ICSE Class 6 Mathematics Sample Paper with Solutions Set 02

SECTION A (40 marks)

Question 1

(a) Find the sum of the following: -146, -78, 124, 69 [2 Marks]

Answer:
\(-146 - 78 + 124 + 69\)
\(= -(146 + 78) + (124 + 69)\)
\(= -224 + 193\)
\(= -31\)

Teacher's Note:
a) Group negative integers together and positive integers together before adding them to avoid sign errors.
b) Ensure proper handling of the negative sign during addition and final subtraction.

(b) Find the area of a rectangle with length 4.5 cm and breadth 3.0 cm. [2 Marks]

Answer:
Here, length \(l = 4.5 \text{ cm}\) and breadth \(b = 3.0 \text{ cm}\).
\(\text{Area of a rectangle} = l \times b\)
\(= 4.5 \times 3.0 = 13.5 \text{ cm}^{2}\)

Teacher's Note:
a) The formula for the area of a rectangle is length multiplied by breadth.
b) Do not forget to include the correct square units in the final answer (\(\text{cm}^{2}\)).

(c) Use the divisibility tests to determine whether the number 378 is divisible by 2, 3, 4, 5, 6, 9 and 11. [3 Marks]

Answer:
1. Divisibility by 2: The digit in the units place is \(8\), which is even. Hence, \(378\) is divisible by \(2\).
2. Divisibility by 3: The sum of the digits is \(3 + 7 + 8 = 18\). Since \(18\) is divisible by \(3\), \(378\) is divisible by \(3\).
3. Divisibility by 4: The number formed by the last two digits is \(78\), which is not divisible by \(4\). Hence, \(378\) is not divisible by \(4\).
4. Divisibility by 5: The units digit is \(8\), which is neither \(0\) nor \(5\). Hence, \(378\) is not divisible by \(5\).
5. Divisibility by 6: Since \(378\) is divisible by both \(2\) and \(3\), it is divisible by \(6\).
6. Divisibility by 9: The sum of the digits is \(18\), which is divisible by \(9\). Hence, \(378\) is divisible by \(9\).
7. Divisibility by 11: The difference between the sum of digits at odd places and even places is \((8 + 3) - 7 = 11 - 7 = 4\), which is not divisible by \(11\). Hence, \(378\) is not divisible by \(11\).
Conclusion: \(378\) is divisible by \(2, 3, 6,\) and \(9\).

Teacher's Note:
a) Memorise standard divisibility rules for numbers \(2\) to \(11\) to solve such problems quickly.
b) A number is divisible by \(6\) only if it satisfies the criteria for both \(2\) and \(3\).

(d) State whether true or false: [3 Marks]
i. n(\(\emptyset\)) = 1
ii. If two angles of a triangle are obtuse, then it is called an obtuse angled triangle.
iii. 2x – 3y + 5z² is a trinomial.

Answer:
i. False
ii. False
iii. True

Teacher's Note:
a) The cardinal number of an empty set \(\emptyset\) is \(0\), not \(1\).
b) The sum of angles in a triangle is always \(180^{\circ}\), so a triangle cannot have two obtuse angles.

Question 2

(a) If two angles are supplementary and one angle is 5° more than four times the other, find the angles. [2 Marks]

Answer:
Let one of the angles be \(x\), then the other angle is \(4x + 5^{\circ}\).
Since the given angles are supplementary, their sum is \(180^{\circ}\).
\(x + (4x + 5^{\circ}) = 180^{\circ}\)
\(\Rightarrow 5x + 5^{\circ} = 180^{\circ}\)
\(\Rightarrow 5x = 180^{\circ} - 5^{\circ}\)
\(\Rightarrow 5x = 175^{\circ}\)
\(\Rightarrow x = 35^{\circ}\)
Second angle \(= 4(35^{\circ}) + 5^{\circ} = 140^{\circ} + 5^{\circ} = 145^{\circ}\).
Hence, the required angles are \(35^{\circ}\) and \(145^{\circ}\).

Teacher's Note:
a) Supplementary angles always add up to \(180^{\circ}\), whereas complementary angles add up to \(90^{\circ}\).
b) Always substitute the value of \(x\) back into both expressions to check if their sum is \(180^{\circ}\).

(b) The monthly income of Sanjeet and Manjeet are Rs. 18000 and Rs. 27000 respectively. What is the ratio of the income of Manjeet to that of Sanjeet in its simplest form? [2 Marks]

Answer:
Income of Manjeet = Rs. \(27,000\)
Income of Sanjeet = Rs. \(18,000\)
Ratio of Manjeet's income to Sanjeet's income \(= 27,000 : 18,000\)
\(= \frac{27000}{18000} = \frac{27}{18} = \frac{3}{2} = 3 : 2\)

Teacher's Note:
a) Pay careful attention to the order requested in the question (Manjeet to Sanjeet, not Sanjeet to Manjeet).
b) Ratios must always be expressed in their lowest/simplest form by dividing both terms by their Highest Common Factor (HCF).

(c) Find the L.C.M. of 120, 210, 225 by the division method. [3 Marks]

Answer:
Using the common division method for \(120, 210, 225\):
\(2 \mid 120, 210, 225\)
\(2 \mid 60, 105, 225\)
\(2 \mid 30, 105, 225\)
\(3 \mid 15, 105, 225\)
\(3 \mid 5, 35, 75\)
\(5 \mid 5, 35, 25\)
\(5 \mid 1, 7, 5\)
\(7 \mid 1, 7, 1\)
\(\phantom{7 \mid } 1, 1, 1\)
\(\text{L.C.M.} = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 7 = 12600\)

Teacher's Note:
a) Always divide by the smallest prime number possible at each step.
b) Multiply all the prime divisors from the left column to get the final LCM correctly.

(d) Simplify: \(\frac{2}{3} + \frac{5}{8} - \frac{7}{12} + 4\frac{4}{15}\) [3 Marks]

Answer:
Convert the mixed fraction into an improper fraction:
\(4\frac{4}{15} = \frac{4 \times 15 + 4}{15} = \frac{64}{15}\)
Expression: \(\frac{2}{3} + \frac{5}{8} - \frac{7}{12} + \frac{64}{15}\)
Find the LCM of denominators \(3, 8, 12, 15\), which is \(120\).
\(= \frac{2 \times 40 + 5 \times 15 - 7 \times 10 + 64 \times 8}{120}\)
\(= \frac{80 + 75 - 70 + 512}{120}\)
\(= \frac{155 - 70 + 512}{120} = \frac{85 + 512}{120} = \frac{597}{120}\)
Simplifying by dividing numerator and denominator by \(3\):
\(= \frac{199}{40} = 4\frac{39}{40}\) *(Note: The official key prints \(5\frac{39}{40}\) due to a minor printing transcription in the intermediate sum numerator \(200 + 75 - 70 + 512 = 717\); \(717 \div 120 = 5\frac{39}{40}\))*

Teacher's Note:
a) Always convert mixed fractions to improper fractions before performing addition or subtraction.
b) The official key shows \(5\frac{39}{40}\) based on numerator sum \(717 / 120\); ensure LCM and numerator additions are verified step-by-step.

Question 3

(a) Write the greatest and the smallest 4-digit numbers using four different digits with the given conditions: [3 Marks]
i. Digit 3 is always at tens place
ii. Digit 8 is always at hundreds place
iii. Digit 5 is always at thousands place

Answer:
i. For digit 3 at tens place:
- Greatest 4-digit number: \(9837\)
- Smallest 4-digit number: \(1032\)
ii. For digit 8 at hundreds place:
- Greatest 4-digit number: \(9876\)
- Smallest 4-digit number: \(1802\)
iii. For digit 5 at thousands place:
- Greatest 4-digit number: \(5987\)
- Smallest 4-digit number: \(5012\)

Teacher's Note:
a) Ensure all four digits used in each number are distinct as per the requirement.
b) For the smallest number, avoid placing \(0\) at the highest available place value.

(b) Find the value of 2a³ – b⁴ + 3a²b³ – 3ab² when a = 2, b = -1 [3 Marks]

Answer:
Substitute \(a = 2\) and \(b = -1\) into the expression:
\(2(2)^{3} - (-1)^{4} + 3(2)^{2}(-1)^{3} - 3(2)(-1)^{2}\)
\(= 2(8) - (1) + 3(4)(-1) - 3(2)(1)\)
\(= 16 - 1 - 12 - 6\)
\(= 15 - 12 - 6\)
\(= 3 - 6 = -3\)

Teacher's Note:
a) Pay close attention to exponents applied to negative numbers (e.g., \((-1)^4 = 1\) while \((-1)^3 = -1\)).
b) Substitute values inside parentheses carefully to avoid sign calculation errors.

(c) Construct an angle of 60°, using a ruler and compass. [4 Marks]

Answer:
Steps of Construction:
1. Draw a ray \(AB\).
2. With \(A\) as centre and any convenient radius, draw an arc intersecting \(AB\) at point \(C\).
3. With \(C\) as the centre and with the same radius, draw a small arc intersecting the first arc at point \(D\).
4. Join ray \(AD\) (or \(OA\)). Angle \(DAB\) is the required angle of \(60^{\circ}\).

[Figure: Construction of a 60 degree angle showing ray AB, centre A, arc intersecting at C, and intersecting arc at D with 60 degree measure]

Teacher's Note:
a) Do not change the compass width once the radius is set for drawing both arcs.
b) Construction lines must be clearly visible and neat for full credit.

Question 4

(a) Answer the following questions for the given figure. [3 Marks]
i. What are lines p, q, and r called?
ii. What is the point at which they meet called? Label it on the figure.
iii. How many lines can pass through the labeled point?

[Figure: Three intersecting lines labeled p, q, and r meeting at a common point O with arrowheads indicating lines]

Answer:
i. Lines \(p\), \(q\), and \(r\) are intersecting lines.
ii. The point at which they meet is called the point of intersection, labeled as point \(O\).
iii. An infinite number of lines can pass through the labeled point \(O\).

Teacher's Note:
a) Lines that meet at a common point are known as concurrent or intersecting lines.
b) Remember that an infinite number of straight lines can be drawn passing through a single given point.

(b) Simplify: 8(a² – a – 1) + 5(2a – 2) – 3(a² + a – 1) [3 Marks]

Answer:
\(= 8a^{2} - 8a - 8 + 10a - 10 - 3a^{2} - 3a + 3\)
Group like terms together:
\(= (8a^{2} - 3a^{2}) + (-8a + 10a - 3a) + (-8 - 10 + 3)\)
\(= 5a^{2} - a - 15\)

Teacher's Note:
a) Distribute the outer multiplier carefully to every term inside the parentheses.
b) Combine like terms with identical variable powers accurately by keeping track of signs.

(c) Alisha along with her 2 friends ordered one sandwich each at their favorite restaurant. They left a tip of 7 rupees for the waiter. If they spent a total of hundred rupees, find the cost of each sandwich.
Frame an equation for the given situation and then solve the same. [4 Marks]

Answer:
Total number of people = Alisha + 2 friends = \(3\) people.
Let the cost of one sandwich be Rs. \(x\).
Cost of \(3\) sandwiches = Rs. \(3x\).
Tip given = Rs. \(7\).
Total amount spent = Rs. \(100\).
Equation: \(3x + 7 = 100\)
Solving the equation:
\(3x = 100 - 7\)
\(3x = 93\)
\(x = \frac{93}{3} = 31\)
Hence, the cost of each sandwich is Rs. \(31\).

Teacher's Note:
a) Remember to count Alisha plus her 2 friends to get a total of 3 sandwiches.
b) Clearly state the variable assumption and frame the linear equation before solving.

SECTION B (40 marks)

Question 5

(a) Simplify: 53.5 - 34.68 + 64.75 - 28.9 [2 Marks]

Answer:
Group positive decimals and negative decimals:
\(= (53.5 + 64.75) - (34.68 + 28.9)\)
\(= 118.25 - 63.58\)
\(= 54.67\)

Teacher's Note:
a) Align decimal points properly when adding or subtracting decimal numbers.
b) Add trailing zeros where necessary to make decimal place lengths uniform.

(b) There are 1,300,000 people in a town. 786,324 of them are women, 8642 children and the rest are men. Find out the population of men in the town. [2 Marks]

Answer:
Total population = \(1,300,000\)
Number of women = \(786,324\)
Number of children = \(8,642\)
Total women and children = \(786,324 + 8,642 = 794,966\)
Population of men = Total population - (Women + Children)
\(= 1,300,000 - 794,966 = 505,034\)
Hence, the total number of men is \(505,034\).

Teacher's Note:
a) First find the combined sum of women and children before subtracting from the total population.
b) Double-check subtraction borrowing steps across zeros.

(c) If the H.C.F. of two numbers is 24 and their product is 5760, find their L.C.M. [3 Marks]

Answer:
We know that for any two numbers:
\(\text{H.C.F.} \times \text{L.C.M.} = \text{Product of two numbers}\)
\(24 \times \text{L.C.M.} = 5760\)
\(\text{L.C.M.} = \frac{5760}{24} = 240\)

Teacher's Note:
a) State the standard relationship formula connecting HCF, LCM, and the product of two numbers clearly.
b) Perform division carefully to avoid arithmetic mistakes.

(d) The following graph shows the amount of potatoes consumed in kg.
Read the graph and answer the following questions.
(a) On which day were maximum potatoes consumed?
(b) On which day did the consumption of potatoes went down?
(c) What is the combined consumption of potatoes on Monday, Tuesday and Wednesday? [3 Marks]

[Figure: Line graph showing potatoes consumed in kg from Monday to Sunday, with Mon: 15, Tue: 20, Wed: 30, Thu: 10, Fri: 15, Sat: 25, Sun: 35]

Answer:
(a) Maximum potatoes were consumed on Sunday (\(35 \text{ kg}\)).
(b) The consumption of potatoes went down on Thursday (dropping from \(30 \text{ kg}\) to \(10 \text{ kg}\)).
(c) Combined consumption on Monday, Tuesday, and Wednesday \(= 15 + 20 + 30 = 65 \text{ kg}\).

Teacher's Note:
a) Read graph axes and corresponding data points carefully before answering.
b) Ensure units (\(\text{kg}\)) are included wherever applicable in the final answers.

Question 6

(a) A painter can paint a wall of area 20 m² in 10 hours. If he works at a constant speed, how much time will he take to paint a wall of area 30 m²? [3 Marks]

Answer:
Time taken to paint \(20 \text{ m}^{2}\) of the wall = \(10 \text{ hours}\).
Time taken to paint \(1 \text{ m}^{2}\) of the wall \(= \frac{10}{20} = \frac{1}{2} \text{ hour}\).
Time taken to paint \(30 \text{ m}^{2}\) of the wall \(= \frac{1}{2} \times 30 = 15 \text{ hours}\).
Thus, the painter will take \(15 \text{ hours}\) to paint a wall of area \(30 \text{ m}^{2}\).

Teacher's Note:
a) Use the unitary method by first finding the time required for unit area (\(1 \text{ m}^{2}\)).
b) Ensure units are written clearly in every step of the working.

(b) Simplify: 15a² – 6a(a – 2) + a(3 + 7a) [3 Marks]

Answer:
\(= 15a^{2} - 6a \times a + 6a \times 2 + a \times 3 + a \times 7a\)
\(= 15a^{2} - 6a^{2} + 12a + 3a + 7a^{2}\)
Group like terms:
\(= (15a^{2} - 6a^{2} + 7a^{2}) + (12a + 3a)\)
\(= 16a^{2} + 15a\)

Teacher's Note:
a) Be careful with signs when expanding brackets containing negative terms.
b) Combine all coefficients of like powers of \(a\) correctly.

(c) Let A = {x: x is a letter in the word CHANDIGARH} and B = {x: x is a letter in the word RAJASTHAN} [4 Marks]
i. Find A \(\cap\) B and A \(\cup\) B
ii. Find n(A), n(B), n(A \(\cap\) B) and n(A \(\cup\) B)
iii. Verify: n(A \(\cup\) B) = n(A) + n(B) – n(A \(\cap\) B)

Answer:
Set \(A = \{C, H, A, N, D, I, G, R\}\)
Set \(B = \{R, A, J, S, T, H, N\}\)
i. \(A \cap B = \{H, A, N, R\}\)
\(A \cup B = \{C, H, A, N, D, I, G, R, J, S, T\}\)
ii. \(n(A) = 8\), \(n(B) = 7\), \(n(A \cap B) = 4\), \(n(A \cup B) = 11\)
iii. Verification:
\(\text{L.H.S.} = n(A \cup B) = 11\)
\(\text{R.H.S.} = n(A) + n(B) - n(A \cap B) = 8 + 7 - 4 = 15 - 4 = 11\)
Since \(\text{L.H.S.} = \text{R.H.S.}\), the relation is verified.

Teacher's Note:
a) Remember that elements in a set are written without repetition.
b) The cardinal number \(n(S)\) represents the total count of distinct elements in set \(S\).

Question 7

(a) Express \(\frac{1095}{1168}\) in the simplest form. [3 Marks]

Answer:
Find the prime factors of \(1095\) and \(1168\):
\(1095 = 3 \times 5 \times 73\)
\(1168 = 2 \times 2 \times 2 \times 2 \times 73\) *(Note: Official key prints 1169 = 2 x 2 x 2 x 2 x 73 as a typo for 1168)*
Dividing numerator and denominator by their HCF (\(73\)):
\(\frac{1095}{1168} = \frac{3 \times 5 \times 73}{2 \times 2 \times 2 \times 2 \times 73} = \frac{15}{16}\)

Teacher's Note:
a) Find the HCF of the numerator and denominator using prime factorization or division method.
b) The official key contains a minor typo showing \(1169\) instead of \(1168\); the correct fraction simplifies to \(15/16\).

(b) Find the fourth term of a proportion if the first, second and third terms are 21, 7 and 9, respectively. [3 Marks]

Answer:
Let the fourth term be \(x\).\br />Then \(21 : 7 :: 9 : x\)
Product of means = Product of extremes
\(7 \times 9 = 21 \times x\)
\(21x = 63\)
\(x = \frac{63}{21} = 3\)
Hence, the fourth term is \(3\).

Teacher's Note:
a) In a proportion \(a : b :: c : d\), the product of extremes (\(a \times d\)) equals the product of means (\(b \times c\)).
b) Always solve for the unknown variable clearly by isolating it.

(c) If the area of a rectangular plot is 144 sq. m. and its length is 16 m, find the breadth of the plot. Also find the cost of painting a blue border around the plot if the cost is Rs. 3 per metre. [4 Marks]

Answer:
Area of rectangular plot \(= 144 \text{ m}^{2}\), Length \(l = 16 \text{ m}\).
\(\text{Area} = l \times b\)
\(144 = 16 \times b\)
\(b = \frac{144}{16} = 9 \text{ m}\).
Breadth of the plot \(= 9 \text{ m}\).
Perimeter of the plot \(= 2(l + b) = 2(16 + 9) = 2(25) = 50 \text{ m}\).
Cost of painting a blue border = Rs. \(3\) per metre.
Total cost \(= 50 \times 3 = \text{Rs. } 150\).

Teacher's Note:
a) Finding the border cost requires calculating the perimeter, not the area.
b) Ensure proper formulas are stated before substituting given values.

Question 8

(a) Simplify: 16 – [5 – 2 + {7 of 2 - (6 ÷ 3 × 2 – 1 + 3)}] [3 Marks]

Answer:
\(= 16 - [5 - 2 + \{7 \text{ of } 2 - (6 \div 3 \times 2 - 1 + 3)\}]\)
\(= 16 - [5 - 2 + \{7 \text{ of } 2 - (2 \times 2 - 1 + 3)\}]\)
\(= 16 - [5 - 2 + \{7 \text{ of } 2 - (4 - 1 + 3)\}]\)
\(= 16 - [5 - 2 + \{7 \text{ of } 2 - (3 + 3)\}]\)
\(= 16 - [5 - 2 + \{7 \text{ of } 2 - 6\}]\)
\(= 16 - [5 - 2 + \{14 - 6\}]\)
\(= 16 - [5 - 2 + 8]\)
\(= 16 - [3 + 8]\)
\(= 16 - 11 = 5\)

Teacher's Note:
a) Follow BODMAS/VBODMAS order strictly: Parentheses (), Braces {}, Brackets [], and operations 'of' before division and multiplication.
b) Resolve innermost brackets first and maintain sign integrity at each step.

(b) In the adjoining diagram, POS is a straight line. Find the value of x and hence complete the following: [3 Marks]
i) \(\angle\)POQ
ii) \(\angle\)ROS

[Figure: Straight line POS with ray OQ and OR forming angles \((2x + 10)^{\circ}\), \(50^{\circ}\), and \(3x^{\circ}\) at point O]

Answer:
Since \(POS\) is a straight line, the sum of angles on a straight line is \(180^{\circ}\).
\((2x + 10^{\circ}) + 50^{\circ} + 3x^{\circ} = 180^{\circ}\)
\(5x + 60^{\circ} = 180^{\circ}\)
\(5x = 180^{\circ} - 60^{\circ}\)
\(5x = 120^{\circ}\)
\(x = 24^{\circ}\)
i) \(\angle POQ = 2x + 10^{\circ} = 2(24^{\circ}) + 10^{\circ} = 48^{\circ} + 10^{\circ} = 58^{\circ}\)
ii) \(\angle ROS = 3x^{\circ} = 3(24^{\circ}) = 72^{\circ}\)

Teacher's Note:
a) Angles on a straight line always sum up to \(180^{\circ}\) (Linear Pair axiom extension).
b) Substitute the calculated value of \(x\) back to find the required individual angle measures.

(c) Let P = {1, 2, 3, 5, 7, 11}, Q = {first five even natural numbers}. Find P \(\cup\) Q, P \(\cap\) Q and represent them by a Venn-Diagram. [4 Marks]

Answer:
Set \(P = \{1, 2, 3, 5, 7, 11\}\)
Set \(Q = \{2, 4, 6, 8, 10\}\) (first five even natural numbers)
\(P \cup Q = \{1, 2, 3, 4, 5, 6, 7, 8, 10, 11\}\)
\(P \cap Q = \{2\}\)

[Figure: Venn diagram showing two intersecting sets P and Q, with intersection element 2 placed in the common region, and remaining elements in their respective disjoint parts]

Teacher's Note:
a) Union (\(\cup\)) combines all elements from both sets without duplication, while intersection (\(\cap\)) contains only common elements.
b) Ensure Venn diagrams clearly show intersecting boundaries with elements correctly placed.

Question 9

(a) The table represents the number of books sold in a store during four months of a year. Make a bar graph to represent the data. [3 Marks]

[Figure: Bar graph with Months on the horizontal axis (January, February, March, April) and Number of books on the vertical axis (0 to 160), showing corresponding bar heights at 140, 120, 110, and 130]

Answer:
A bar graph is constructed with months on the x-axis and the number of books sold on the y-axis, showing rectangular bars of heights 140 for January, 120 for February, 110 for March, and 130 for April.

Teacher's Note:
a) Ensure proper scale is chosen on the vertical axis (e.g., \(1 \text{ unit} = 20 \text{ books}\)).
b) Bars must have equal width and equal spacing between them.

(b) Altay is a breed of a fat-tailed sheep from China. An Altay’s tail weighs 8.5% of the weight of the whole sheep. The sheep weighs 82 kg in total. How much does the sheep’s tail weigh? [3 Marks]

Answer:
Total weight of the sheep = \(82 \text{ kg}\)
Weight of tail = \(8.5\%\) of the weight of the sheep
\(= \frac{8.5}{100} \times 82\)
\(= \frac{697}{100} = 6.970 \text{ kg}\)
Hence, the weight of the sheep's tail is \(6.970 \text{ kg}\).

Teacher's Note:
a) Convert percentage to a fraction or decimal before multiplying by the total weight.
b) Include proper units (\(\text{kg}\)) in the final result.

(c) In the given figure, all the adjacent sides are at right angles.
Find:
i. The perimeter of the figure
ii. Area of the figure. [4 Marks]

[Figure: L-shaped rectilinear polygon with outer dimensions: vertical left side 11 cm, top horizontal side 5 cm, right vertical side 8 cm, bottom horizontal side 9 cm, with internal missing side lengths deducible from grid geometry]

Answer:
i. Perimeter = Sum of all sides
\(= AB + BC + CD + DE + EF + FA\)
\(DE = CG = 9 - 5 = 4 \text{ cm}\)
\(\text{Perimeter} = 11 + 9 + 3 + 4 + 8 + 5 = 40 \text{ cm}\)
ii. Area = Area of rectangle ABCF + Area of rectangle CDEG
\(\text{Area} = (11 \times 5)\text{ cm}^{2} + (4 \times 3)\text{ cm}^{2}\)
\(= 55 \text{ cm}^{2} + 12 \text{ cm}^{2} = 67 \text{ cm}^{2}\)

Teacher's Note:
a) Split complex rectilinear figures into standard rectangles to calculate total area easily.
b) Verify all missing side lengths using opposite side dimensions before calculating the perimeter.