CBSE Class 9 Mathematics Herons Formula Assignment Set 09

IX - Mathematics Assignment No - 03 - Heron's Formula - Area

Question 1. The area of a rectangular field is \(3584\text{ m}^2\) and its length is \(64\text{ m}\). A boy runs round the field at the rate of \(8\text{ km/hr}\). How long will he take to go \(5\) times round it.
Answer: Area of the rectangular field = \(3584\text{ m}^2\)
Length of the field \((l) = 64\text{ m}\)
Breadth of the field \((b) = \frac{\text{Area}}{\text{Length}} = \frac{3584}{64} = 56\text{ m}\)
Perimeter of the rectangular field \(= 2(l + b) = 2(64 + 56) = 2 \times 120 = 240\text{ m}\)
Total distance covered in \(5\) rounds \(= 5 \times 240 = 1200\text{ m} = 1.2\text{ km}\)
Speed of the boy \(= 8\text{ km/hr}\)
Time taken \(= \frac{\text{Distance}}{\text{Speed}} = \frac{1.2}{8}\text{ hours} = 0.15\text{ hours} = 0.15 \times 60\text{ minutes} = 9\text{ minutes}\)
In simple words: First find the width of the rectangle, then calculate the total distance for 5 perimeters and divide by speed to get the time.

Exam Tip: Always convert units consistently (meters to kilometers or hours to minutes) before performing final time and distance calculations.

Question 2. Find the length of altitude of an equilateral triangle of side \(36\text{ cm}\) each.
Answer: Let the side of the equilateral triangle be \(a = 36\text{ cm}\)
The altitude of an equilateral triangle is given by the formula:
Altitude \(= \frac{\sqrt{3}}{2}a\)
Altitude \(= \frac{\sqrt{3}}{2} \times 36 = 18\sqrt{3}\text{ cm}\)
In simple words: Substitute the side length into the standard altitude formula for an equilateral triangle.

Exam Tip: Memorize the altitude formula \(\frac{\sqrt{3}}{2}a\) and area formula \(\frac{\sqrt{3}}{4}a^2\) for quick solutions in equilateral triangle problems.

Question 3. The base of a parallelogram is twice its height. If the area is \(512\text{ cm}^2\), find the base and the height.
Answer: Let the height of the parallelogram be \(h\text{ cm}\)
Then, the base \((b) = 2h\text{ cm}\)
Area of the parallelogram \(=\text{base} \times \text{height}\)
\(512 = (2h) \times h\)
\(2h^2 = 512\)
\(h^2 = \frac{512}{2} = 256\)
\(h = \sqrt{256} = 16\text{ cm}\)
Height \((h) = 16\text{ cm}\)
Base \((b) = 2 \times 16 = 32\text{ cm}\)
In simple words: Set up an algebraic equation using the area formula of a parallelogram and solve for the height, then find the base.

Exam Tip: Clearly state the variables at the beginning and include proper square units for area and linear units for dimensions.

Question 4. The area of a trapezium is \(1080\text{ cm}^2\). If the lengths of its parallel sides are \(55.6\text{ cm}\) and \(34.4\text{ cm}\). Find the distance between the parallel sides.
Answer: Area of the trapezium \(= 1080\text{ cm}^2\)
Length of parallel sides, \(a = 55.6\text{ cm}\) and \(b = 34.4\text{ cm}\)
Let the distance between the parallel sides be \(h\text{ cm}\)
Area of trapezium \(= \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height}\)
\(1080 = \frac{1}{2} \times (55.6 + 34.4) \times h\)
\(1080 = \frac{1}{2} \times 90 \times h\)
\(1080 = 45 \times h\)
\(h = \frac{1080}{45} = 24\text{ cm}\)
In simple words: Add the two parallel sides, multiply by half, and divide the total area by this value to find the height.

Exam Tip: Adding decimal numbers carefully \(55.6 + 34.4 = 90.0\) simplifies the division significantly.

Question 5. The area of a trapezium is \(180\text{ cm}^2\) and its height is \(9\text{ cm}\). If one of the parallel side is \(6\text{ cm}\) long than the other. Find the two parallel sides.
Answer: Let one parallel side be \(x\text{ cm}\)
Then the other parallel side \(= (x + 6)\text{ cm}\)
Height \((h) = 9\text{ cm}\)
Area \(= 180\text{ cm}^2\)
Area of trapezium \(= \frac{1}{2} \times (x + x + 6) \times h\)
\(180 = \frac{1}{2} \times (2x + 6) \times 9\)
\(180 \times 2 = (2x + 6) \times 9\)
\(360 = 9(2x + 6)\)
\(2x + 6 = \frac{360}{9} = 40\)
\(2x = 40 - 6 = 34\)
\(x = \frac{34}{2} = 17\text{ cm}\)
First parallel side \(= 17\text{ cm}\)
Second parallel side \(= 17 + 6 = 23\text{ cm}\)
In simple words: Express both parallel sides in terms of a single variable, substitute into the trapezium area formula, and solve for the variable.

Question 6. The cost of carpeting a room \(15\text{ m}\) long with a carpet \(75\text{ cm}\) wide at Rs \(8\) per metre is Rs \(1920\text{ cm}\). [Note: transcribed verbatim as written in paper, representing Rs 1920 total cost]. Find the width of the room.
Answer: Total cost of carpeting = Rs \(1920\)
Rate of carpet = Rs \(8\) per metre
Total length of carpet required \(= \frac{\text{Total Cost}}{\text{Rate}} = \frac{1920}{8} = 240\text{ m}\)
Width of the carpet \(= 75\text{ cm} = 0.75\text{ m}\)
Area of the carpet = Area of the room
Area of carpet \(= \text{Length of carpet} \times \text{Width of carpet} = 240 \times 0.75 = 180\text{ m}^2\)
Length of the room \((l) = 15\text{ m}\)
Let the width of the room be \(b\)
Area of room \(= l \times b\)
\(180 = 15 \times b\)
\(b = \frac{180}{15} = 12\text{ m}\)
In simple words: Find the total length of the carpet by dividing total cost by rate, compute total area, and then divide by the room length to find the room width.

Exam Tip: Always convert centimeters to meters (\(75\text{ cm} = 0.75\text{ m}\)) before calculating areas involving meters.

Question 7. The area of trapezium is \(352\text{ cm}^2\). The distance between parallel sides is \(16\text{ cm}\). If one of the parallel side is \(25\text{ cm}\) find the other.
Answer: Area of trapezium \(= 352\text{ cm}^2\)
Height \((h) = 16\text{ cm}\)
Let one parallel side be \(a = 25\text{ cm}\) and the other be \(b\)
Area \(= \frac{1}{2} \times (a + b) \times h\)
\(352 = \frac{1}{2} \times (25 + b) \times 16\)
\(352 = (25 + b) \times 8\)
\(25 + b = \frac{352}{8} = 44\)
\(b = 44 - 25 = 19\text{ cm}\)
In simple words: Substitute the known values into the area formula and solve a simple linear equation for the unknown side.

Question 8. The length and width of a rectangular field are in the ratio \(3:2\). If the area of the field is \(3456\text{ m}^2\), find the cost of fencing the field at Rs \(3.50\) per metre.
Answer: Let the length \((l) = 3x\) and width \((w) = 2x\)
Area \(= l \times w = 3x \times 2x = 6x^2\)
\(6x^2 = 3456\)
\(x^2 = \frac{3456}{6} = 576\)
\(x = \sqrt{576} = 24\text{ m}\)
Length \((l) = 3 \times 24 = 72\text{ m}\)
Width \((w) = 2 \times 24 = 48\text{ m}\)
Perimeter of the field \(= 2(l + w) = 2(72 + 48) = 2 \times 120 = 240\text{ m}\)
Cost of fencing \(= \text{Perimeter} \times \text{Rate} = 240 \times 3.50 = \text{Rs } 840\)
In simple words: Use the ratio variables to find dimensions, calculate the perimeter, and multiply by the fencing rate.

Exam Tip: When given ratios, always introduce a common variable \(x\) (e.g., \(3x\) and \(2x\)) to set up equations correctly.

Question 9. The cost of cultivating a triangular field \(ABC\) at Rs \(30\) a hectare is Rs \(132\). The measure of the side \(AC\) is \(1760\) metres. Find the measure of the perpendicular drawn from \(B\) to \(AC\).
Answer: Total cost of cultivation = Rs \(132\)
Rate = Rs \(30\) per hectare
Total Area in hectares \(= \frac{132}{30} = 4.4\text{ hectares}\)
Since \(1\text{ hectare} = 10,000\text{ m}^2\),
Total Area in \(\text{m}^2 = 4.4 \times 10,000 = 44,000\text{ m}^2\)
Base \((AC) = 1760\text{ m}\)
Area of triangle \(= \frac{1}{2} \times \text{Base} \times \text{Height}\)
\(44000 = \frac{1}{2} \times 1760 \times h\)
\(44000 = 880 \times h\)
\(h = \frac{44000}{880} = 50\text{ m}\)
In simple words: Find total area in square meters by dividing total cost by rate and converting hectares, then use the triangle area formula to find the perpendicular height.

Exam Tip: Remember the crucial unit conversion: \(1\text{ hectare} = 10,000\text{ m}^2\).

Question 10. From a point in the interior of an equilateral triangle, perpendiculars are drawn to the three sides of \(8\text{ cm}\), \(10\text{ cm}\), \(11\text{ cm}\) respectively. Find the area of triangle.
Answer: Let the side of the equilateral triangle be \(a\)
Area of an equilateral triangle with side \(a\) is given by \(\frac{\sqrt{3}}{4}a^2\)
Alternatively, joining the interior point to the three vertices divides the equilateral triangle into three smaller triangles with base \(a\) and heights \(p_1 = 8\text{ cm}\), \(p_2 = 10\text{ cm}\), and \(p_3 = 11\text{ cm}\).
Total Area \(=\frac{1}{2}a p_1 + \frac{1}{2}a p_2 + \frac{1}{2}a p_3 = \frac{1}{2}a(p_1 + p_2 + p_3)\)
Sum of perpendiculars \(= 8 + 10 + 11 = 29\text{ cm}\)
Total Area \(= \frac{1}{2}a(29) = 14.5a\text{ cm}^2\)
Also, Area of equilateral triangle \(= \frac{\sqrt{3}}{4}a^2\)
\(\frac{\sqrt{3}}{4}a^2 = 14.5a\)
\(\frac{\sqrt{3}}{4}a = 14.5\)
\(a = \frac{14.5 \times 4}{\sqrt{3}} = \frac{58}{\sqrt{3}} = \frac{58 \times 1.732}{3} = 33.518\text{ cm}\)
Now, calculating the area \(= 14.5 \times 33.518 = 486.0\text{ cm}^2\) (or using exact value \(\frac{\sqrt{3}}{4} \times \left(\frac{58}{\sqrt{3}}\right)^2 = \frac{3364\sqrt{3}}{12} = \frac{841\sqrt{3}}{3} = \frac{841 \times 1.732}{3} \approx 485.6\text{ cm}^2\))
In simple words: Equate the sum of the areas of the three triangles formed by the interior point to the standard area formula of the equilateral triangle to find the side length, then compute the total area.

Exam Tip: Use \(\sqrt{3} \approx 1.732\) for final numerical approximations when required.