Q1. Area of a square with side \(5\) cm?

Ans: \(A=5^2=25\) sq.cm.

Q2. Area formula of a rectangle?

Ans: \(A=\text{length}\times\text{breadth}.\)

Q3. Area of a right triangle with perpendicular sides 6 and 8?

Ans: \(A=\tfrac12\times6\times8=24\) sq.units.

Q4. Area of parallelogram with base 10 and height 4?

Ans: \(A=10\times4=40\) sq.units.

Q5. Area of rhombus with diagonals 10 and 8?

Ans: \(A=\tfrac12\times10\times8=40\) sq.units.

Q6. Area of trapezium with parallel sides 7 and 9 and height 5?

Ans: \(A=\tfrac12(7+9)\times5=40\) sq.units.

Q7. Semiperimeter \(s\) formula for triangle sides \(a,b,c\)?

Ans: \(s=\dfrac{a+b+c}{2}\).

Q8. Area of a circle of radius 7 (use \(\pi=\tfrac{22}{7}\))?

Ans: \(A=\pi r^2=\tfrac{22}{7}\times7^2=22\times7=154\) sq.units.

Q9. If area of parallelogram is 112 and base 10, find height.

Ans: \(h=A/b=112/10=11.2\) units.

Q10. Area of triangle base 12 and height 5?

Ans: \(A=\tfrac12\times12\times5=30\) sq.units.

Q11. Units of area when sides in cm?

Ans: square centimetres (sq.cm).

Q12. Area of rectangle 8 by 6?

Ans: \(8\times6=48\) sq.units.

Q13. Area of rhombus with diagonals 11.2 and 7.5?

Ans: \(A=\tfrac12\times11.2\times7.5=42\) sq.units.

Q14. If area of circle is \(3850\) and \(\pi=\tfrac{22}{7}\), radius is?

Ans: \(r^2=\dfrac{3850\times7}{22}=1225\Rightarrow r=35.\)

Q15. Area formula for parallelogram (in words)?

Ans: base times corresponding height.

Q16. If diagonals of a rhombus are equal, what is the shape?

Ans: If diagonals equal and all sides equal → square.

Q17. Heron's formula expressed?

Ans: \(A=\sqrt{s(s-a)(s-b)(s-c)}\) where \(s=\tfrac{a+b+c}{2}\).

Q18. Circumference formula used when deriving circle area?

Ans: \(C=2\pi r\).

Q19. For trapezium, if parallel sides are equal, it reduces to?

Ans: rectangle (area = side × height).

Q20. Quick: area of triangle with sides 17,25,26 (use Heron)?

Ans: \(s=(17+25+26)/2=34\). \(A=\sqrt{34(34-17)(34-25)(34-26)}=\sqrt{34\cdot17\cdot9\cdot8}=204\) sq.units.

Q1. Find area of parallelogram with base 18 cm and height 11 cm.

Ans: \(A=18\times11=198\) sq.cm.

Q2. Area of rhombus with diagonals 15 cm and 24 cm?

Ans: \(A=\tfrac12\times15\times24=180\) sq.cm.

Q3. If area of trapezium with parallel sides 7 and 8 and height 6, compute A.

Ans: \(A=\tfrac12(7+8)\times6=45\) sq.cm.

Q4. A rhombus has area 96 and one diagonal 12. Find the other diagonal.

Ans: \(96=\tfrac12\times12\times d_2\Rightarrow d_2=16\) cm.

Q5. Area of triangle with base 96 and height 70?

Ans: \(A=\tfrac12\times96\times70=3360\) sq.m.

Q6. Compute area of triangle with sides 60,60,96 using Heron.

Ans: \(s=(60+60+96)/2=108.\) \(A=\sqrt{108(108-60)(108-60)(108-96)}=\sqrt{108\cdot48\cdot48\cdot12}=1728\) sq.m.

Q7. Area of circle: radius 21 cm, using \(\pi=\tfrac{22}{7}\)?

Ans: \(A=\pi r^2=\tfrac{22}{7}\times21^2=1386\) sq.cm.

Q8. Diameters from areas: if area = 176 sq.cm and \(\pi=\tfrac{22}{7}\), diameter?

Ans: \(r^2=\dfrac{176\times7}{22}=56\Rightarrow r=\sqrt{56}\Rightarrow d=2\sqrt{56}\) (approx \(2\times7.483=14.966\) cm).

Q9. A circular garden diameter 42 m with a 3.5 m wide road around it: area of road?

Ans: Outer radius = \(21+3.5=24.5\). \(A_{\text{road}}=\pi(24.5^2-21^2)=\pi(600.25-441)=\pi\times159.25\). Using \(\pi\approx3.1416\), approx \(500.1\) sq.m.

Q10. Area of trapezium with parallel sides 8.5 and 11.5 and height 4.2?

Ans: \(A=\tfrac12(8.5+11.5)\times4.2=10\times4.2=42\) sq.cm.

Q11. Find height if area of parallelogram is 29.6 and base is 8.

Ans: \(h=A/b=29.6/8=3.7\) units.

Q12. Sides of triangle 45,39,42 — find area (Heron).

Ans: \(s=(45+39+42)/2=63\). \(A=\sqrt{63(18)(24)(21)}=\sqrt{63\cdot18\cdot24\cdot21}=1188\) sq.units (calculation gives \(1188\)).

Q13. Plot LMNO found earlier: total area 5088 sq.m — how computed?

Ans: area = area triangle LMN (Heron = 1728) + area triangle LNO (\(\tfrac12\times96\times70=3360\)) = 5088 sq.m.

Q14. If rhombus perimeter 100 and one diagonal 48, find area.

Ans: side \(=100/4=25\). Half diagonal \(=24\). Use Pythagoras: other half diagonal \(= \sqrt{25^2-24^2}=\sqrt{625-576}=\sqrt{49}=7\Rightarrow\) other diagonal \(=14\). Area \(=\tfrac12\times48\times14=336\) sq.units.

Q15. Area of trapezium PQRS where PQ=7, distance between parallels=4, SM=3 (isosceles)?

Ans sketch: if given PQ=7 and height 4 → use \(A=\tfrac12(PQ+SR)\times4\). Need SR; if symmetric info gives SR= ? (textbook problem requires SR compute via geometry). (If SR known, plug values.)

Q16. Area of irregular polygon: explain method.

Ans: Divide polygon into triangles and simple quadrilaterals, compute each area, then sum.

Q17. Approximate area via graph paper counting — how?

Ans: Count full squares + squares >1/2 + ignore <1/2; multiply by area per square.

Q18. Area of circle if circumference is 88 cm (\(\pi=\tfrac{22}{7}\))?

Ans: circumference \(=2\pi r=88\Rightarrow r=88/(2\pi)=88/(44/7)=14.\) Area \(=\pi r^2=\tfrac{22}{7}\times14^2=616\) sq.cm.

Q19. Area of triangle with base 12+15 and height 13? (from figure)

Ans: base = 27, so \(A=\tfrac12\times27\times13=175.5\) sq.m.

Q20. Area of a circle radius 28 cm?

Ans: \(A=\pi r^2=\pi\times28^2\). With \(\pi=\tfrac{22}{7}\): \(=\tfrac{22}{7}\times784=2464\) sq.cm.

Q1. Using Heron, find area of triangle with sides \(17,25,26\) (show steps).

Ans: \(s=(17+25+26)/2=34.\) Then \(A=\sqrt{34(34-17)(34-25)(34-26)}=\sqrt{34\cdot17\cdot9\cdot8}=\sqrt{83616}=204\) sq.units.

Q2. Plot LMNO: LM=60, MN=60, LN=96, OP=70; compute area (shown earlier).

Ans: triangle LMN: \(s=108\), \(A_{LMN}=\sqrt{108\cdot48\cdot48\cdot12}=1728\). triangle LNO: base LN=96, height OP=70 → \(A=\tfrac12\times96\times70=3360\). Total 5088 sq.m.

Q3. A polygon composed of triangles and trapezium (example): compute total area = 871 sq.m — outline steps.

Ans: Compute each part separately: triangles via \(\tfrac12 bh\) and trapezium via \(\tfrac12(h)(sum\;parallel\;sides)\); then sum: 65 + 405 + 136 + 265 = 871 sq.m.

Q4. A trapezium has parallel sides 13 and 25 and height 12: find area.

Ans: \(A=\tfrac12(13+25)\times12=\tfrac12\times38\times12=228\) sq.units.

Q5. Prove area of rhombus = \(\tfrac12 d_1 d_2\).

Ans: Diagonals perpendicular, divide rhombus into 4 right triangles each area \(\tfrac12\times(d_1/2)\times(d_2/2)\). Sum = \(4\times\tfrac12\times d_1/2\times d_2/2=\tfrac12 d_1 d_2\).

Q6. Triangle area by coordinates? (basic idea)

Ans: Divide into right triangles or use formula; (textbook avoids coordinates), but standard formula: \(A=\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|\).

Q7. A circular ground area 3850 sq.m → find radius using \(\pi=\tfrac{22}{7}\).

Ans: \(r^2=3850\times7/22=1225\Rightarrow r=35\) m.

Q8. Find area difference between two concentric circles radii 24.5 and 21.

Ans: \(A=\pi(24.5^2-21^2)=\pi(600.25-441)=\pi(159.25)\approx500.1\) sq.m.

Q9. Given triangle sides 45,39,42 compute area stepwise.

Ans: \(s=(45+39+42)/2=63.\) \(A=\sqrt{63(63-45)(63-39)(63-42)}=\sqrt{63\cdot18\cdot24\cdot21}=1188\) sq.units.

Q10. Using graph paper counts: 13 full squares and 11 >1/2 squares. Area per small square 1 sq.cm. Total area?

Ans: \(13+11=24\) sq.cm.

Q11. Find area of circle if diameter = 42 and road width 3.5 → area road numeric (use \(\pi=\tfrac{22}{7}\)).

Ans: inner r=21; outer r=24.5; \(A_{\text{road}}=\pi(24.5^2-21^2)=\tfrac{22}{7}\times159.25\approx500.1\) sq.m.

Q12. If triangle base 96 and area 3360, compute height.

Ans: \(3360=\tfrac12\times96\times h\Rightarrow h=70.\)

Q13. A plot consists of triangle + trapezium + two right triangles sum = 871. Show decomposition approach.

Ans: Decompose into parts whose area formulas are known, compute each, sum — example in textbook yields 871 sq.m.

Q14. If area of circle = 394.24 sq.cm, find diameter (use \(\pi=3.1416\)).

Ans: \(r^2=A/\pi=394.24/3.1416\approx125.5\Rightarrow r\approx11.2\Rightarrow d\approx22.4\) cm.

Q15. Area of polygon with given sides (Practice 15.5): method summary.

Ans: Break polygon into triangles/trapeziums using diagonals and perpendiculars; compute areas and sum.

Q16. Area of circle with radius 10.5 cm (use \(\pi=\tfrac{22}{7}\)).

Ans: \(A=\tfrac{22}{7}\times10.5^2=\tfrac{22}{7}\times110.25=22\times15.75=346.5\) sq.cm.

Q17. Using Heron: triangle 17,25,26 produced integer area — why?

Ans: because \(s=34\) and factors etc. produce perfect square under radical; some integer sides give integer area.

Q18. If circle circumference 88 (π=22/7) find area stepwise.

Ans: \(2\pi r=88\Rightarrow r=14\). Area \(=\pi r^2=\tfrac{22}{7}\times14^2=616.\)

Q19. If trapezium has parallel sides 30 and 25 height 30, area?

Ans: \(A=\tfrac12(30+25)\times30=825\) sq.units.

Q20. Area question: composite figure of rectangles and circles — outline approach.

Ans: Compute areas of individual simple shapes (rectangles, semicircles, etc.), add or subtract depending on shading, ensure consistent units.

1. Base = 18 cm, height = 11 cm. Find area.

Ans: \(A=18\times11=198\) sq.cm.

2. Area = 29.6 sq.cm, base = 8 cm. Find height.

Ans: \(h=A/b=29.6/8=3.7\) cm.

3. Area = 83.2 sq.cm, height = 6.4 cm. Find base.

Ans: \(b=A/h=83.2/6.4=13\) cm.

1. Diagonals 15 and 24 cm. Area?

Ans: \(A=\tfrac12\times15\times24=180\) sq.cm.

2. Diagonals 16.5 and 14.2 cm. Area?

Ans: \(A=\tfrac12\times16.5\times14.2=0.5\times234.3=117.15\) sq.cm.

3. Perimeter 100 cm, diagonal 48 cm. Area?

Ans: side \(=100/4=25.\) half diagonal = 24. Other half diagonal half = \(\sqrt{25^2-24^2}=7\Rightarrow d_2=14.\) Area \(=\tfrac12\times48\times14=336\) sq.cm.

4. Diagonal = 30, area = 240. Find perimeter.

Ans: Let other diagonal \(d\). \(240=\tfrac12\times30\times d\Rightarrow d=16.\) Half-diagonals 15 and 8; side \(=\sqrt{15^2+8^2}=\sqrt{225+64}= \sqrt{289}=17\). Perimeter \(=4\times17=68\) cm.

1. ABCD where AB=13, DC=9, AD=8 — find area (assume AD perpendicular to DC as implied)?

Ans: If AD is height h=8 and parallel sides AB and DC given, area \(=\tfrac12(13+9)\times8=88\) sq.units. (If geometry different, use given heights.)

2. Parallel sides 8.5 & 11.5, height 4.2 → area?

Ans: \(A=\tfrac12(8.5+11.5)\times4.2=10\times4.2=42\) sq.cm.

3. PQRS is isosceles trapezium with PQ=7, distance between parallels 4, SM=3, SM perpendicular — find area.

Ans: If SM=3 is part of slanted side, SR length compute by geometry (not enough data here). If SR known, apply \(A=\tfrac12(PQ+SR)\times4\).

1. Triangle sides 45,39,42 → area?

Ans: \(s=(45+39+42)/2=63.\) \(A=\sqrt{63(63-45)(63-39)(63-42)}=\sqrt{63\cdot18\cdot24\cdot21}=1188\) sq.units.

2. For the figure PQRS with given 36,56,25,15, find area (split into triangles/trapezium).

Ans: Decompose as per textbook: triangle and trapezium areas computed and summed. (Detailed numeric steps depend on exact drawing coordinates; follow sample in chapter.)

3. In figure ABCD with given sides 40,9,13, find area via decomposition.

Ans: Divide into triangles/rectangles using perpendiculars; compute each area and sum. (Textbook example — compute numbers following given heights.)

1. Given plot diagrams (measures in metres) — find area by splitting into triangles/trapezia (examples in textbook).

Ans: Follow decomposition shown in chapter: compute each piece (rectangle: \(lw\), triangle: \(\tfrac12 bh\), trapezium: \(\tfrac12(h)(sum\;parallel\;sides)\)), then add. (See textbook worked solutions for each specific diagram.)

1. Radii 28, 10.5, 17.5 — areas?

Ans: \(A=\pi r^2\). Using \(\pi=\tfrac{22}{7}\): (1) \(r=28\Rightarrow A=\tfrac{22}{7}\times28^2=2464\) sq.cm. (2) \(r=10.5\Rightarrow A=346.5\) sq.cm. (3) \(r=17.5\Rightarrow A=\tfrac{22}{7}\times17.5^2=962.5\) sq.cm.

2. Areas given 176, 394.24, 12474 → diameters?

Ans: \(r=\sqrt{A/\pi}\), \(d=2r\). Compute numerically with \(\pi\). Example (2): \(r=\sqrt{394.24/3.1416}\approx11.2\Rightarrow d\approx22.4\) cm.

3. Circular garden d=42, road width=3.5 → area of road?

Ans: inner r=21, outer r=24.5, area road \(=\pi(24.5^2-21^2)=\pi\times159.25\approx500.1\) sq.m.

4. Circle circumference 88 cm — area?

Ans: \(2\pi r=88\Rightarrow r=14\). Area \(=\pi\times14^2=\tfrac{22}{7}\times196=616\) sq.cm.

Q: If 13 full squares (1 sq.cm) and 11 parts >1/2 counted, total area?

Ans: \(13 + 11 = 24\) sq.cm (approx).