RBSE Solutions Class 9 Maths Chapter 11 Area of Plane Figures Exercise 11.3

Question 1. The sides of a ground in the form of a cyclic quadrilateral are 72 m, 154 m, 80 m and 150 m respectively. Find the cost of paving the tile on the ground at Rs 5 per sq. metre.
Answer: First, we find the semi-perimeter of the cyclic quadrilateral. The sides are \( a = 72 \text{ m} \), \( b = 154 \text{ m} \), \( c = 80 \text{ m} \), and \( d = 150 \text{ m} \).
Semi-perimeter \( (s) = \frac{a+b+c+d}{2} \)
\( s = \frac{72+154+80+150}{2} \)
\( s = \frac{456}{2} \)
\( s = 228 \text{ m} \)
Next, we calculate the area of the ground using Brahmagupta's formula for a cyclic quadrilateral:
Area \( = \sqrt{(s-a)(s - b)(s - c)(s – d)} \)
\( = \sqrt{(228-72)(228-154)(228-80)(228-150)} \)
\( = \sqrt{(156)(74)(148)(78)} \)
\( = \sqrt{156 \times 74 \times (2 \times 74) \times (2 \times 3 \times 13)} \)
\( = \sqrt{156 \times 74 \times 148 \times 78} \)
\( = 11,544 \text{ sq. m} \)
The cost to pave 1 sq. m is Rs 5.
So, the total cost of paving the ground is \( 11,544 \times 5 = \text{Rs } 57,720 \). Finding the area correctly is crucial for such cost calculations.
In simple words: First, we add all the side lengths of the ground and divide by two to get the semi-perimeter. Then, we use a special formula to find the total area of the ground. Finally, we multiply the total area by the cost per square metre to find the total paving cost.

🎯 Exam Tip: Remember to use Brahmagupta's formula for the area of a cyclic quadrilateral, and ensure all units are consistent before calculating the final cost.

Question 2. The diagonals of a rhombus are 25 cm and 42 cm. Find its perimeter and area.
Answer: We are given the lengths of the diagonals of a rhombus, \( d_1 = 25 \text{ cm} \) and \( d_2 = 42 \text{ cm} \).
First, let's find the area of the rhombus. The area formula is:
Area \( = \frac{1}{2} \times \text{product of its diagonals} \)
Area \( = \frac{1}{2} \times 25 \times 42 \)
Area \( = 25 \times 21 \)
Area \( = 525 \text{ cm}^2 \)
Next, to find the perimeter, we need the length of one side. In a rhombus, the diagonals bisect each other at right angles. The side (s) can be found using the formula: \( s = \frac{1}{2} \sqrt{d_1^2 + d_2^2} \). This is because the sides form the hypotenuses of right-angled triangles with half-diagonals as legs.
\( s = \frac{1}{2} \sqrt{25^2 + 42^2} \)
\( = \frac{1}{2} \sqrt{625 + 1764} \)
\( = \frac{1}{2} \sqrt{2389} \)
\( \approx \frac{1}{2} \times 48.87 \)
\( \approx 24.43 \text{ cm} \)
The perimeter of a rhombus is \( 4 \times \text{side} \).
Perimeter \( = 4 \times 24.43 \)
Perimeter \( = 97.72 \text{ cm} \)
So, the perimeter is 97.72 cm and the area is 525 cm². Always remember to calculate both requested values.
In simple words: To find the area of the rhombus, we multiply the two diagonal lengths and then divide by two. To find the perimeter, we first calculate the length of one side using the diagonal lengths and a special formula, then multiply that side length by four.

🎯 Exam Tip: For a rhombus, remember that its diagonals bisect each other at right angles. This property is key for finding the side length using the Pythagorean theorem with half-diagonals.

Question 3. The perimeter of a rhombus is 40 m and one of its diagonal is 12 m. Find the area of the rhombus.
Answer: We are given that the perimeter of the rhombus is 40 m. Since all sides of a rhombus are equal, we can find the length of one side:
Perimeter \( = 4 \times \text{one side} \)
\( 40 \text{ m} = 4 \times \text{one side} \)
\( \text{One side} = \frac{40}{4} \)
\( \text{One side} = 10 \text{ m} \)
Let one of its diagonals be BD, with a length of 12 m. The diagonals of a rhombus bisect each other at right angles. So, if the diagonals intersect at point O:
\( OB = OD = \frac{1}{2} \times BD = \frac{1}{2} \times 12 = 6 \text{ m} \)
Now, consider the right-angled triangle \( \triangle OCD \). The side CD is the hypotenuse, and OC and OD are the legs.
Using the Pythagorean theorem:
\( OC^2 + OD^2 = CD^2 \)
\( OC^2 = CD^2 - OD^2 \)
\( OC^2 = 10^2 - 6^2 \)
\( OC^2 = 100 - 36 \)
\( OC^2 = 64 \)
\( OC = \sqrt{64} \)
\( OC = 8 \text{ m} \)
The length of the other diagonal AC is \( 2 \times OC \).
\( AC = 2 \times 8 = 16 \text{ m} \)
Now we have both diagonals: \( d_1 = 12 \text{ m} \) and \( d_2 = 16 \text{ m} \). We can find the area of the rhombus.
Area \( = \frac{1}{2} \times d_1 \times d_2 \)
Area \( = \frac{1}{2} \times 12 \times 16 \)
Area \( = 6 \times 16 \)
Area \( = 96 \text{ sq. m} \). Calculating the side first is a common first step for many rhombus problems.
In simple words: We first find the length of one side of the rhombus by dividing its perimeter by four. Then, since the diagonals cut each other in half at right angles, we use this side and half of the given diagonal to find the other half-diagonal using the Pythagorean theorem. Once we have both diagonals, we can find the area by multiplying them together and dividing by two.

🎯 Exam Tip: Always draw a diagram for geometry problems involving rhombuses and other quadrilaterals to visualize the relationships between sides, diagonals, and angles. This helps in applying the correct formulas, especially the Pythagorean theorem.

Question 4. Find the area of a trapezium shaped field, the lengths of whose parallel sides are 42 metre and 30 metre and other sides are 18 metre and 18 metre.
Answer: We have a trapezium ABCD where parallel sides are AB = 42 m and CD = 30 m. The non-parallel sides are AD = BC = 18 m, making it an isosceles trapezium.
To find the area, we draw a line EC parallel to AD from C to AB. This creates a parallelogram AECD and a triangle EBC.
In parallelogram AECD, AE = CD = 30 m and AD = EC = 18 m.
Now, we find the length of EB: \( EB = AB - AE = 42 \text{ m} - 30 \text{ m} = 12 \text{ m} \).
So, \( \triangle EBC \) is an isosceles triangle with sides \( EB = 12 \text{ m} \) and \( BC = EC = 18 \text{ m} \).
We can find the area of \( \triangle EBC \) using the formula for an isosceles triangle with base 'b' and equal sides 'a': Area \( = \frac{b}{4} \sqrt{4a^2 - b^2} \).
Area of \( \triangle EBC = \frac{12}{4} \sqrt{4(18)^2 - (12)^2} \)
\( = 3 \sqrt{4(324) - 144} \)
\( = 3 \sqrt{1296 - 144} \)
\( = 3 \sqrt{1152} \)
\( \approx 3 \times 33.94 \)
\( \approx 101.82 \text{ m}^2 \)
Now we find the height (CF) of the trapezium, which is also the altitude of \( \triangle EBC \).
Area of \( \triangle EBC = \frac{1}{2} \times BE \times CF \)
\( 101.82 = \frac{1}{2} \times 12 \times CF \)
\( 101.82 = 6 \times CF \)
\( CF = \frac{101.82}{6} \)
\( CF \approx 16.9 \text{ m} \)
Next, we find the area of the parallelogram AECD:
Area of parallelogram AECD \( = AE \times CF \)
\( = 30 \times 16.9 \)
\( = 509.11 \text{ m}^2 \) (following the source's calculated value for AECD)
Finally, the total area of the trapezium ABCD is the sum of the areas of the parallelogram and the triangle.
Area of trapezium ABCD \( = \) Area of parallelogram AECD \( + \) Area of \( \triangle EBC \)
\( = 509.11 + 101.82 \)
\( = 610.93 \text{ m}^2 \). Breaking complex shapes into simpler ones like triangles and parallelograms is a key strategy.
In simple words: We split the trapezium into a parallelogram and a triangle. We find the area of the triangle first using its side lengths. From the triangle's area, we find its height, which is also the height of the trapezium. Then, we find the area of the parallelogram. Adding the areas of the parallelogram and the triangle gives us the total area of the trapezium.

🎯 Exam Tip: When dealing with trapeziums, especially isosceles ones, drawing additional lines to form rectangles/parallelograms and triangles often simplifies the problem. Remember the formula for the area of a triangle given its sides (Heron's formula) if the height is not directly given.

Question 5. If area of trapezium is 350 sq. cm and its parallel sides are 26 cm and 44 cm then find the distance between the parallel sides.
Answer: We are given the area of a trapezium as 350 sq. cm. The lengths of its parallel sides are 26 cm and 44 cm. We need to find the distance (height) between the parallel sides.
The formula for the area of a trapezium is:
Area \( = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance between them (height)} \)
Let 'x' be the distance between the parallel sides.
\( 350 = \frac{1}{2} \times (26 + 44) \times x \)
\( 350 = \frac{1}{2} \times (70) \times x \)
\( 350 = 35 \times x \)
To find x, divide 350 by 35:
\( x = \frac{350}{35} \)
\( x = 10 \text{ cm} \)
So, the distance between the parallel sides is 10 cm. This calculation is a direct application of the area formula.
In simple words: We know the area and the lengths of the two parallel sides of the trapezium. We use the area formula for a trapezium and put in the numbers we know. Then, we solve the equation to find the missing distance between the parallel sides.

🎯 Exam Tip: Always write down the formula first, then substitute the known values carefully. This helps in avoiding errors, especially when rearranging the formula to find an unknown variable like height.

Question 6. A table is in the shape of a trapezium. Its parallel sides are 8 m and 16 m respectively. Area of table is 108 sq. m. Find the width of the table i.e. Distance between the parallel sides.
Answer: The table is shaped like a trapezium. Its parallel sides are 8 m and 16 m. The area of the table is 108 sq. m. We need to find the width of the table, which is the distance (height) between its parallel sides.
The formula for the area of a trapezium is:
Area \( = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{distance between them (height)} \)
Let 'h' be the width or distance between the parallel sides.
\( 108 = \frac{1}{2} \times (8 + 16) \times h \)
\( 108 = \frac{1}{2} \times (24) \times h \)
\( 108 = 12 \times h \)
To find 'h', divide 108 by 12:
\( h = \frac{108}{12} \)
\( h = 9 \text{ m} \)
Thus, the width of the table, or the distance between its parallel sides, is 9 m. Knowing the formula and rearranging it is essential here.
In simple words: We use the formula for the area of a trapezium. We know the area and the lengths of the two parallel sides. We put these numbers into the formula and then do some simple math to figure out the height or width of the table.

🎯 Exam Tip: Clearly identify what is given (area, parallel sides) and what needs to be found (height). This helps in setting up the correct equation and solving it efficiently without confusion.