Samacheer Kalvi Class 9 Maths Solutions Chapter 7 Mensuration Exercise 7.1

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.1

 

Question 1. Using Heron's formula, find the area of a triangle whose sides are
(i) 10 cm, 24 cm, 26 cm
(ii) 1.8 m, 8 m, 8.2 m
Answer:
(i) Given sides: \(a = 10 \text{ cm}\), \(b = 24 \text{ cm}\) and \(c = 26 \text{ cm}\)
First, calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{10 + 24 + 26}{2}\)
\(s = \frac{60}{2}\)
\(s = 30 \text{ cm}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 30 - 10 = 20 \text{ cm}\)
\(s-b = 30 - 24 = 6 \text{ cm}\)
\(s-c = 30 - 26 = 4 \text{ cm}\)
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{30 \times 20 \times 6 \times 4} \)
Area \( = \sqrt{(2 \times 3 \times 5) \times (2^2 \times 5) \times (2 \times 3) \times 2^2} \)
Area \( = \sqrt{2^6 \times 3^2 \times 5^2} \)
Area \( = 2^3 \times 3 \times 5 \)
Area \( = 8 \times 3 \times 5 \)
Area \( = 120 \text{ cm}^2 \)
The area of the triangle is \(120 \text{ cm}^2\).

(ii) Given sides: \(a = 1.8 \text{ m}\), \(b = 8 \text{ m}\), \(c = 8.2 \text{ m}\)
First, calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{1.8 + 8 + 8.2}{2}\)
\(s = \frac{18}{2}\)
\(s = 9 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 9 - 1.8 = 7.2 \text{ m}\)
\(s-b = 9 - 8 = 1 \text{ m}\)
\(s-c = 9 - 8.2 = 0.8 \text{ m}\)
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{9 \times 7.2 \times 1 \times 0.8} \)
Area \( = \sqrt{9 \times 5.76} \)
Area \( = 3 \times \sqrt{5.76} \)
Area \( = 3 \times 2.4 \)
Area \( = 7.2 \text{ m}^2 \)
The area of the triangle is \(7.2 \text{ sq. m}\). Heron's formula is very useful for finding the area when only the side lengths are known, without needing angles.
In simple words: First, add up all the sides and divide by two to get the half-perimeter. Then, subtract each side from this half-perimeter. Finally, multiply these four numbers together and find the square root to get the triangle's area.

🎯 Exam Tip: Remember to use the correct units (cm² or m²) for the final area calculation and ensure all intermediate calculations are precise.

 

Question 2. The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area and cost of levelling the ground at the rate of Rs 20 per m².
Answer: Given sides of the triangular ground: \(a = 22 \text{ m}\), \(b = 120 \text{ m}\), and \(c = 122 \text{ m}\)
First, calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{22 + 120 + 122}{2}\)
\(s = \frac{264}{2}\)
\(s = 132 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 132 - 22 = 110 \text{ m}\)
\(s-b = 132 - 120 = 12 \text{ m}\)
\(s-c = 132 - 122 = 10 \text{ m}\)
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{132 \times 110 \times 12 \times 10} \)
Area \( = \sqrt{(2^2 \times 3 \times 11) \times (10 \times 11) \times (2^2 \times 3) \times 10} \)
Area \( = \sqrt{2^4 \times 3^2 \times 10^2 \times 11^2} \)
Area \( = 2^2 \times 3 \times 10 \times 11 \)
Area \( = 4 \times 3 \times 10 \times 11 \)
Area \( = 1320 \text{ sq.m} \)
The area of the ground is \(1320 \text{ sq.m}\).
The cost of levelling for one square metre is Rs 20.
Total cost of levelling the ground \( = \text{Area} \times \text{Rate per sq.m} \)
Total cost \( = 1320 \times 20 \)
Total cost \( = \text{Rs } 26400 \)
The total cost to level the triangular ground is Rs 26400.
In simple words: First, calculate the area of the triangular ground using its side lengths. Once you have the total area, multiply it by the cost to level one square meter to find the total expense.

🎯 Exam Tip: For problems involving cost, always make sure to calculate the area first, then multiply by the given rate per unit area. Pay attention to units for both area and cost.

 

Question 3. The perimeter of a triangular plot is 600 m. If the sides are in the ratio 5 : 12 : 13, then find the area of the plot.
Answer: Let the sides of the triangle be \(a = 5x\), \(b = 12x\), and \(c = 13x\).
The perimeter of the triangular plot is 600 m.
Perimeter \( = a + b + c \)
\(5x + 12x + 13x = 600\)
\(30x = 600\)
\(x = \frac{600}{30}\)
\(x = 20\)
Now, find the actual lengths of the sides:
\(a = 5x = 5 \times 20 = 100 \text{ m}\)
\(b = 12x = 12 \times 20 = 240 \text{ m}\)
\(c = 13x = 13 \times 20 = 260 \text{ m}\)
The semi-perimeter \(s\) is half of the perimeter:
\(s = \frac{600}{2}\)
\(s = 300 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 300 - 100 = 200 \text{ m}\)
\(s-b = 300 - 240 = 60 \text{ m}\)
\(s-c = 300 - 260 = 40 \text{ m}\)
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{300 \times 200 \times 60 \times 40} \)
Area \( = \sqrt{(3 \times 100) \times (2 \times 100) \times (6 \times 10) \times (4 \times 10)} \)
Area \( = \sqrt{3 \times 2 \times 6 \times 4 \times 100 \times 100 \times 10 \times 10} \)
Area \( = \sqrt{3 \times 2 \times (2 \times 3) \times 2^2 \times (10^2)^3} \)
Area \( = \sqrt{2^4 \times 3^2 \times (10^2)^3} \)
Area \( = 2^2 \times 3 \times 10^3 \)
Area \( = 4 \times 3 \times 1000 \)
Area \( = 12000 \text{ m}^2 \)
The area of the triangular plot is \(12000 \text{ sq.m}\). This specific ratio (5:12:13) indicates a right-angled triangle, making calculations like this often easier.
In simple words: First, use the perimeter and the ratio of sides to find the actual length of each side. Then, use Heron's formula with these side lengths to calculate the area of the triangular plot.

🎯 Exam Tip: When given side ratios and a perimeter, always find the value of 'x' first to determine the actual side lengths before applying any area formulas.

 

Question 4. Find the area of an equilateral triangle whose perimeter is 180 cm.
Answer: For an equilateral triangle, all three sides are equal. Let each side be \(a\).
Perimeter of an equilateral triangle \( = 3a \)
Given perimeter \( = 180 \text{ cm}\)
So, \(3a = 180\)
\(a = \frac{180}{3}\)
\(a = 60 \text{ cm}\)
The formula for the area of an equilateral triangle is:
Area \( = \frac{\sqrt{3}}{4} a^2 \text{ sq.unit}\)
Area \( = \frac{\sqrt{3}}{4} \times (60)^2 \)
Area \( = \frac{\sqrt{3}}{4} \times 3600 \)
Area \( = \sqrt{3} \times 900 \)
Using the approximate value \( \sqrt{3} \approx 1.732 \):
Area \( = 1.732 \times 900 \)
Area \( = 1558.8 \text{ sq.m} \)
The area of the equilateral triangle is \(1558.8 \text{ sq.m}\). Equilateral triangles are a special case of triangles where all properties are symmetrical.
In simple words: First, find the length of one side of the equilateral triangle by dividing its perimeter by three. Then, use the special formula for an equilateral triangle's area, which involves the square of its side length and the square root of 3.

🎯 Exam Tip: Remember the specific formula for the area of an equilateral triangle (\( \frac{\sqrt{3}}{4}a^2 \)) as it saves time compared to using Heron's formula.

 

Question 5. An advertisement board is in the form of an isosceles triangle with perimeter 36 m and each of the equal sides are 13 m. Find the cost of painting it at Rs 17.50 per square metre.
Answer: For an isosceles triangle, two sides are equal.
Given equal sides \( = 13 \text{ m}\)
Let \(a = 13 \text{ m}\) and \(b = 13 \text{ m}\).
Perimeter of the isosceles triangle \( = 36 \text{ m}\)
The third side \(c = \text{Perimeter} - (a + b)\)
\(c = 36 - (13 + 13)\)
\(c = 36 - 26\)
\(c = 10 \text{ m}\)
Now we have the three sides of the triangle: \(a = 13 \text{ m}\), \(b = 13 \text{ m}\), and \(c = 10 \text{ m}\).
First, calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{13 + 13 + 10}{2}\)
\(s = \frac{36}{2}\)
\(s = 18 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 18 - 13 = 5 \text{ m}\)
\(s-b = 18 - 13 = 5 \text{ m}\)
\(s-c = 18 - 10 = 8 \text{ m}\)
Using Heron's formula for the area of a triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{18 \times 5 \times 5 \times 8} \)
Area \( = \sqrt{(2 \times 3^2) \times 5^2 \times 2^3} \)
Area \( = \sqrt{2^4 \times 3^2 \times 5^2} \)
Area \( = 2^2 \times 3 \times 5 \)
Area \( = 4 \times 3 \times 5 \)
Area \( = 60 \text{ sq.m} \)
The area of the advertisement board is \(60 \text{ sq.m}\).
The cost of painting for one square metre is Rs 17.50.
Total cost of painting \( = \text{Area} \times \text{Rate per sq.m} \)
Total cost \( = 60 \times 17.50 \)
Total cost \( = \text{Rs } 1050 \)
The total cost of painting the board is Rs 1050.
In simple words: First, figure out the length of the third side of the isosceles triangle. Then, use Heron's formula to find the total area of the board. Finally, multiply this area by the given cost per square meter to get the total painting cost.

🎯 Exam Tip: When dealing with isosceles triangles, remember to correctly identify the unique side before applying Heron's formula, and don't forget the final step of calculating the total cost.

 

Question 6. Find the area of the unshaded region.
Answer:

The unshaded region is triangle ABC minus triangle ABD.
First, consider the right-angled triangle ABD (from the diagram, angle D appears to be the right angle, and AD and BD are legs).
Given: \(AD = 16 \text{ cm}\) and \(BD = 12 \text{ cm}\).
Using the Pythagorean theorem, find the hypotenuse AB:
\(AB^2 = AD^2 + BD^2\)
\(AB^2 = 12^2 + 16^2\)
\(AB^2 = 144 + 256\)
\(AB^2 = 400\)
\(AB = \sqrt{400}\)
\(AB = 20 \text{ cm}\)
Area of the right-angled triangle ABD \( = \frac{1}{2} \times \text{base} \times \text{height} \)
Area of \( \triangle ABD = \frac{1}{2} \times BD \times AD \)
Area of \( \triangle ABD = \frac{1}{2} \times 12 \times 16 \)
Area of \( \triangle ABD = 6 \times 16 \)
Area of \( \triangle ABD = 96 \text{ cm}^2 \)

Next, find the Area of the larger triangle ABC.
The sides of \( \triangle ABC \) are: \(a = BC = 34 \text{ cm}\), \(b = AC = 42 \text{ cm}\), and \(c = AB = 20 \text{ cm}\).
First, calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{34 + 42 + 20}{2}\)
\(s = \frac{96}{2}\)
\(s = 48 \text{ cm}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 48 - 34 = 14 \text{ cm}\)
\(s-b = 48 - 42 = 6 \text{ cm}\)
\(s-c = 48 - 20 = 28 \text{ cm}\)
Using Heron's formula for the area of \( \triangle ABC \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{48 \times 14 \times 6 \times 28} \)
Area \( = \sqrt{(2^4 \times 3) \times (2 \times 7) \times (2 \times 3) \times (2^2 \times 7)} \)
Area \( = \sqrt{2^8 \times 3^2 \times 7^2} \)
Area \( = 2^4 \times 3 \times 7 \)
Area \( = 16 \times 3 \times 7 \)
Area \( = 336 \text{ cm}^2 \)
Finally, calculate the area of the unshaded region:
Area of unshaded region \( = \text{Area of } \triangle ABC - \text{Area of } \triangle ABD \)
Area of unshaded region \( = 336 - 96 \)
Area of unshaded region \( = 240 \text{ cm}^2 \)
The unshaded region has an area of \(240 \text{ cm}^2\). When a larger figure contains a smaller one, subtracting the smaller area from the larger area gives the area of the remaining part.
In simple words: First, find the area of the smaller, shaded triangle. Then, calculate the area of the larger triangle that contains it. To find the unshaded part, simply subtract the small triangle's area from the big triangle's area.

🎯 Exam Tip: When a figure is composed of simpler shapes, break it down into those shapes. For right-angled triangles, use \( \frac{1}{2} \times \text{base} \times \text{height} \) for area; for others, Heron's formula is reliable.

 

Question 7. Find the area of a quadrilateral ABCD whose sides are AB = 13 cm, BC = 12 cm, CD = 9 cm, AD = 14 cm and diagonal BD = 15 cm.
Answer:

To find the area of the quadrilateral ABCD, we can divide it into two triangles, \( \triangle ABD \) and \( \triangle BCD \), using the diagonal BD.

First, consider \( \triangle ABD \):
The sides are \(a = BD = 15 \text{ cm}\), \(b = AD = 14 \text{ cm}\), and \(c = AB = 13 \text{ cm}\).
Calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{15 + 14 + 13}{2}\)
\(s = \frac{42}{2}\)
\(s = 21 \text{ cm}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 21 - 15 = 6 \text{ cm}\)
\(s-b = 21 - 14 = 7 \text{ cm}\)
\(s-c = 21 - 13 = 8 \text{ cm}\)
Using Heron's formula for the area of \( \triangle ABD \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{21 \times 6 \times 7 \times 8} \)
Area \( = \sqrt{(3 \times 7) \times (2 \times 3) \times 7 \times (2^3)} \)
Area \( = \sqrt{2^4 \times 3^2 \times 7^2} \)
Area \( = 2^2 \times 3 \times 7 \)
Area \( = 4 \times 3 \times 7 \)
Area \( = 84 \text{ cm}^2 \)

Next, consider \( \triangle BCD \):
The sides are \(a = BD = 15 \text{ cm}\), \(b = CD = 9 \text{ cm}\), and \(c = BC = 12 \text{ cm}\).
Calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{15 + 9 + 12}{2}\)
\(s = \frac{36}{2}\)
\(s = 18 \text{ cm}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 18 - 15 = 3 \text{ cm}\)
\(s-b = 18 - 9 = 9 \text{ cm}\)
\(s-c = 18 - 12 = 6 \text{ cm}\)
Using Heron's formula for the area of \( \triangle BCD \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{18 \times 3 \times 9 \times 6} \)
Area \( = \sqrt{(2 \times 3^2) \times 3 \times 3^2 \times (2 \times 3)} \)
Area \( = \sqrt{2^2 \times 3^6} \)
Area \( = 2 \times 3^3 \)
Area \( = 2 \times 27 \)
Area \( = 54 \text{ cm}^2 \)

Finally, the total area of the quadrilateral ABCD is the sum of the areas of the two triangles:
Area of quadrilateral ABCD \( = \text{Area of } \triangle ABD + \text{Area of } \triangle BCD \)
Area of quadrilateral ABCD \( = 84 + 54 \)
Area of quadrilateral ABCD \( = 138 \text{ cm}^2 \)
The area of the quadrilateral ABCD is \(138 \text{ cm}^2\). Quadrilaterals can often be split into simpler triangles for area calculations.
In simple words: Break the quadrilateral into two triangles using the given diagonal. Calculate the area of each triangle using Heron's formula. Then, add the areas of the two triangles together to get the total area of the quadrilateral.

🎯 Exam Tip: Always draw the quadrilateral and its diagonal to clearly identify the sides of the two resulting triangles before applying Heron's formula.

 

Question 8. A park is in the shape of a quadrilateral. The sides of the park are 15 m, 20 m, 26 m and 17 m and the angle between the first two sides is a right angle. Find the area of the park.
Answer:

Let the quadrilateral be ABCD with sides \(AB = 15 \text{ m}\), \(BC = 20 \text{ m}\), \(CD = 26 \text{ m}\), and \(AD = 17 \text{ m}\). The angle between the first two sides (AB and BC) is a right angle, so \( \angle B = 90^\circ \).
We can divide the quadrilateral into two triangles, \( \triangle ABC \) (a right-angled triangle) and \( \triangle ACD \), using the diagonal AC.

First, consider the right-angled \( \triangle ABC \):
Using the Pythagorean theorem to find the diagonal AC:
\(AC^2 = AB^2 + BC^2\)
\(AC^2 = 15^2 + 20^2\)
\(AC^2 = 225 + 400\)
\(AC^2 = 625\)
\(AC = \sqrt{625}\)
\(AC = 25 \text{ m}\)
Area of \( \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} \)
Area of \( \triangle ABC = \frac{1}{2} \times AB \times BC \)
Area of \( \triangle ABC = \frac{1}{2} \times 15 \times 20 \)
Area of \( \triangle ABC = 15 \times 10 \)
Area of \( \triangle ABC = 150 \text{ sq.m} \)

Next, consider \( \triangle ACD \):
The sides are \(a = AC = 25 \text{ m}\), \(b = AD = 17 \text{ m}\), and \(c = CD = 26 \text{ m}\).
Calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{25 + 17 + 26}{2}\)
\(s = \frac{68}{2}\)
\(s = 34 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 34 - 25 = 9 \text{ m}\)
\(s-b = 34 - 17 = 17 \text{ m}\)
\(s-c = 34 - 26 = 8 \text{ m}\)
Using Heron's formula for the area of \( \triangle ACD \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{34 \times 9 \times 17 \times 8} \)
Area \( = \sqrt{(2 \times 17) \times 3^2 \times 17 \times 2^3} \)
Area \( = \sqrt{2^4 \times 3^2 \times 17^2} \)
Area \( = 2^2 \times 3 \times 17 \)
Area \( = 4 \times 3 \times 17 \)
Area \( = 204 \text{ sq.m} \)

Finally, the total area of the quadrilateral park is the sum of the areas of the two triangles:
Area of quadrilateral \( = \text{Area of } \triangle ABC + \text{Area of } \triangle ACD \)
Area of quadrilateral \( = 150 + 204 \)
Area of quadrilateral \( = 354 \text{ sq.m} \)
The area of the park is \(354 \text{ sq.m}\). Knowing one angle is a right angle simplifies one part of the calculation considerably.
In simple words: First, use the right angle and the two sides that form it to find the area of the first triangle and the length of the diagonal. Then, use this diagonal and the other two given sides to find the area of the second triangle. Add both areas to get the total area of the park.

🎯 Exam Tip: Always look for right angles in quadrilateral problems, as they allow for direct calculation of one triangle's area using \( \frac{1}{2}bh \) and finding a diagonal with Pythagoras theorem.

 

Question 9. A land is in the shape of rhombus. The perimeter of the land is 160 m and one of the diagonal is 48 m. Find the area of the land.
Answer:

A rhombus has four equal sides.
Given perimeter of the rhombus \( = 160 \text{ m}\)
Side of the rhombus \( = \frac{\text{Perimeter}}{4} = \frac{160}{4} = 40 \text{ m}\)
So, each side of the rhombus (AB, BC, CD, DA) is 40 m.
One of the diagonals is given as 48 m. Let's assume this is diagonal AC \( = 48 \text{ m}\).
A rhombus is divided into two congruent triangles by a diagonal. We can find the area of one triangle, say \( \triangle ABC \), and then multiply by 2 to get the area of the rhombus.

Consider \( \triangle ABC \):
The sides are \(a = AB = 40 \text{ m}\), \(b = BC = 40 \text{ m}\), and \(c = AC = 48 \text{ m}\).
Calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{40 + 40 + 48}{2}\)
\(s = \frac{128}{2}\)
\(s = 64 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 64 - 40 = 24 \text{ m}\)
\(s-b = 64 - 40 = 24 \text{ m}\)
\(s-c = 64 - 48 = 16 \text{ m}\)
Using Heron's formula for the area of \( \triangle ABC \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{64 \times 24 \times 24 \times 16} \)
Area \( = \sqrt{8^2 \times 24^2 \times 4^2} \)
Area \( = 8 \times 24 \times 4 \)
Area \( = 768 \text{ sq.m} \)

Since the rhombus ABCD consists of two congruent triangles (\( \triangle ABC \) and \( \triangle ADC \)), the total area of the rhombus is twice the area of one triangle.
Area of rhombus ABCD \( = 2 \times \text{Area of } \triangle ABC \)
Area of rhombus ABCD \( = 2 \times 768 \)
Area of rhombus ABCD \( = 1536 \text{ sq.m} \)
The area of the land in the shape of a rhombus is \(1536 \text{ sq.m}\). A rhombus can be seen as two identical triangles joined along a common diagonal.
In simple words: First, find the length of one side of the rhombus by dividing its perimeter by four. Then, consider one of the two identical triangles formed by a diagonal and use Heron's formula to find its area. Double this triangle's area to get the total area of the rhombus.

🎯 Exam Tip: Remember that the diagonals of a rhombus bisect each other at right angles, which can also be used to find the area if both diagonals are known (\( \frac{1}{2} d_1 d_2 \)).

 

Question 10. The adjacent sides of a parallelogram measures 34 m, 20 m and the measure of the diagonal is 42 m. Find the area of parallelogram.
Answer:

Let the parallelogram be ABCD. The adjacent sides are \(AB = 34 \text{ m}\) and \(AD = 20 \text{ m}\). The diagonal is \(AC = 42 \text{ m}\).
In a parallelogram, opposite sides are equal, so \(BC = AD = 20 \text{ m}\) and \(CD = AB = 34 \text{ m}\).
A diagonal divides a parallelogram into two congruent triangles. We can find the area of one triangle, say \( \triangle ABC \), and then multiply by 2 to get the area of the parallelogram.

Consider \( \triangle ABC \):
The sides are \(a = AB = 34 \text{ m}\), \(b = BC = 20 \text{ m}\), and \(c = AC = 42 \text{ m}\).
Calculate the semi-perimeter \(s\):
\(s = \frac{a + b + c}{2}\)
\(s = \frac{34 + 20 + 42}{2}\)
\(s = \frac{96}{2}\)
\(s = 48 \text{ m}\)
Now, find \(s-a\), \(s-b\), and \(s-c\):
\(s-a = 48 - 34 = 14 \text{ m}\)
\(s-b = 48 - 20 = 28 \text{ m}\)
\(s-c = 48 - 42 = 6 \text{ m}\)
Using Heron's formula for the area of \( \triangle ABC \):
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{48 \times 14 \times 28 \times 6} \)
Area \( = \sqrt{(2^4 \times 3) \times (2 \times 7) \times (2^2 \times 7) \times (2 \times 3)} \)
Area \( = \sqrt{2^8 \times 3^2 \times 7^2} \)
Area \( = 2^4 \times 3 \times 7 \)
Area \( = 16 \times 3 \times 7 \)
Area \( = 336 \text{ sq.m} \)

Since ABCD is a parallelogram, the Area of \( \triangle ABC \) is equal to the Area of \( \triangle ADC \).
Area of parallelogram ABCD \( = 2 \times \text{Area of } \triangle ABC \)
Area of parallelogram ABCD \( = 2 \times 336 \)
Area of parallelogram ABCD \( = 672 \text{ sq.m} \)
The area of the parallelogram is \(672 \text{ sq.m}\). Parallelograms, like rhombuses, can be easily divided into two identical triangles.
In simple words: Use the given adjacent sides and the diagonal to find the area of one of the two identical triangles that make up the parallelogram. Then, multiply this triangle's area by two to get the total area of the parallelogram.

🎯 Exam Tip: Remember that a diagonal divides a parallelogram into two triangles of equal area, so you only need to calculate the area of one triangle and then double it.