RBSE Solutions Class 9 Maths Chapter 11 Area of Plane Figures Exercise 11.2

Question 1. Find the area of a quadrilateral when one of the diagonal measure 12 cm and the length of the perpendicular drawn from the opposite vertices to the diagonal are 7 cm and 8 cm respectively.
Answer: To find the area of a quadrilateral when a diagonal and perpendiculars from opposite vertices to that diagonal are given, we use a specific formula. We sum the lengths of the two perpendiculars, then multiply by half of the diagonal's length.
Area of quadrilateral \( = \frac { 1 }{ 2 } \times \text{diagonal} \times (\text{sum of offsets}) \)
\( = \frac { 1 }{ 2 } \times 12 \times (7 + 8) \)
\( = \frac { 1 }{ 2 } \times 12 \times 15 \)
\( = 6 \times 15 \)
\( = 90 \text{ cm}^2 \)
The area of the quadrilateral is 90 square centimeters.
In simple words: Imagine cutting the quadrilateral into two triangles using the diagonal. The area is found by adding the areas of these two triangles. We use the diagonal as the base and the perpendiculars as their heights.

🎯 Exam Tip: Remember that "offsets" refer to the perpendicular distances from the vertices to the diagonal. Ensure you add these distances before multiplying.

Question 2. The area of a parallelogram shaped field is 2000 sq. m. If its base is 50 m then find its height.
Answer: The area of a parallelogram is calculated by multiplying its base by its corresponding height (or altitude). We can use this formula to find the unknown height when the area and base are known.
Area of parallelogram shaped field \( = \text{base} \times \text{corresponding altitude} \)
Given, Area \( = 2000 \text{ m}^2 \)
Given, Base \( = 50 \text{ m} \)
So, \( 2000 \text{ m}^2 = 50 \text{ m} \times h \)
To find the height \( h \), we divide the area by the base:
\( h = \frac{2000 \text{ m}^2}{50 \text{ m}} \)
\( h = 40 \text{ m} \)
Therefore, the height of the field is 40 meters.
In simple words: If you know how much space a parallelogram takes up (its area) and how long its bottom side is (its base), you can find its height by dividing the area by the base.

🎯 Exam Tip: Always ensure the units are consistent. Here, both area and base are in meters, so the height will also be in meters.

Question 3. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 6 cm and DA = 5 cm and diagonal AC = 5 cm.
Answer: We can find the area of the quadrilateral ABCD by dividing it into two triangles, \( \triangle \text{ABC} \) and \( \triangle \text{ACD} \), using the diagonal AC. Then, we find the area of each triangle and add them together. This method is useful for any quadrilateral.
Area of quadrilateral ABCD \( = \text{area of } \triangle \text{ABC} + \text{area of } \triangle \text{ACD} \)

For \( \triangle \text{ABC} \): Sides are AB = 3 cm, BC = 4 cm, and AC = 5 cm.
Since \( 3^2 + 4^2 = 9 + 16 = 25 = 5^2 \), \( \triangle \text{ABC} \) is a right-angled triangle with the right angle at B.
Area of \( \triangle \text{ABC} = \frac { 1 }{ 2 } \times \text{base} \times \text{height} \)
\( = \frac { 1 }{ 2 } \times \text{AB} \times \text{BC} \)
\( = \frac { 1 }{ 2 } \times 3 \text{ cm} \times 4 \text{ cm} \)
\( = 6 \text{ cm}^2 \)

For \( \triangle \text{ACD} \): Sides are AC = 5 cm, CD = 6 cm, and DA = 5 cm.
We will use Heron's formula to find its area. First, calculate the semi-perimeter (s):
\( s = \frac{\text{AC} + \text{CD} + \text{DA}}{2} = \frac{5 + 6 + 5}{2} = \frac{16}{2} = 8 \text{ cm} \)
Area of \( \triangle \text{ACD} = \sqrt{s(s-\text{AC})(s-\text{CD})(s-\text{DA})} \)
\( = \sqrt{8(8-5)(8-6)(8-5)} \)
\( = \sqrt{8 \times 3 \times 2 \times 3} \)
\( = \sqrt{144} \)
\( = 12 \text{ cm}^2 \)

Now, sum the areas of the two triangles:
Area of quadrilateral ABCD \( = 6 \text{ cm}^2 + 12 \text{ cm}^2 \)
\( = 18 \text{ cm}^2 \)
The total area of the quadrilateral ABCD is 18 square centimeters.In simple words: We split the four-sided shape into two triangles using its diagonal. Then we calculate the area of each triangle separately. For the first triangle, we found it was a special type (right-angled). For the second, we used a formula that works for any triangle (Heron's formula). Finally, we added both triangle areas to get the total area of the quadrilateral.

🎯 Exam Tip: When a quadrilateral is given with all sides and one diagonal, it's often easiest to split it into two triangles and use Heron's formula for each, or the \( \frac{1}{2} \times \text{base} \times \text{height} \) formula if one is a right-angled triangle.

Question 4. Find the area of a quadrilateral ABCD whose sides are 9 cm, 40 cm, 28 cm and 15 cm respectively and the angle between the first two sides is a right angle.
Answer: To find the area of this quadrilateral, we can divide it into two triangles by drawing a diagonal. Since the angle between the first two sides (AB and BC) is a right angle, \( \triangle \text{ABC} \) is a right-angled triangle. This makes calculating its area straightforward, and then we use its hypotenuse as a side for the second triangle.
Given sides are AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm.
The angle between AB and BC is 90°, so \( \angle \text{B} = 90^\circ \).

First, find the area of \( \triangle \text{ABC} \):
In right-angled \( \triangle \text{ABC} \), the base is AB = 9 cm and the height is BC = 40 cm.
Area of \( \triangle \text{ABC} = \frac { 1 }{ 2 } \times \text{base} \times \text{height} \)
\( = \frac { 1 }{ 2 } \times 9 \text{ cm} \times 40 \text{ cm} \)
\( = 9 \times 20 \text{ cm}^2 \)
\( = 180 \text{ cm}^2 \)

Next, find the length of the diagonal AC using the Pythagorean theorem:
\( \text{AC}^2 = \text{AB}^2 + \text{BC}^2 \)
\( \text{AC}^2 = 9^2 + 40^2 \)
\( \text{AC}^2 = 81 + 1600 \)
\( \text{AC}^2 = 1681 \)
\( \text{AC} = \sqrt{1681} = 41 \text{ cm} \)

Now, find the area of \( \triangle \text{ACD} \): The sides are AC = 41 cm, CD = 28 cm, and DA = 15 cm.
We will use Heron's formula. First, calculate the semi-perimeter (s):
\( s = \frac{\text{AC} + \text{CD} + \text{DA}}{2} = \frac{41 + 28 + 15}{2} = \frac{84}{2} = 42 \text{ cm} \)
Area of \( \triangle \text{ACD} = \sqrt{s(s-\text{AC})(s-\text{CD})(s-\text{DA})} \)
\( = \sqrt{42(42-41)(42-28)(42-15)} \)
\( = \sqrt{42 \times 1 \times 14 \times 27} \)
\( = \sqrt{(2 \times 3 \times 7) \times (1) \times (2 \times 7) \times (3 \times 3 \times 3)} \)
\( = \sqrt{2^2 \times 3^4 \times 7^2} \)
\( = 2 \times 3^2 \times 7 \)
\( = 2 \times 9 \times 7 \)
\( = 126 \text{ cm}^2 \)

Finally, the total area of the quadrilateral ABCD is the sum of the areas of the two triangles:
Area of quadrilateral ABCD \( = \text{Area of } \triangle \text{ABC} + \text{Area of } \triangle \text{ACD} \)
\( = 180 \text{ cm}^2 + 126 \text{ cm}^2 \)
\( = 306 \text{ cm}^2 \)
The total area of the quadrilateral is 306 square centimeters.
In simple words: We split the four-sided shape into two triangles. One triangle had a right angle, so its area was easy to find. We also found the diagonal that connected the two triangles. For the second triangle, we used a special formula (Heron's) with all three of its side lengths. Adding these two triangle areas gave us the total area of the quadrilateral.

🎯 Exam Tip: Always draw a rough sketch of the quadrilateral and the diagonal to visualize the two triangles. This helps identify right-angled triangles and apply the correct formulas.

Question 5. The length of two adjacent sides of a parallelogram are 50 cm and 40 cm respectively and its diagonal is 30 cm. Find the area of the parallelogram.
Answer: A parallelogram can be divided into two congruent (identical) triangles by any of its diagonals. If we find the area of one of these triangles, we can simply double it to get the total area of the parallelogram. Here, the diagonal forms a triangle with the two adjacent sides.
Let the adjacent sides of the parallelogram be a = 50 cm and b = 40 cm, and the diagonal be d = 30 cm.
These three lengths form a triangle, say \( \triangle \text{ABC} \).
First, we find the area of this triangle using Heron's formula.
Calculate the semi-perimeter (s) of \( \triangle \text{ABC} \):
\( s = \frac{a + b + d}{2} = \frac{50 + 40 + 30}{2} = \frac{120}{2} = 60 \text{ cm} \)
Now, calculate the area of \( \triangle \text{ABC} \):
Area \( = \sqrt{s(s-a)(s-b)(s-d)} \)
\( = \sqrt{60(60-50)(60-40)(60-30)} \)
\( = \sqrt{60 \times 10 \times 20 \times 30} \)
\( = \sqrt{360000} \)
\( = 600 \text{ cm}^2 \)
Since the parallelogram is made up of two identical triangles (formed by the diagonal), its area is twice the area of one triangle.
Area of parallelogram ABCD \( = 2 \times \text{Area of } \triangle \text{ABC} \)
\( = 2 \times 600 \text{ cm}^2 \)
\( = 1200 \text{ cm}^2 \)
The area of the parallelogram is 1200 square centimeters.
In simple words: We know the three sides of a triangle that makes up half of the parallelogram. We use a formula (Heron's) to find the area of this triangle. Then, we just double that area to get the total area of the whole parallelogram.

🎯 Exam Tip: Always remember that a diagonal divides a parallelogram into two congruent triangles. This means their areas are equal, simplifying the calculation significantly.

Question 6. Find the area of a parallelogram, whose one diagonal is 5.2 cm and the perpendicular distance on this diagonal from opposite vertices is 3.5 cm.
Answer: To find the area of a parallelogram using a diagonal and the perpendicular distances (offsets) from the other two vertices to this diagonal, we can think of the parallelogram as two triangles. Each triangle shares the diagonal as its base, and the perpendicular distances serve as their heights. The area is half the product of the diagonal and the sum of the two perpendiculars. For a parallelogram, these two perpendiculars are equal.
Given, diagonal \( = 5.2 \text{ cm} \)
Given, perpendicular distance (offset) from opposite vertices \( = 3.5 \text{ cm} \)
Since there are two opposite vertices, the sum of offsets \( = 3.5 \text{ cm} + 3.5 \text{ cm} = 7 \text{ cm} \)
Area of parallelogram \( = \frac { 1 }{ 2 } \times \text{diagonal} \times (\text{sum of offsets}) \)
\( = \frac { 1 }{ 2 } \times 5.2 \text{ cm} \times (3.5 \text{ cm} + 3.5 \text{ cm}) \)
\( = \frac { 1 }{ 2 } \times 5.2 \text{ cm} \times 7 \text{ cm} \)
\( = 2.6 \text{ cm} \times 7 \text{ cm} \)
\( = 18.2 \text{ cm}^2 \)
The area of the parallelogram is 18.2 square centimeters.
In simple words: Imagine the parallelogram is cut into two triangles along its diagonal. The area is half of the diagonal's length multiplied by the total height from both corners to that diagonal.

🎯 Exam Tip: For a parallelogram, the perpendicular distances from both opposite vertices to a diagonal are equal. Therefore, the "sum of offsets" is simply twice the given perpendicular distance.

Question 7. The adjacent sides of a plot in the form of parallelogram are 39 meters and 25 meters and its diagonal is 56 metre. Find the cost of levelling the plot at the rate of Rs 100 per sq. m.
Answer: First, we need to find the total area of the parallelogram-shaped plot. A diagonal divides a parallelogram into two identical triangles. We can find the area of one triangle using Heron's formula and then double it for the parallelogram's area. Once we have the area, we can calculate the total cost of levelling.
Let the adjacent sides of the parallelogram be a = 39 m and b = 25 m, and the diagonal be d = 56 m.
These three lengths form a triangle, say \( \triangle \text{ABC} \).
First, calculate the semi-perimeter (s) of \( \triangle \text{ABC} \):
\( s = \frac{a + b + d}{2} = \frac{39 + 25 + 56}{2} = \frac{120}{2} = 60 \text{ m} \)
Now, calculate the area of \( \triangle \text{ABC} \) using Heron's formula:
Area \( = \sqrt{s(s-a)(s-b)(s-d)} \)
\( = \sqrt{60(60-39)(60-25)(60-56)} \)
\( = \sqrt{60 \times 21 \times 35 \times 4} \)
\( = \sqrt{(2 \times 2 \times 3 \times 5) \times (3 \times 7) \times (5 \times 7) \times (2 \times 2)} \)
\( = \sqrt{2^4 \times 3^2 \times 5^2 \times 7^2} \)
\( = 2^2 \times 3 \times 5 \times 7 \)
\( = 4 \times 3 \times 5 \times 7 \)
\( = 420 \text{ m}^2 \)
The area of the parallelogram ABCD is twice the area of \( \triangle \text{ABC} \).
Area of parallelogram ABCD \( = 2 \times 420 \text{ m}^2 = 840 \text{ m}^2 \)

Next, calculate the cost of levelling the plot:
Rate of levelling \( = \text{Rs } 100 \text{ per sq. m} \)
Total cost of levelling \( = \text{Area} \times \text{Rate} \)
\( = 840 \text{ m}^2 \times \text{Rs } 100/\text{m}^2 \)
\( = \text{Rs } 84,000 \)
The total cost of levelling the plot is Rs 84,000.In simple words: We treated the parallelogram as two triangles put together. We used the side lengths and diagonal to find the area of one triangle with a special formula (Heron's). Then, we doubled that area to get the total area of the plot. Finally, we multiplied the total area by the cost per square meter to find the total levelling cost.

🎯 Exam Tip: Pay close attention to units! If sides are in meters, the area will be in square meters. Also, remember that levelling costs usually apply per unit area, so multiply the total area by the rate.