — Trapeziums, sides $3.6,5.76$; $A_2=18$

Find $A_1$.

1. $\frac{3.6}{5.76}=\frac{5}{8}$; ${\left(\frac{5}{8}\right)}^2=\frac{25}{64}=\frac{A_1}{18}$.

Answer: $A_1=\frac{450}{64}=7.03125$ cm${}^2$

Q9(iv) — Cones, slant lengths $15,12$; $A_2=96$

Find $A_1$.

1. ${\left(\frac{15}{12}\right)}^2=\frac{25}{16}=\frac{A_1}{96}$.

Answer: $A_1=150$ cm${}^2$

Q9(v) — Triangles, ${\ell }_1=3$; $A_1=3\frac{4}{7},A_2=63$: find ${\ell }_2$

Find ${\ell }_2$.

1. $A_1=\frac{25}{7}$; ${\left(\frac{3}{{\ell }_2}\right)}^2=\frac{25/7}{63}=\frac{25}{441}$.
2. ${\ell }_2^2=\frac{9\times 441}{25}=\frac{3969}{25}\Rightarrow {\ell }_2=\frac{63}{5}$.

Answer: ${\ell }_2=12.6$ cm

Q10(i) — Area of $△ADE$: $BC\parallel DE$, area $ABC=36$, $AB=6,BD=4$

$D,E$ lie on extensions of $AB,AC$. Find the area of $△ADE$.

1. $AD=AB+BD=10$; $△ABC\sim △ADE$ with scale $\frac{AD}{AB}=\frac{10}{6}=\frac{5}{3}$.
2. Area ratio $={\left(\frac{5}{3}\right)}^2=\frac{25}{9}$; area $ADE=36\times \frac{25}{9}$.

Answer: $100$ cm${}^2$

Q10(ii) — Area of trapezium $BCED$

Using the same data, find the area of trapezium $BCED$.

1. Area $BCED=$ area $ADE-$ area $ABC=100-36$.

Answer: $64$ cm${}^2$

Q11(i) — $△ABC\sim △DEF$, $k=3$, area $ABC=50$

Find the area of $△DEF$.

1. Area $DEF=$ area $ABC\times k^2=50\times 9$.

Answer: $450$ cm${}^2$

Q11(ii) — Quadrilaterals, $k=\frac{1}{4}$, area $ABCD=64$

Find the area of $EFGH$.

1. Area $EFGH=64\times {\left(\frac{1}{4}\right)}^2=\frac{64}{16}$.

Answer: $4$ cm${}^2$

Q12(i) — Areas $16,25$: find the side ratio

Find the ratio of corresponding sides.

1. Side ratio $=\sqrt{16}:\sqrt{25}$.

Answer: $4:5$

Q12(ii) — Areas $144,81$, large base $=30$: find the small base

Find the corresponding base of the smaller triangle.

1. Side ratio $=\sqrt{144}:\sqrt{81}=12:9=4:3$.
2. $\frac{30}{\text{small}}=\frac{4}{3}\Rightarrow \text{small}=\frac{30\times 3}{4}$.

Answer: $22.5$ cm

Q13 — Larger heptagon side $=1.7\times$ smaller; smaller area $=100$

Find the larger heptagon's area.

1. Area ratio $=1.7^2=2.89$.
2. Larger area $=100\times 2.89$.

Answer: $289$ cm${}^2$

Q14(i) — Sphere radii ratio $3:4$: find the volume ratio

Find the ratio of volumes.

1. Volume ratio $=3^3:4^3$.

Answer: $27:64$

Q14(ii) — Tetrahedron volume ratio $8:27$: find the side ratio

Find the ratio of sides.

1. Side ratio $=\sqrt[3]{8}:\sqrt[3]{27}$.

Answer: $2:3$

Q14(iii) — Cone volume ratio $64:125$: height ratio

Find the ratio of heights.

1. Height ratio $=\sqrt[3]{64}:\sqrt[3]{125}$.

Answer: $4:5$

Q14(iv) — Same cones: base-area ratio

Find the ratio of base areas.

1. Length ratio $=4:5$; area ratio $=4^2:5^2$.

Answer: $16:25$

Q15(i) — Pyramids, edges $12,16$; $V_2=1536$

Find $V_1$.

1. ${\left(\frac{12}{16}\right)}^3={\left(\frac{3}{4}\right)}^3=\frac{27}{64}=\frac{V_1}{1536}$.

Answer: $V_1=648$ cm${}^3$

Q15(ii) — Cylinders, heights $2.5,8.75$; $V_2=171.5$

Find $V_1$.

1. $\frac{2.5}{8.75}=\frac{2}{7}$; ${\left(\frac{2}{7}\right)}^3=\frac{8}{343}=\frac{V_1}{171.5}$.

Answer: $V_1=4$ cm${}^3$

Q15(iii) — Boxes, $A_1=392,A_2=162$; $V_2=729$

Find $V_1$.

1. Linear ratio $=\sqrt{\frac{392}{162}}=\sqrt{\frac{196}{81}}=\frac{14}{9}$.
2. $V_1=729\times {\left(\frac{14}{9}\right)}^3=729\times \frac{2744}{729}$.

Answer: $V_1=2744$ cm${}^3$

Q15(iv) — Spheres, $V_1=64,V_2=216,r_2=12$

Find $r_1$.

1. $\frac{r_1^3}{12^3}=\frac{64}{216}=\frac{8}{27}$.
2. $r_1^3=1728\times \frac{8}{27}=512\Rightarrow r_1=8$.

Answer: $r_1=8$ cm

WP3 — Cans, length ratio $3:2$, large surface $=96$

Find the small can's surface area.

1. Area ratio $=3^2:2^2=9:4$.
2. Small $=96\times \frac{4}{9}$.

Answer: $\frac{384}{9}\approx 42.67$ m${}^2$

WP4 — Same cans, small volume $=240$

Find the large can's volume.

1. Volume ratio $=3^3:2^3=27:8$.
2. Large $=240\times \frac{27}{8}$.

Answer: $1012.5$ m${}^3$

WP5 — Tanks, height ratio $5:3$, large surface $=250$

Find the small tank's surface area.

1. Area ratio $=25:9$.
2. Small $=250\times \frac{9}{25}$.

Answer: $90$ m${}^2$

WP6 — Same tanks, small volume $=270$

Find the large tank's volume.

1. Volume ratio $=125:27$.
2. Large $=270\times \frac{125}{27}$.

Answer: $1250$ m${}^3$

Q16 — Parallelogram $AB=10,AD=6,\angle BAD=45°$

Find the area.

1. Area $=AB\times AD\times \sin (\angle BAD)=10\times 6\times \frac{\sqrt{2}}{2}$.

Answer: $30\sqrt{2}\approx 42.43$ cm${}^2$

WP7 — Reflections to tile a $3600$-unit square with a 3-4-5 triangle

Find the minimum number of reflections needed.

1. Area of one triangle $=\frac{1}{2}(3)(4)=6$.
2. Triangles needed $=\frac{3600}{6}=600$; the first tile isn't a reflection, so $600-1$ reflections produce the rest.

Answer: $599$ reflections

WP8 — Hexagon tessellation: $25$ hexagons, side $5$ cm

Find the total area and the outer perimeter (treating the whole tessellation as one larger similar hexagon).

1. One hexagon's area $=\frac{3\sqrt{3}}{2}(5{)}^2=\frac{75\sqrt{3}}{2}$; total $=25\times \frac{75\sqrt{3}}{2}=\frac{1875\sqrt{3}}{2}$.
2. Treating total area as $25\times$ one hexagon means the equivalent big hexagon's side is $5\times \sqrt{25}=25$ cm; perimeter $=6\times 25$.

Answer: Total area $\approx 1623.8$ cm${}^2$; outer perimeter $=150$ cm

WP9 — Tiles for a $12$ m $\times$ $15$ m floor

Find the number of $1$ m $\times$ $1$ m tiles needed.

1. Number of tiles $=12\times 15$.

Answer: $180$ tiles

WP10 — Paint (gallons) for a $10$ m $\times$ $120$ m wall

One gallon covers $35$ m${}^2$. Find the gallons needed.

1. Wall area $=10\times 120=1200$ m${}^2$.
2. Gallons $=\frac{1200}{35}\approx 34.3$, rounded up since partial gallons aren't sold.

Answer: $35$ gallons (exactly $\frac{240}{7}\approx 34.29$)

WP11 — Paint (litres) for a $10$ m $\times$ $4$ m wall

One litre covers $7$ m${}^2$. Find the litres needed.

1. Wall area $=10\times 4=40$ m${}^2$.
2. Litres $=\frac{40}{7}$.

Answer: $\approx 5.71$ litres

Q17 — Trapezoidal window: parallel sides $3,1.5$, height $2$

Find the area.

1. Area $=\frac{1}{2}(3+1.5)(2)$.

Answer: $4.5$ m${}^2$

Q18 — Polygon with interior-angle sum $1080°$

Find the number of sides.

1. $(n-2)\times 180=1080\Rightarrow n-2=6$.

Answer: $n=8$ sides

WP12 — Bottle heights ratio $2:1$: surface area and capacity ratios

Find both ratios.

1. Surface area ratio $=2^2:1^2$.
2. Capacity (volume) ratio $=2^3:1^3$.

Answer: Surface area $=4:1$; capacity $=8:1$

WP13(i) — Model car scale $\frac{1}{10}$: windscreen area ratio

Find the ratio of windscreen areas.

1. Area ratio $=1^2:10^2$.

Answer: $1:100$

WP13(ii) — Boot capacity ratio

Find the ratio of boot capacities.

1. Volume ratio $=1^3:10^3$.

Answer: $1:1000$

WP13(iii) — Width ratio

Find the ratio of widths.

1. Widths are linear measurements.

Answer: $1:10$

WP13(iv) — Number-of-wheels ratio

Find the ratio of the number of wheels.

1. Wheel count is a discrete quantity, not a scaled geometric measure — both have $4$ wheels regardless of size.

Answer: $1:1$ (does not scale with size)

WP14 — Jugs, heights $8,12,16$; smallest holds $\frac{1}{2}$ L

Find the capacities of the other two.

1. For $h=12$: ${\left(\frac{12}{8}\right)}^3={\left(\frac{3}{2}\right)}^3=\frac{27}{8}$; capacity $=500\times \frac{27}{8}=1687.5$ mL.
2. For $h=16$: ${\left(\frac{16}{8}\right)}^3=8$; capacity $=500\times 8=4000$ mL.

Answer: $1.6875$ L and $4$ L

WP15 — Glasses, heights $7.5,9,10.5$; tallest holds $343$ mL

Find the capacities of the other two.

1. For $h=7.5$: ${\left(\frac{7.5}{10.5}\right)}^3={\left(\frac{5}{7}\right)}^3=\frac{125}{343}$; capacity $=343\times \frac{125}{343}=125$ mL.
2. For $h=9$: ${\left(\frac{9}{10.5}\right)}^3={\left(\frac{6}{7}\right)}^3=\frac{216}{343}$; capacity $=343\times \frac{216}{343}=216$ mL.

Answer: $125$ mL and $216$ mL

WP16(i) — Ratio of lengths

Door area ratio $1:2500$ (an area ratio). Find the ratio of lengths.

1. Linear ratio $=\sqrt{1}:\sqrt{2500}$.

Answer: $1:50$

WP16(ii) — Ratio of petrol-tank capacities

Find the ratio of volumes.

1. Volume ratio $=1^3:50^3$.

Answer: $1:125000$

WP16(iii) — Model width, actual car $150$ cm wide

Find the model's width.

1. Model width $=150\times \frac{1}{50}$.

Answer: $3$ cm

WP16(iv) — Actual rear-window area, model $=3$ cm${}^2$

Find the actual car's rear-window area.

1. Actual $=$ model $\times 2500=3\times 2500$.

Answer: $7500$ cm${}^2$

WP17(i) — Ratio of heights

Find the ratio of the jars' heights.

1. Linear ratio $=\sqrt{144}:\sqrt{169}$.

Answer: $12:13$

WP17(ii) — Ratio of capacities

Find the ratio of capacities.

1. Volume ratio $=12^3:13^3$.

Answer: $1728:2197$

WP18 — Hexagon + 6 squares + 6 triangles, side $\frac{1}{2}$ m each

Find the total area of one pattern.

1. Hexagon: $\frac{3\sqrt{3}}{2}(0.5{)}^2=0.375\sqrt{3}$.
2. 6 squares: $6\times (0.5{)}^2=1.5$.
3. 6 triangles: $6\times \frac{\sqrt{3}}{4}(0.5{)}^2=0.375\sqrt{3}$.
4. Total $=0.375\sqrt{3}+1.5+0.375\sqrt{3}=1.5+0.75\sqrt{3}$.

Answer: $1.5+0.75\sqrt{3}\approx 2.80$ m${}^2$