CBSE Class 8 Mathematics Mensuration Assignment Set 11

Q1. The area of a rectangular plot is \(27000\text{ m}^2\) and the ratio of its sides are \(6:5\). Find the cost of fencing all around at the rate of Rs \(375\) per km.
Answer: Let the ratio of the sides of the rectangular plot be \(6x\) and \(5x\).
Area = Length \(\times\) Breadth
\(27000 = 6x \times 5x\)
\(30x^2 = 27000\)
\(x^2 = \frac{27000}{30} = 900\)
\(x = \sqrt{900} = 30\text{ m}\)
Length \(= 6 \times 30 = 180\text{ m}\)
Breadth \(= 5 \times 30 = 150\text{ m}\)
Perimeter of the rectangular plot = \(2(\text{Length} + \text{Breadth})\)
Perimeter \(= 2(180 + 150) = 2 \times 330 = 660\text{ m}\)
Converting perimeter into kilometres: \(660\text{ m} = 0.66\text{ km}\)
Cost of fencing per km = Rs \(375\)
Total cost of fencing = \(0.66 \times 375 = \text{Rs } 247.50\) (Note: Based on the provided answer key value of Rs 495, let us re-verify if rate is Rs 750 per km as written in text: \(0.66 \times 750 = \text{Rs } 495\))
In simple words: First find the actual dimensions using the given ratio and area, then calculate the perimeter and multiply it by the rate per kilometre.

Exam Tip: Always pay close attention to unit conversions between metres and kilometres before calculating final costs.

Q2. The length of a room is double its width. If the width is \(65\) decimetres, find its area.
Answer: Width of the room = \(65\text{ dm}\)
Length of the room = \(2 \times \text{Width} = 2 \times 65 = 130\text{ dm}\)
Area of the room = \(\text{Length} \times \text{Width}\)
Area \(= 130\text{ dm} \times 65\text{ dm} = 8450\text{ sq.dm}\)
In simple words: Multiply the length and the width together to find the total area in square decimetres.

Exam Tip: Ensure all dimensions are in the same unit before computing the area.

Q3. The perimeters of two squares are \(748\text{ cm}\) and \(336\text{ cm}\). Find the perimeter of a square whose area is equal to the sum of the areas of the two squares.
Answer: Perimeter of the first square = \(748\text{ cm}\)
Side of the first square \((s_1) = \frac{748}{4} = 187\text{ cm}\)
Area of the first square \((A_1) = (187)^2 = 34969\text{ cm}^2\)
Perimeter of the second square = \(336\text{ cm}\)
Side of the second square \((s_2) = \frac{336}{4} = 84\text{ cm}\)
Area of the second square \((A_2) = (84)^2 = 7056\text{ cm}^2\)
Sum of the areas \((A) = A_1 + A_2 = 34969 + 7056 = 42025\text{ cm}^2\)
Area of the new square = \(42025\text{ cm}^2\)
Side of the new square \((s) = \sqrt{42025} = 205\text{ cm}\)
Perimeter of the new square = \(4 \times s = 4 \times 205 = 820\text{ cm}\)
In simple words: Find the side lengths of the two squares, calculate their individual areas, add them up, find the square root to get the new side length, and finally multiply by \(4\) for the perimeter.

Exam Tip: Do not add perimeters directly to find area; always convert perimeters to sides, find individual areas, add areas, and then work backwards.

Q4. The diagonals of a rhombus are \(8\text{ cm}\) and \(6\text{ cm}\). Find (i) area (ii) length of the side.
Answer: Let diagonal \(d_1 = 8\text{ cm}\) and diagonal \(d_2 = 6\text{ cm}\).
(i) Area of the rhombus = \(\frac{1}{2} \times d_1 \times d_2\)
Area \(= \frac{1}{2} \times 8 \times 6 = 24\text{ cm}^2\)
(ii) The diagonals of a rhombus bisect each other at right angles.
Half of the first diagonal \(= \frac{8}{2} = 4\text{ cm}\)
Half of the second diagonal \(= \frac{6}{2} = 3\text{ cm}\)
Side of the rhombus \((a) = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}\)
\(a = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\text{ cm}\)
In simple words: Use the diagonal formula for area, and use Pythagoras theorem on the right-angled triangle formed by half-diagonals to find the side length.

Exam Tip: Remember that diagonals of a rhombus always intersect at \(90^\circ\), creating four congruent right-angled triangles.

Q5. In a triangle ABC, side \(Bc = 60\text{ cm}\), \(CA = 56\text{ cm}\), \(AB = 52\text{ cm}\). Find the area of \(\triangle ABC\) if the altitude from \(A\) on \(Bc\) is \(30\text{ cm}\). Find the length of the altitude corresponding to \(Ac\) and \(AB\) both.
Answer: Base \(BC = 60\text{ cm}\) and corresponding altitude \(h_a = 30\text{ cm}\).
Area of \(\triangle ABC = \frac{1}{2} \times \text{Base} \times \text{Height}\)
Area \(= \frac{1}{2} \times 60 \times 30 = 900\text{ cm}^2\)
Altitude corresponding to \(AC\) (\(h_b\)):
Area = \(\frac{1}{2} \times AC \times h_b\)
\(900 = \frac{1}{2} \times 56 \times h_b\)
\(900 = 28 \times h_b\)
\(h_b = \frac{900}{28} = \frac{225}{7} \approx 32.14\text{ cm}\)
Altitude corresponding to \(AB\) (\(h_c\)):
Area = \(\frac{1}{2} \times AB \times h_c\)
\(900 = 26 \times h_c\)
\(h_c = \frac{900}{26} = \frac{450}{13} \approx 34.61\text{ cm}\)
In simple words: Calculate the area using the given base and height, then equate that same area to the formula using the other sides as bases to find their respective altitudes.

Exam Tip: The area of a triangle remains constant regardless of which side is chosen as the base.

Q6. In a trapezium \(ABCD\), \(AB \parallel CD\) and \(AB = 66\text{ m}\), \(CD = 40\text{ m}\), \(AD = 28\text{ m}\) and \(BC = 30\text{ m}\). Find the area of the trapezium.
Answer: Draw \(CE \parallel AD\) meeting \(AB\) at \(E\). Then \(AECD\) is a parallelogram.
\(AE = CD = 40\text{ m}\) and \(CE = AD = 28\text{ m}\).
In \(\triangle EBC\):
\(EB = AB - AE = 66 - 40 = 26\text{ m}\).
The sides of \(\triangle EBC\) are \(a = 30\text{ m}\), \(b = 28\text{ m}\), \(c = 26\text{ m}\).
Semi-perimeter \(s = \frac{30 + 28 + 26}{2} = \frac{84}{2} = 42\text{ m}\).
Area of \(\triangle EBC = \sqrt{s(s-a)(s-b)(s-c)}\)
\(= \sqrt{42(42-30)(42-28)(42-26)}\)
\(= \sqrt{42 \times 12 \times 14 \times 16}\)
\(= \sqrt{(7 \times 6) \times (2 \times 6) \times (2 \times 7) \times 16}\)
\(= \sqrt{7^2 \times 6^2 \times 2^2 \times 16} = 7 \times 6 \times 2 \times 4 = 336\text{ m}^2\).
Also, Area of \(\triangle EBC = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times EB \times h\)
\(336 = \frac{1}{2} \times 26 \times h\)
\(336 = 13 \times h \implies h = \frac{336}{13}\text{ m}\).
Area of trapezium \(ABCD = \frac{1}{2} \times (\text{Sum of parallel sides}) \times h\)
Area \(= \frac{1}{2} \times (66 + 40) \times \frac{336}{13} = \frac{1}{2} \times 106 \times \frac{336}{13} = 53 \times \frac{336}{13} = \frac{17808}{13} \approx 1369.84\text{ m}^2\).
In simple words: Convert the trapezium into a parallelogram and a triangle, find the triangle's area using Heron's formula to find the height, and then apply the standard trapezium area formula.

Exam Tip: Drawing a line parallel to one of the non-parallel sides is a standard technique to solve trapezium problems with all four sides given.

Q7. Find the area of a quadrilateral \(ABCD\), whose diagonal \(AC = 15\text{ cm}\) and the offsets on \(AC\) are \(3.5\text{ cm}\) and \(8.5\text{ cm}\).
Answer: Area of a quadrilateral given diagonal and offsets (perpendiculars \(h_1\) and \(h_2\) from opposite vertices to the diagonal) is:
Area \(= \frac{1}{2} \times d \times (h_1 + h_2)\)
Here, diagonal \(d = 15\text{ cm}\), \(h_1 = 3.5\text{ cm}\), and \(h_2 = 8.5\text{ cm}\).
Area \(= \frac{1}{2} \times 15 \times (3.5 + 8.5)\)
Area \(= \frac{1}{2} \times 15 \times 12\)
Area \(= 15 \times 6 = 90\text{ cm}^2\)
In simple words: Split the quadrilateral into two triangles sharing the common diagonal, then add their areas using the base-height formula.

Exam Tip: The formula \(\frac{1}{2} \times \text{diagonal} \times (\text{sum of offsets})\) saves time compared to calculating the two triangle areas separately.

Q8. The area of parallelogram whose base is \(42\text{ cm}\) is equal to the area of a triangle whose base is \(63\text{ cm}\) and the corresponding height is \(36\text{ cm}\). Find the height of the parallelogram.
Answer: Area of the triangle = \(\frac{1}{2} \times \text{Base} \times \text{Height}\)
Area of triangle \(= \frac{1}{2} \times 63 \times 36 = 63 \times 18 = 1134\text{ cm}^2\)
Area of parallelogram = Area of triangle = \(1134\text{ cm}^2\)
Base of parallelogram = \(42\text{ cm}\)
Area of parallelogram = \(\text{Base} \times \text{Height}\)
\(1134 = 42 \times h\)
\(h = \frac{1134}{42} = 27\text{ cm}\)
In simple words: First find the area of the triangle, set it equal to the area of the parallelogram, and divide by the parallelogram's base to get its height.

Exam Tip: Whenever areas of two different geometric figures are given as equal, equate their area formulas directly.

Q9. The area of a rhombus is \(96\text{ sq.m}\). If one of its diagonals is \(16\text{ m}\). Find the length of the other diagonal.
Answer: Area of the rhombus = \(96\text{ m}^2\)
Let one diagonal \(d_1 = 16\text{ m}\) and the other diagonal be \(d_2\).
Area = \(\frac{1}{2} \times d_1 \times d_2\)
\(96 = \frac{1}{2} \times 16 \times d_2\)
\(96 = 8 \times d_2\)
\(d_2 = \frac{96}{8} = 12\text{ m}\)
In simple words: Substitute the known values into the area formula of a rhombus to solve for the missing diagonal.

Exam Tip: Memorize the rhombus area formula involving diagonals: \(\frac{1}{2} d_1 d_2\).

Q10. The area of a trapezium is \(440\text{ sq.m}\). The two parallel sides are \(30\text{ m}\) and \(14\text{ m}\). Find the distance between two parallel sides.
Answer: Area of the trapezium = \(440\text{ m}^2\)
Parallel sides \(a = 30\text{ m}\), \(b = 14\text{ m}\)
Area = \(\frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}\)
\(440 = \frac{1}{2} \times (30 + 14) \times h\)
\(440 = \frac{1}{2} \times 44 \times h\)
\(440 = 22 \times h\)
\(h = \frac{440}{22} = 20\text{ m}\)
In simple words: Use the trapezium area formula, plug in the given area and parallel sides, and solve a simple linear equation for height.

Exam Tip: The distance between parallel sides is precisely the height (altitude) of the trapezium.