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Exam practice

Every question from the end of every unit, with hints and answers to open. After you check an answer, mark how you did, and come back to the ones you didn’t get.

107 questions

Unit 5 · Area

Exercise 5.1

Question 1

a.

What is a scalene triangle?

Answer

A triangle whose three sides are all different in length (and so its three angles are all different too).

b.

The sides of a triangle are p cm, q cm and r cm. What is its perimeter?

Answer

(p + q + r) cm, and its semi-perimeter is p+q+r2 \frac{p + q + r}{2} cm.

c.

Write Heron’s formula for the area of a triangle.

Answer

With sides a, b, c and s=a+b+c2 s = \frac{a + b + c}{2} : Area =s(s−a)(s−b)(s−c) = \sqrt{s(s - a)(s - b)(s - c)} .

Question 2

a.
20 cm12 cm16 cmABC
Figure 5.5
Answer

s=12+20+162=24 s = \frac{12 + 20 + 16}{2} = 24 ; area =24×12×4×8=9216= = \sqrt{24 \times 12 \times 4 \times 8} = \sqrt{9216} = 96 cm².

b.
8 cm6 cm10 cmMNO
Figure 5.6
Answer

s=6+8+102=12 s = \frac{6 + 8 + 10}{2} = 12 ; area =12×6×4×2=576= = \sqrt{12 \times 6 \times 4 \times 2} = \sqrt{576} = 24 cm².

c.
15 cm13 cm14 cmPQR
Figure 5.7
Answer

s=13+15+142=21 s = \frac{13 + 15 + 14}{2} = 21 ; area =21×8×6×7=7056= = \sqrt{21 \times 8 \times 6 \times 7} = \sqrt{7056} = 84 cm².

d.
20 cm13 cm21 cmDEF
Figure 5.8
Answer

s=20+21+132=27 s = \frac{20 + 21 + 13}{2} = 27 ; area =27×7×6×14=15876= = \sqrt{27 \times 7 \times 6 \times 14} = \sqrt{15876} = 126 cm².

Question 3

a.
5 cm3 cm4 cm4 cm7 cmABCD
Figure 5.9
Answer

△ABC (3, 5, 7): s = 7.5, 7.5×4.5×2.5×0.5=42.1875≈6.50 \sqrt{7.5 \times 4.5 \times 2.5 \times 0.5} = \sqrt{42.1875} \approx 6.50 .
△ACD (7, 4, 4): s = 7.5, 7.5×0.5×3.5×3.5=45.9375≈6.78 \sqrt{7.5 \times 0.5 \times 3.5 \times 3.5} = \sqrt{45.9375} \approx 6.78 .
Area ≈6.495+6.778= \approx 6.495 + 6.778 = 13.27 cm².

b.
12 cm10 cm14 cm6 cm16 cmPQRS
Figure 5.10
Answer

△PQS (6, 16, 14): s = 18, 18×12×2×4=1728≈41.57 \sqrt{18 \times 12 \times 2 \times 4} = \sqrt{1728} \approx 41.57 .
△QRS (12, 10, 16): s = 19, 19×7×9×3=3591≈59.92 \sqrt{19 \times 7 \times 9 \times 3} = \sqrt{3591} \approx 59.92 .
Area ≈41.569+59.925= \approx 41.569 + 59.925 = 101.49 cm².

Question 4

4.

The sides of a triangular field are in the ratio 3 : 5 : 7, and its perimeter is 300 m. Find its area.

Answer

3x + 5x + 7x = 300, so x = 20: sides 60, 100 and 140 m; s = 150.
Area =150×90×50×10=6750000=15003≈ = \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6750000} = 1500\sqrt{3} \approx 2,598.08 m².

Question 5

5.

The sides of a triangular field are in the ratio 12 : 17 : 25, and its perimeter is 540 ft. Find its area.

Answer

54x = 540, x = 10: sides 120, 170 and 250 ft; s = 270.
Area =270×150×100×20=81000000= = \sqrt{270 \times 150 \times 100 \times 20} = \sqrt{81000000} = 9,000 ft².

Question 6

6.

The sides of a triangular garden are in the ratio 12:13:15 \frac{1}{2} : \frac{1}{3} : \frac{1}{5} , and its perimeter is 62 m. Find its area.

Hint

Multiply each part of the ratio by 30, the smallest number all three denominators divide.

Answer

12:13:15=15:10:6 \frac{1}{2} : \frac{1}{3} : \frac{1}{5} = 15 : 10 : 6 . 31x = 62, x = 2: sides 30, 20 and 12 m; s = 31.
Area =31×1×11×19=6479≈ = \sqrt{31 \times 1 \times 11 \times 19} = \sqrt{6479} \approx 80.49 m².

Question 7

7.

A triangular field has perimeter 20 m and area 65 6\sqrt{5} m², and one side is 9 m. Find the other two sides.

Answer

s = 10, b + c = 11, so c = 11 − b and 10 − c = b − 1.
65=10×1×(10−b)(b−1)180=10(10−b)(b−1)18=−b2+11b−10b2−11b+28=0(b−4)(b−7)=0 \begin{aligned} 6\sqrt{5} &= \sqrt{10 \times 1 \times (10 - b)(b - 1)} \\ 180 &= 10(10 - b)(b - 1) \\ 18 &= -b^2 + 11b - 10 \\ b^2 - 11b + 28 &= 0 \\ (b - 4)(b - 7) &= 0 \end{aligned}
The other sides are 4 m and 7 m.

Question 8

8.

A “school ahead” traffic sign is an equilateral triangle with sides a cm. Its perimeter is 360 cm. Find its area with Heron’s formula.

a cma cma cmABC
Figure 5.11
Answer

3a = 360, so a = 120 cm and s = 180.
Area =180×60×60×60=38880000=36003≈ = \sqrt{180 \times 60 \times 60 \times 60} = \sqrt{38880000} = 3600\sqrt{3} \approx 6,235.38 cm².
Check with the equilateral triangle’s own formula: 34×1202=36003 \frac{\sqrt{3}}{4} \times 120^2 = 3600\sqrt{3} . ✓

Question 9

9.

The scale drawing shows a four-sided piece of land to be divided between two brothers along the diagonal AC: the elder gets △ABC and the younger △ACD. Whose land is bigger, and by how much? What was the area before it was divided?

28 m9 m40 m15 mABCD
Figure 5.12
Hint

Find AC by Pythagoras’ theorem in △ABC first.

Answer

△ABC is right-angled at B: area =12×40×9=180 = \frac{1}{2} \times 40 \times 9 = 180 m², and AC=92+402=41 AC = \sqrt{9^2 + 40^2} = 41 m.
△ACD (28, 15, 41): s = 42, area =42×14×27×1=15876=126 = \sqrt{42 \times 14 \times 27 \times 1} = \sqrt{15876} = 126 m².
The elder brother’s land (180 m²) is 54 m² more than the younger’s (126 m²); the whole land was 306 m².

Question 10

10.

A garden ABCD has ∠C=90° \angle C = 90° , AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. Find its area.

12 m5 m8 m9 mABCD
Figure 5.13
Answer

BD=122+52=13 BD = \sqrt{12^2 + 5^2} = 13 m; △BCD=12×12×5=30 \triangle BCD = \frac{1}{2} \times 12 \times 5 = 30 m².
△ABD (9, 8, 13): s = 15, area =15×6×7×2=1260≈35.50 = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} \approx 35.50 m².
Garden ≈30+35.50= \approx 30 + 35.50 = 65.50 m².

Question 11

11.

A triangle and a parallelogram on the same base have equal areas. The triangle’s sides are 26 cm, 28 cm and 30 cm, and the parallelogram’s base is 28 cm. Find the parallelogram’s height.

Answer

s = 42; triangle =42×16×14×12=112896=336 = \sqrt{42 \times 16 \times 14 \times 12} = \sqrt{112896} = 336 cm².
Parallelogram: 28 × h = 336, so h = 12 cm.

Exercise 5.2

Question 1

a.

How are the areas of opposite walls of a rectangular room related?

Answer

They are equal: each pair is the same rectangle, l × h or b × h.

b.

In a cubical room, how are the four walls related to each other and to the floor?

Answer

All four walls are equal, and each is equal to the floor (l2 l^2 ); the four walls together are 4l2 4l^2 , four times the floor.

c.

Are the floor and ceiling of a rectangular room equal?

Answer

Yes, both are l × b.

d.

Which surfaces of a rectangular room are equal?

Answer

The floor and the ceiling; the two walls along the length; the two walls along the breadth.

Question 2

a.
PQRSWNMT
Figure 5.19
Answer

Floor PQRS; ceiling WNMT; four walls PQNW, QRMN, RSTM and SPWT.

b.
ADEFBCHG
Figure 5.20
Answer

Floor ADEF; ceiling BCHG; four walls ADCB, DEHC, EFGH and FABG.

Question 3

3.

A room is 8 m long, 5 m wide and 2.5 m high. Find a. the area of its four walls, b. the area of its floor and ceiling.

Answer
  • a.2 × 2.5 × (8 + 5) = 65 m².
  • b.2 × 8 × 5 = 80 m².

Question 4

a.

the area of the floor.

Answer

8.5 × 6 = 51 m².

b.

the area of the ceiling.

Answer

The same as the floor: 51 m².

c.

the area of the two walls along its length.

Answer

2 × 8.5 × 2.4 = 40.8 m².

d.

the area of the two walls along its breadth.

Answer

2 × 6 × 2.4 = 28.8 m².

e.

the area of the four walls.

Answer

40.8 + 28.8 = 69.6 m² (or 2 × 2.4 × 14.5).

f.

the area of the four walls, floor and ceiling.

Answer

69.6 + 51 + 51 = 171.6 m².

Question 5

5.

The four walls of a meeting hall have an area of 135 m². The hall is 15 m long and 12 m wide. Find its height.

Answer

2h(15 + 12) = 135, so 54h = 135 and h = 2.5 m.

Question 6

6.

The four walls of a room have an area of 432 ft². The room is 14 ft wide and 9 ft high. Find its length.

Answer

2 × 9 × (l + 14) = 432, so l + 14 = 24 and l = 10 ft.

Question 7

7.

The perimeter of Ishan’s bedroom floor is 60 m, and its four walls have an area of 144 m². Find the room’s height.

Answer

Four walls = floor’s perimeter × height: 60h = 144, so h = 2.4 m.

Question 8

a.

Find the area of its floor and ceiling.

Answer

2 × 30 × 25 = 1,500 m².

b.

Find the area of its four walls without the doors and windows.

Answer

Four walls = 2 × 4 × 55 = 440 m². Windows = 5 × 3.75 = 18.75 m²; doors = 2 × 15 = 30 m².
Without them: 440 − 18.75 − 30 = 391.25 m².

c.

The ceiling is decorated with 3 patterns in every square meter. How many patterns are there in all?

Answer

Ceiling = 750 m², so 750 × 3 = 2,250 patterns.

d.

Three national flags are painted for every 15 m² of the four walls, measured whole (doors and windows included). How many flags are painted?

Answer

44015×3= \frac{440}{15} \times 3 = 88 flags.
(If only the 391.25 m² of wall without the openings counted, there would be room for 391.2515×3=78.25 \frac{391.25}{15} \times 3 = 78.25 , so 78 flags.)

Question 9

a.

Find the area of its four walls without the doors and windows.

Answer

2 × 3 × 14 = 84 m²; doors 6 m², windows 3 m²; 84 − 6 − 3 = 75 m².

b.

The students paste one sheet with a picture on every square meter of that wall. How many sheets do they paste?

Answer

75 sheets.

Question 10

a.

Find the area of its four walls without the door.

Answer

4×2.72=4×7.29=29.16 4 \times 2.7^2 = 4 \times 7.29 = 29.16 m²; 29.16 − 2 = 27.16 m².

b.

One photo frame is hung for every 4 m² of that wall. How many frames can be hung?

Answer

27.16 ÷ 4 = 6.79: 6 frames (a seventh would need 28 m²).

Exercise 5.3

Question 1

1.

Find the cost of laying carpet at Rs 130 a square foot in a room 12 ft long and 11 ft wide.

Answer

12 × 11 = 132 ft²; 132 × 130 = Rs 17,160.

Question 2

2.

A room is 15 ft long, 12 ft wide and 8 ft high inside. It has two windows of 6 ft × 4 ft 6 in and a door of 3 ft × 6 ft 6 in. Find the cost of painting the four walls (not the windows, the door or the ceiling) at Rs 125 a square foot.

Answer

Four walls = 2 × 8 × 27 = 432 ft². Windows = 2 × 6 × 4.5 = 54 ft²; door = 3 × 6.5 = 19.5 ft² (6 in = 0.5 ft).
Area = 432 − 54 − 19.5 = 358.5 ft²; cost = 358.5 × 125 = Rs 44,812.50.

Question 3

3.

A square room has sides 15 ft and height 8 ft. Find the cost of plastering its floor, four walls and ceiling at Rs 120 a square foot.

Answer

Floor and ceiling =2×152=450 = 2 \times 15^2 = 450 ft²; four walls = 4 × 15 × 8 = 480 ft²; total 930 ft².
Cost = 930 × 120 = Rs 1,11,600.

Question 4

a.

Find the total area of its floor and ceiling.

Answer

2 × 32 × 32 = 2,048 m².

b.

Find the total area of its windows and doors.

Answer

6 × 4.5 + 2 × 12 = 27 + 24 = 51 m².

c.

Find the area of its four walls without the windows and doors.

Answer

4 × 32 × 4 − 51 = 512 − 51 = 461 m².

d.

One chair needs 2 m² of floor. How many such chairs fit in the hall?

Answer

1024 ÷ 2 = 512 chairs.

e.

Find the cost of plastering the four walls (without the openings) at Rs 350 a square meter.

Answer

461 × 350 = Rs 1,61,350.

Question 5

5.

The floor of the largest room in Simran’s house is 500 ft². It is 25 ft long and 9 ft high, with three windows of 6 ft × 5 ft and a door of 3 ft × 6.5 ft. Find the cost of coloring its four walls, without the windows and door, at Rs 300 a square foot.

Answer

Breadth = 500 ÷ 25 = 20 ft. Four walls = 2 × 9 × 45 = 810 ft². Openings = 3 × 30 + 19.5 = 109.5 ft².
Area = 700.5 ft²; cost = 700.5 × 300 = Rs 2,10,150.

Question 6

6.

A square room is 8 ft high. Carpeting it costs Rs 39,200 at Rs 200 a square foot. Find the cost of coloring its four walls at Rs 250 a square foot.

Answer

Floor = 39200 ÷ 200 = 196 ft², so each side is 14 ft. Four walls = 4 × 14 × 8 = 448 ft².
Cost = 448 × 250 = Rs 1,12,000.

Question 7

7.

A school pays Rs 7,056 to carpet a square room for its teachers and staff at Rs 144 a square meter. Pasting paper on the four walls, without the windows and doors, at Rs 400 a square meter costs Rs 30,000. The windows and doors have a total area of 9 m². Find the room’s height.

Answer

Floor = 7056 ÷ 144 = 49 m², so each side is 7 m. Walls without openings = 30000 ÷ 400 = 75 m²; whole walls = 84 m².
4 × 7 × h = 84, so h = 3 m.

Question 8

8.

A room in Nima’s house is twice as long as it is wide and 2.8 m high. Painting pictures on its four walls, without the windows and door, costs Rs 54,000 at Rs 900 a square meter, and the windows and door have an area of 7.2 m². Find the cost of tiling its floor at Rs 200 a square meter.

Answer

Walls without openings = 54000 ÷ 900 = 60 m²; whole walls = 67.2 m².
With breadth b and length 2b: 2 × 2.8 × 3b = 16.8b = 67.2, so b = 4 m and l = 8 m.
Floor = 32 m²; cost = 32 × 200 = Rs 6,400.

Question 9

9.

A school hall is twice as long as it is wide, and twice as wide as it is high. Coloring its four walls, windows and doors included, costs Rs 43,200 at Rs 225 a square meter. Find the cost of tiling its floor at Rs 250 a square meter.

Answer

Walls = 43200 ÷ 225 = 192 m². With height h, breadth 2h, length 4h: 2h(4h+2h)=12h2=192 2h(4h + 2h) = 12h^2 = 192 , so h = 4 m; the hall is 16 m by 8 m.
Floor = 128 m²; cost = 128 × 250 = Rs 32,000.

Question 10

a.

How many pieces of turf are needed?

Answer

Park = 2000 m²; one piece = 0.2 × 0.2 = 0.04 m²; 2000 ÷ 0.04 = 50,000 pieces.

b.

Laying one piece costs Rs 225. Find the total cost.

Answer

50000 × 225 = Rs 1,12,50,000.

c.

Some years later the park is made 15 m longer and 10 m wider. How many more pieces are needed for the new part, and what do they cost at the same rate?

Answer

New park = 65 × 50 = 3250 m²; the new part is 3250 − 2000 = 1250 m².
Pieces = 1250 ÷ 0.04 = 31,250; cost = 31250 × 225 = Rs 70,31,250.

Exercise 5.4

Question 1

a.

How is land measured in the mountains and hills of Nepal?

Answer

In ropani, aana, paisa and daam.

b.

Which units are used where you live?

Answer

In the hills, mountains and Kathmandu Valley, ropani-aana-paisa-daam; in the Terai, bigha-kattha-dhur. Square feet and square meters are understood everywhere.

c.

How many square feet are in one ropani?

Answer

5,476 ft² (a square 74 ft on each side).

d.

How many aana are in one ropani, and how many square feet in one aana?

Answer

16 aana; one aana is 342.25 ft².

e.

How many square meters are in one bigha?

Answer

6,772.63 m².

f.

How many kattha are in one bigha, and how many square feet in one kattha?

Answer

20 kattha; one kattha is 3,645 ft².

g.

Which is bigger, an aana or a kattha?

Answer

A kattha (338.63 m², against 31.79 m² for an aana): more than ten aana.

h.

A rectangular plot 37 ft wide has an area of one ropani. How long is it?

Answer

5476 ÷ 37 = 148 ft.

i.

A landowner in the Terai has 10 bigha. How many ropani is that?

Answer

10×72900÷5476≈ 10 \times 72900 \div 5476 \approx 133.13 ropani.

Question 2

a.

2 kattha

Answer

2 × 3645 = 7,290 ft².

b.

4 aana

Answer

4 × 342.25 = 1,369 ft².

c.

8 dhur

Answer

8 × 182.25 = 1,458 ft².

d.

5 aana

Answer

5 × 342.25 = 1,711.25 ft².

e.

6 aana 3 paisa

Answer

6 × 342.25 + 3 × 85.5625 = 2053.5 + 256.69 = 2,310.19 ft².

f.

4 kattha 10 dhur

Answer

4 × 3645 + 10 × 182.25 = 14580 + 1822.5 = 16,402.5 ft².

Question 3

a.

2 ropani

Answer

2 × 508.72 = 1,017.44 m².

b.

1 bigha

Answer

6,772.63 m².

c.

16 dhur

Answer

16 × 16.93 = 270.88 m².

d.

3 kattha

Answer

3 × 338.63 = 1,015.89 m².

e.

5 kattha 10 dhur

Answer

5 × 338.63 + 10 × 16.93 = 1693.15 + 169.30 = 1,862.45 m².

Question 4

a.

16 dhur

Answer

16 × 182.25 = 2916 ft²; 2916÷342.25≈ 2916 \div 342.25 \approx 8.52 aana.

b.

4 kattha

Answer

14580÷342.25≈ 14580 \div 342.25 \approx 42.60 aana.

c.

3 kattha 2 dhur

Answer

10935 + 364.5 = 11299.5 ft²; ÷342.25≈ \div 342.25 \approx 33.02 aana.

d.

6 kattha 10 dhur

Answer

21870 + 1822.5 = 23692.5 ft²; ÷342.25≈ \div 342.25 \approx 69.23 aana.

e.

2 ropani

Answer

2 × 16 = 32 aana.

f.

1 bigha

Answer

72900÷342.25≈ 72900 \div 342.25 \approx 213.00 aana (213.0003…).

Question 5

a.

3 aana

Answer

3 × 342.25 = 1026.75 ft²; ÷182.25≈ \div 182.25 \approx 5.63 dhur.

b.

6 aana 3 paisa

Answer

2310.1875÷182.25≈ 2310.1875 \div 182.25 \approx 12.68 dhur.

c.

16 kattha 5 dhur

Answer

16 × 20 + 5 = 325 dhur.

d.

1 bigha 4 kattha

Answer

24 × 20 = 480 dhur.

e.

2 bigha

Answer

40 × 20 = 800 dhur.

Question 6

6.

Aitaman has a plot of one ropani facing a road 20 ft wide; its side along the road is 148 ft. He divides it into four plots of equal area, each with its own side on the road. What is the area of each plot, and its length and breadth?

Answer

The plot is 148 × 37 ft (5476 ÷ 148 = 37). Each part: 5476 ÷ 4 = 1,369 ft² (4 aana), 37 ft along the road and 37 ft deep: a square.

Question 7

i.

3,056 ft²

Answer
  • a.3056 ÷ 5476 = 0.55807; × 16 = 8.929: 8 aana; 0.929 × 4 = 3.717: 3 paisa; 0.717 × 4 = 2.87 daam. 0-8-3-2.87.
  • b.3056 ÷ 72900 × 20 = 0.838: 0 kattha; 0.838 × 20 = 16.77. 0-0-16.77.
ii.

1,426.85 m²

Answer
  • a.1426.85 ÷ 508.72 = 2.80478: 2 ropani; 0.80478 × 16 = 12.876: 12 aana; 0.876 × 4 = 3.506: 3 paisa; 0.506 × 4 = 2.02. 2-12-3-2.02.
  • b.1426.85 ÷ 6772.63 = 0.21068; × 20 = 4.2136: 4 kattha; 0.2136 × 20 = 4.27. 0-4-4.27.
iii.

5,026.80 ft²

Answer
  • a.5026.8 ÷ 5476 = 0.91797; × 16 = 14.6875: 14 aana; 0.6875 × 4 = 2.75: 2 paisa; 0.75 × 4 = 3. 0-14-2-3.
  • b.5026.8 ÷ 72900 × 20 = 1.3791: 1 kattha; 0.3791 × 20 = 7.58. 0-1-7.58.
iv.

2,167.24 m²

Answer
  • a.2167.24 ÷ 508.72 = 4.26018: 4 ropani; × 16: 4.163, 4 aana; × 4: 0.653, 0 paisa; × 4 = 2.61. 4-4-0-2.61.
  • b.2167.24 ÷ 6772.63 × 20 = 6.4000: 6 kattha; the rest × 20 = 8.00. 0-6-8.00.
v.

80,736.75 ft²

Answer
  • a.80736.75 ÷ 5476 = 14.74375: 14 ropani; 0.74375 × 16 = 11.9: 11 aana; 0.9 × 4 = 3.6: 3 paisa; 0.6 × 4 = 2.40. 14-11-3-2.40.
  • b.80736.75 ÷ 72900 = 1.1075: 1 bigha; 0.1075 × 20 = 2.15: 2 kattha; 0.15 × 20 = 3. 1-2-3.

Question 8

8.

A Lalpurja gives a plot’s area as 0-5-3-0 (B-K-D-kanwa) and 1,743.70 m². Check it with the conversion table.

Answer

5 × 338.63 + 3 × 16.93 = 1693.15 + 50.79 = 1,743.94 m². It agrees with the Lalpurja’s 1,743.70 m² to within a quarter of a square meter, less than a sixtieth of a dhur: the record checks.

Question 9

a.

in R-A-P-D.

Answer

677.26 ÷ 508.72 = 1.33130: 1 ropani; 0.3313 × 16 = 5.3008: 5 aana; 0.3008 × 4 = 1.2032: 1 paisa; 0.2032 × 4 = 0.81. 1-5-1-0.81.

b.

in B-K-D.

Answer

677.26 ÷ 338.63 = 2 exactly: 0-2-0, 2 kattha.

c.

in square feet.

Answer

2 kattha = 2 × 3645 = 7,290 ft² (or 677.26×3.28082≈7,290 677.26 \times 3.2808^2 \approx 7{,}290 ).

Question 10

a.

Change the areas of plots 1, 2, 6, 8 and 9 into square meters.

Answer

1 aana = 31.79, 1 paisa = 7.95, 1 daam = 1.99 m².
Plot 1: 3 × 31.79 + 2 × 7.95 + 3 × 1.99 = 117.24 m².
Plot 2: 4 × 31.79 + 2 × 1.99 = 131.14 m².
Plot 6: 4 × 31.79 = 127.16 m².
Plot 8: 3 × 31.79 + 7.95 = 103.32 m².
Plot 9: 5 × 31.79 + 7.95 + 3 × 1.99 = 172.87 m².

b.

Change the areas of plots 4, 18, 20 and 22 into square feet.

Answer

Plot 4: 4 × 342.25 + 2 × 85.5625 = 1,540.13 ft².
Plot 18: 3 × 342.25 + 2 × 85.5625 + 2 × 21.390625 = 1,240.66 ft².
Plot 20: 4 × 342.25 = 1,369 ft².
Plot 22: 7 × 342.25 = 2,395.75 ft².

c.

Change the areas of plots 9 and 22 into B-K-D.

Answer

Plot 9: 172.87÷16.93≈10.21 172.87 \div 16.93 \approx 10.21 : 0-0-10.21.
Plot 22: 2395.75÷182.25≈13.15 2395.75 \div 182.25 \approx 13.15 : 0-0-13.15.

Question 11

a.
25 m16 m24 m4 m wide road
Figure 5.24
Answer

s=25+16+242=32.5 s = \frac{25 + 16 + 24}{2} = 32.5 ; area =32.5×7.5×16.5×8.5≈184.89 = \sqrt{32.5 \times 7.5 \times 16.5 \times 8.5} \approx 184.89 m².

  • a.184.89 ÷ 508.72 = 0.36344; 5 aana, 3 paisa, 1.04 daam: 0-5-3-1.04.
  • b.184.89÷16.93≈10.92 184.89 \div 16.93 \approx 10.92 : 0-0-10.92.
b.
34 ft61 ft40 ft58 ft73 ft20 ft wide road
Figure 5.25
Answer

Triangle (34, 61, 73): s = 84, 84×50×23×11≈1030.82 \sqrt{84 \times 50 \times 23 \times 11} \approx 1030.82 ft².
Triangle (73, 40, 58): s = 85.5, 85.5×12.5×45.5×27.5≈1156.41 \sqrt{85.5 \times 12.5 \times 45.5 \times 27.5} \approx 1156.41 ft².
Area ≈2187.23 \approx 2187.23 ft².

  • a.2187.23 ÷ 5476 = 0.39942: 0-6-1-2.25.
  • b.2187.23÷182.25≈12.00 2187.23 \div 182.25 \approx 12.00 : 0-0-12.00.
c.
35.9 ft75.2 ft36.4 ft76.8 ft84.5 ft13 ft wide road
Figure 5.26
Answer

Triangle (35.9, 75.2, 84.5): s = 97.8, area ≈1348.95 \approx 1348.95 ft².
Triangle (84.5, 36.4, 76.8): s = 98.85, area ≈1397.61 \approx 1397.61 ft².
Area ≈2746.55 \approx 2746.55 ft².

  • a.2746.55 ÷ 5476 = 0.50156: 0-8-0-0.40.
  • b.2746.55÷182.25≈15.07 2746.55 \div 182.25 \approx 15.07 : 0-0-15.07.
d.
80 ft75 ft81 ft90 ft108 ft13 ft wide road
Figure 5.27
Answer

Triangle (80, 75, 108): s = 131.5, area ≈2998.64 \approx 2998.64 ft².
Triangle (108, 81, 90): s = 139.5, area ≈3567.16 \approx 3567.16 ft².
Area ≈6565.80 \approx 6565.80 ft².

  • a.6565.80 ÷ 5476 = 1.19901: 1-3-0-2.95.
  • b.6565.80 ÷ 3645 = 1.8013: 1 kattha; 0.8013 × 20 = 16.03. 0-1-16.03.
e.
20 m22 m28 m23 m31.5 m4 m wide road
Figure 5.28
Answer

Triangle (20, 22, 31.5): s = 36.75, area ≈218.33 \approx 218.33 m².
Triangle (31.5, 28, 23): s = 41.25, area ≈311.86 \approx 311.86 m².
Area ≈530.18 \approx 530.18 m².

  • a.530.18 ÷ 508.72 = 1.04218: 1-0-2-2.80.
  • b.530.18 ÷ 338.63 = 1.5657: 1 kattha; 0.5657 × 20 = 11.31. 0-1-11.31.

Question 12

12.

A triangular plot has a perimeter of 160 ft, and its sides are in the ratio 5 : 7 : 8. Find its area in dhur.

Answer

20x = 160, x = 8: sides 40, 56 and 64 ft; s = 80.
Area =80×40×24×16=1228800≈1108.51 = \sqrt{80 \times 40 \times 24 \times 16} = \sqrt{1228800} \approx 1108.51 ft²; ÷182.25≈ \div 182.25 \approx 6.08 dhur.