CBSE Class 9 Mathematics Herons Formula Assignment Set 03

Question 1. Find the area of a triangle, two sides of which are 18cm, and 10cm and the perimeter is 42cm
Answer: Let the three sides of the triangle be \(a\), \(b\), and \(c\).
Given: \(a = 18\text{ cm}\), \(b = 10\text{ cm}\), and Perimeter \(P = 42\text{ cm}\).
Third side \(c = \text{Perimeter} - (a + b) = 42 - (18 + 10) = 42 - 28 = 14\text{ cm}\).
Semi-perimeter \(s = \frac{\text{Perimeter}}{2} = \frac{42}{2} = 21\text{ cm}\).
Using Heron's formula, Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Area = \(\sqrt{21(21-18)(21-10)(21-14)}\)
Area = \(\sqrt{21 \times 3 \times 11 \times 7}\)
Area = \(\sqrt{(3 \times 7) \times 3 \times 11 \times 7}\)
Area = \(\sqrt{3^2 \times 7^2 \times 11} = 3 \times 7 \times \sqrt{11} = 21\sqrt{11}\text{ cm}^2\).
In simple words: First find the third side by subtracting the given two sides from the perimeter, then apply Heron's formula using the semi-perimeter.
Exam Tip: Always find the semi-perimeter \(s\) first, then compute \(s-a\), \(s-b\), and \(s-c\) carefully before multiplying.

Question 2. Find the area of an equilateral triangle one of whose sides measure a cm, using heron’s formula
Answer: For an equilateral triangle, all sides are equal: \(a = b = c = a\text{ cm}\).
Perimeter \(P = a + a + a = 3a\text{ cm}\).
Semi-perimeter \(s = \frac{3a}{2}\text{ cm}\).
Using Heron's formula, Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)
Area = \(\sqrt{\frac{3a}{2}\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)\left(\frac{3a}{2}-a\right)}\)
Area = \(\sqrt{\frac{3a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}\)
Area = \(\sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}a^2}{4}\text{ cm}^2\).
In simple words: Substitute the side length 'a' into Heron's formula to derive the standard area formula for an equilateral triangle.

Exam Tip: Memorize the standard result \(\frac{\sqrt{3}}{4}a^2\) for quick verification in objective questions.

Question 3. An isosceles triangle has perimeter 30cm and each of the equal sides is 12cm. Find the area of the triangle
Answer: Given: Perimeter \(P = 30\text{ cm}\) and each equal side \(a = b = 12\text{ cm}\).
Third side \(c = \text{Perimeter} - (a + b) = 30 - (12 + 12) = 30 - 24 = 6\text{ cm}\).
Semi-perimeter \(s = \frac{30}{2} = 15\text{ cm}\).
Using Heron's formula, Area = \(\sqrt{15(15-12)(15-12)(15-6)}\)
Area = \(\sqrt{15 \times 3 \times 3 \times 9}\)
Area = \(\sqrt{(3 \times 5) \times 3 \times 3 \times 3^2}\)
Area = \(\sqrt{3^4 \times 5} = 3^2 \times \sqrt{5} = 9\sqrt{5}\text{ cm}^2\).
In simple words: Find the unequal base by subtracting the sum of the two equal sides from the perimeter, then apply Heron's formula.

Exam Tip: Factorize numbers under the square root completely to easily take out pairs of identical factors.

Question 4. Sides of a triangle are in the ratio 12: 17: 25 and its perimeter is 540 cm. Find its area
Answer: Let the sides of the triangle be \(12x\), \(17x\), and \(25x\).
Given Perimeter = \(540\text{ cm}\).
\(12x + 17x + 25x = 540\)
\(54x = 540\)
\(x = 10\)
Therefore, the sides are:
\(a = 12 \times 10 = 120\text{ cm}\)
\(b = 17 \times 10 = 170\text{ cm}\)
\(c = 25 \times 10 = 250\text{ cm}\)
Semi-perimeter \(s = \frac{540}{2} = 270\text{ cm}\).
Using Heron's formula, Area = \(\sqrt{270(270-120)(270-170)(270-250)}\)
Area = \(\sqrt{270 \times 150 \times 100 \times 20}\)
Area = \(\sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times 20}\)
Area = \(\sqrt{(9 \times 3 \times 10) \times (3 \times 5 \times 10) \times 100 \times (2 \times 10)}\)
Let us rewrite using prime factors or groupings: \(270 = 27 \times 10 = 3^3 \times 2 \times 5\)
\(150 = 15 \times 10 = 3 \times 5^2 \times 2\)
\(100 = 10^2 = 2^2 \times 5^2\)
\(20 = 2^2 \times 5\)
Combining powers: Total powers of \(2\): \(2^1 \times 2^1 \times 2^2 \times 2^2 = 2^6\)
Total powers of \(3\): \(3^3 \times 3^1 = 3^4\)
Total powers of \(5\): \(5^1 \times 5^2 \times 5^2 \times 5^1 = 5^6\)
Area = \(\sqrt{2^6 \times 3^4 \times 5^6} = 2^3 \times 3^2 \times 5^3 = 8 \times 9 \times 125 = 9000\text{ cm}^2\).
In simple words: Use a variable 'x' for the ratio, find the actual side lengths using the perimeter, and then apply Heron's formula.

Exam Tip: When dealing with large numbers under a square root, prime factorization is much faster and less error-prone than standard multiplication.

Question 5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be grazing
Answer: A rhombus is divided by its diagonals into 4 congruent right-angled triangles, or by one diagonal into 2 congruent triangles.
Consider the triangle formed by two sides of the rhombus and its longer diagonal: sides are \(a = 30\text{ m}\), \(b = 30\text{ m}\), and \(c = 48\text{ m}\).
Semi-perimeter of this triangle \(s = \frac{30 + 30 + 48}{2} = \frac{108}{2} = 54\text{ m}\).
Area of one triangle = \(\sqrt{54(54-30)(54-30)(54-48)}\)
Area = \(\sqrt{54 \times 24 \times 24 \times 6}\)
Area = \(\sqrt{(9 \times 6) \times 24 \times 24 \times 6}\)
Area = \(\sqrt{9 \times 6^2 \times 24^2} = 3 \times 6 \times 24 = 432\text{ m}^2\).
Since the rhombus consists of 2 such triangles:
Total area of the rhombus = \(2 \times 432 = 864\text{ m}^2\).
Area for 18 cows = \(864\text{ m}^2\).
Area of grass field each cow will graze = \(\frac{864}{18} = 48\text{ m}^2\).
In simple words: Find the area of one half of the rhombus using Heron's formula, double it to get the total area, and divide by the number of cows.

Exam Tip: Recognize that a diagonal divides a rhombus into two identical triangles with sides equal to the rhombus side and the diagonal length.

Question 6. Two parallel sides of a trapezium are 60 cm and 77 cm and the other sides are 25 cm 26 cm. find the area of the trapezium
Answer: Let the parallel sides be \(AB = 77\text{ cm}\) and \(DC = 60\text{ cm}\), and non-parallel sides be \(AD = 25\text{ cm}\) and \(BC = 26\text{ cm}\).
Draw a line \(CE \parallel AD\) meeting \(AB\) at \(E\). Then \(AECD\) is a parallelogram.
Therefore, \(AE = DC = 60\text{ cm}\) and \(CE = AD = 25\text{ cm}\).
In triangle \(EBC\):
\(EB = AB - AE = 77 - 60 = 17\text{ cm}\).
The sides of triangle \(EBC\) are \(17\text{ cm}\), \(25\text{ cm}\), and \(26\text{ cm}\).
Semi-perimeter of triangle \(EBC\) (\(s\)) = \(\frac{17 + 25 + 26}{2} = \frac{68}{2} = 34\text{ cm}\).
Area of triangle \(EBC\) = \(\sqrt{34(34-17)(34-25)(34-26)}\)
Area = \(\sqrt{34 \times 17 \times 9 \times 8}\)
Area = \(\sqrt{(17 \times 2) \times 17 \times 9 \times (4 \times 2)}\)
Area = \(\sqrt{17^2 \times 2^2 \times 3^2 \times 2^2}\) -> wait, let's simplify carefully:
\(34 = 17 \times 2\), \(8 = 2^3\), so \(34 \times 8 = 17 \times 2 \times 2^3 = 17 \times 2^4\).
Area = \(\sqrt{17 \times 2^4 \times 17 \times 9} = \sqrt{17^2 \times (2^2)^2 \times 3^2} = 17 \times 4 \times 3 = 204\text{ cm}^2\).
Also, Area of triangle \(EBC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times EB \times h\)
\(204 = \frac{1}{2} \times 17 \times h\)
\(h = \frac{204 \times 2}{17} = 12 \times 2 = 24\text{ cm}\).
Area of trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
Area = \(\frac{1}{2} \times (77 + 60) \times 24 = \frac{1}{2} \times 137 \times 24 = 137 \times 12 = 1644\text{ cm}^2\).
In simple words: Construct a parallelogram inside the trapezium to form a triangle, find its height using Heron's formula, and then use the standard trapezium area formula.

Exam Tip: The standard construction for any trapezium with known non-parallel sides is to draw a line parallel to one non-parallel side to form a triangle.

Question 7. Two parallel sides of a trapezium are 120cm and 154cm and other sides are 50cm and 52cm. Find the area of the trapezium
Answer: Let parallel sides be \(AB = 154\text{ cm}\), \(DC = 120\text{ cm}\), and non-parallel sides be \(AD = 50\text{ cm}\), \(BC = 52\text{ cm}\).
Draw \(CE \parallel AD\) to form parallelogram \(AECD\).
\(AE = 120\text{ cm}\) and \(CE = 50\text{ cm}\).
In triangle \(EBC\):
\(EB = AB - AE = 154 - 120 = 34\text{ cm}\).
Sides of triangle \(EBC\) are \(34\text{ cm}\), \(50\text{ cm}\), and \(52\text{ cm}\).
Semi-perimeter \(s = \frac{34 + 50 + 52}{2} = \frac{136}{2} = 68\text{ cm}\).
Area of triangle \(EBC\) = \(\sqrt{68(68-34)(68-50)(68-52)}\)
Area = \(\sqrt{68 \times 34 \times 18 \times 16}\)
Area = \(\sqrt{(34 \times 2) \times 34 \times (9 \times 2) \times 16}\)
Area = \(\sqrt{34^2 \times 2^2 \times 3^2 \times 4^2} = 34 \times 2 \times 3 \times 4 = 816\text{ cm}^2\).
Height \(h\) of triangle \(EBC\) (and trapezium):
\(\frac{1}{2} \times \text{base} \times h = 816\)
\(\frac{1}{2} \times 34 \times h = 816\)
\(17h = 816 \implies h = \frac{816}{17} = 48\text{ cm}\).
Area of trapezium = \(\frac{1}{2} \times (154 + 120) \times 48 = \frac{1}{2} \times 274 \times 48 = 274 \times 24 = 6576\text{ cm}^2\).
In simple words: Form a triangle by translating one non-parallel side, find its area and height using Heron's formula, and calculate the total trapezium area.

Exam Tip: Match your final calculated area with the given answer key \((6576\text{ cm}^2)\) to verify your calculation steps.

Question 8. The perimeter of a right triangle is 12 cm and its hypotenuse is 5 cm. Find its area
Answer: Let the two perpendicular sides of the right-angled triangle be \(a\) and \(b\).
Given hypotenuse \(c = 5\text{ cm}\).
Perimeter = \(a + b + c = 12\text{ cm}\)
\(a + b + 5 = 12 \implies a + b = 7\text{ cm}\).
By Pythagoras theorem: \(a^2 + b^2 = c^2 = 5^2 = 25\).
Using algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\):
\(7^2 = 25 + 2ab\)
\(49 = 25 + 2ab \implies 2ab = 24 \implies ab = 12\).
Area of a right-angled triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2}ab\)
Area = \(\frac{1}{2} \times 12 = 6\text{ cm}^2\).
In simple words: Use the perimeter and Pythagoras theorem to find the product of the two sides, then apply the right-triangle area formula.

Exam Tip: You do not need to find the individual values of \(a\) and \(b\); finding the product \(ab\) directly gives the area.

Question 9. The lengths of two adjacent sides of a parallelogram are 51cm and 37cm and one of its diagonal is 20cm.Find Its area
Answer: A diagonal divides a parallelogram into two congruent triangles.
Consider the triangle formed by the two adjacent sides and the diagonal: sides are \(a = 51\text{ cm}\), \(b = 37\text{ cm}\), and \(c = 20\text{ cm}\).
Semi-perimeter \(s = \frac{51 + 37 + 20}{2} = \frac{108}{2} = 54\text{ cm}\).
Area of one triangle = \(\sqrt{54(54-51)(54-37)(54-20)}\)
Area = \(\sqrt{54 \times 3 \times 17 \times 34}\)
Area = \(\sqrt{(27 \times 2) \times 3 \times 17 \times (17 \times 2)}\)
Area = \(\sqrt{(9 \times 3) \times 2 \times 3 \times 17^2 \times 2}\)
Area = \(\sqrt{9 \times 3^2 \times 2^2 \times 17^2} = 3 \times 3 \times 2 \times 17 = 306\text{ cm}^2\).
Area of the parallelogram = \(2 \times \text{Area of one triangle} = 2 \times 306 = 612\text{ cm}^2\).
In simple words: A diagonal splits the parallelogram into two identical triangles. Find the area of one triangle using Heron's formula and multiply by 2.

Exam Tip: Always double the triangle area since a parallelogram consists of two congruent triangles sharing the diagonal.

Question 10. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26cm, 28cm and 30cm and the parallelogram stands on the same base 28cm, find the height of the parallelogram
Answer: For the triangle, let sides be \(a = 26\text{ cm}\), \(b = 28\text{ cm}\), and \(c = 30\text{ cm}\).
Semi-perimeter \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42\text{ cm}\).
Area of the triangle = \(\sqrt{42(42-26)(42-28)(42-30)}\)
Area = \(\sqrt{42 \times 16 \times 14 \times 12}\)
Area = \(\sqrt{(14 \times 3) \times 16 \times 14 \times (4 \times 3)}\)
Area = \(\sqrt{14^2 \times 16 \times 3^2 \times 4} = 14 \times 4 \times 3 \times 2 = 336\text{ cm}^2\).
Since the parallelogram and the triangle have the same area and the same base (\(28\text{ cm}\)):
Area of parallelogram = \(\text{Base} \times \text{Height}\)
\(336 = 28 \times h\)
\(h = \frac{336}{28} = 12\text{ cm}\).
In simple words: Calculate the triangle's area using Heron's formula, then equate it to the parallelogram's area formula (\(\text{base} \times \text{height}\)) to find the height.

Exam Tip: Read carefully which side is chosen as the base for the parallelogram when applying \(\text{Area} = \text{base} \times \text{height}\).

Question 11. Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 6cm, DA = 5cm and diagonal AC = 5cm
Answer: The diagonal \(AC\) divides quadrilateral \(ABCD\) into two triangles: \(\triangle ABC\) and \(\triangle ADC\).
For \(\triangle ABC\): sides are \(AB = 3\text{ cm}\), \(BC = 4\text{ cm}\), and \(AC = 5\text{ cm}\).
Notice that \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\), so \(\triangle ABC\) is a right-angled triangle at \(B\).
Area of \(\triangle ABC\) = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2\).
For \(\triangle ADC\): sides are \(AD = 5\text{ cm}\), \(CD = 6\text{ cm}\), and \(AC = 5\text{ cm}\) (an isosceles triangle).
Semi-perimeter \(s = \frac{5 + 6 + 5}{2} = \frac{16}{2} = 8\text{ cm}\).
Area of \(\triangle ADC\) = \(\sqrt{8(8-5)(8-6)(8-5)}\)
Area = \(\sqrt{8 \times 3 \times 2 \times 3} = \sqrt{16 \times 9} = 4 \times 3 = 12\text{ cm}^2\).
Total Area of quadrilateral \(ABCD\) = \(\text{Area}(\triangle ABC) + \text{Area}(\triangle ADC) = 6 + 12 = 18\text{ cm}^2\).
In simple words: Split the quadrilateral into two triangles using the diagonal, find the area of each triangle separately, and add them together.

Exam Tip: Always check if any of the sub-triangles is a right-angled triangle to save time instead of using Heron's formula.

Question 12. Find the area of a quadrilateral PQRS if PQ = 8cm, QR = 6cm, RS = 14cm, PS = 16cm and diagonal PR= 10cm
Answer: Diagonal \(PR\) divides quadrilateral \(PQRS\) into two triangles: \(\triangle PQR\) and \(\triangle PRS\).
For \(\triangle PQR\): sides are \(PQ = 8\text{ cm}\), \(QR = 6\text{ cm}\), and \(PR = 10\text{ cm}\).
Since \(8^2 + 6^2 = 64 + 36 = 100 = 10^2\), \(\triangle PQR\) is a right-angled triangle at \(Q\).
Area of \(\triangle PQR\) = \(\frac{1}{2} \times 8 \times 6 = 24\text{ cm}^2\).
For \(\triangle PRS\): sides are \(PR = 10\text{ cm}\), \(RS = 14\text{ cm}\), and \(PS = 16\text{ cm}\).
Semi-perimeter \(s = \frac{10 + 14 + 16}{2} = \frac{40}{2} = 20\text{ cm}\).
Area of \(\triangle PRS\) = \(\sqrt{20(20-10)(20-14)(20-16)}\)
Area = \(\sqrt{20 \times 10 \times 6 \times 4}\)
Area = \(\sqrt{(10 \times 2) \times 10 \times (2 \times 3) \times 4}\)
Area = \(\sqrt{10^2 \times 2^2 \times 3 \times 4} = 10 \times 2 \times 2 \times \sqrt{3} = 40\sqrt{3}\text{ cm}^2\).
Total Area of quadrilateral \(PQRS\) = \(24 + 40\sqrt{3}\text{ cm}^2\).
In simple words: Divide into two triangles, use Pythagoras theorem for the first right triangle and Heron's formula for the second, then sum both areas.

Exam Tip: Leave radical terms like \(40\sqrt{3}\) in exact form unless decimal approximation is specifically requested.

Question 13. Find the semi-perimeter of a triangle whose sides are 26cm, 28cm and 30cm
Answer: Given sides of the triangle: \(a = 26\text{ cm}\), \(b = 28\text{ cm}\), and \(c = 30\text{ cm}\).
Semi-perimeter \(s = \frac{a + b + c}{2}\)
\(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42\text{ cm}\).
In simple words: Add all three side lengths together and divide the sum by 2.

Exam Tip: Semi-perimeter is simply half of the perimeter.

Question 14. The perimeter of a triangle is 240cm. If two of its sides are 78cm and 50cm. Find the length of perpendicular On the sides of length 50cm from opposite vertex
Answer: Given Perimeter \(P = 240\text{ cm}\), side \(a = 78\text{ cm}\), and side \(b = 50\text{ cm}\).
Third side \(c = \text{Perimeter} - (a + b) = 240 - (78 + 50) = 240 - 128 = 112\text{ cm}\).
Semi-perimeter \(s = \frac{240}{2} = 120\text{ cm}\).
Area of the triangle = \(\sqrt{120(120-78)(120-50)(120-112)}\)
Area = \(\sqrt{120 \times 42 \times 70 \times 8}\)
Area = \(\sqrt{(12 \times 10) \times (6 \times 7) \times (7 \times 10) \times 8}\)
Area = \(\sqrt{(4 \times 3 \times 10) \times (6 \times 7) \times (7 \times 10) \times (4 \times 2)}\) - let's use full prime factorization:
\(120 = 2^3 \times 3 \times 5\)
\(42 = 2 \times 3 \times 7\)
\(70 = 2 \times 5 \times 7\)
\(8 = 2^3\)
Combining powers:
Powers of \(2\): \(3 + 1 + 1 + 3 = 2^8\)
Powers of \(3\): \(1 + 1 = 3^2\)
Powers of \(5\): \(1 + 1 = 5^2\)
Powers of \(7\): \(1 + 1 = 7^2\)
Area = \(\sqrt{2^8 \times 3^2 \times 5^2 \times 7^2} = 2^4 \times 3 \times 5 \times 7 = 16 \times 105 = 1680\text{ cm}^2\).
We need the length of the perpendicular (height \(h\)) on the side of length \(50\text{ cm}\).
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(1680 = \frac{1}{2} \times 50 \times h\)
\(1680 = 25 \times h \implies h = \frac{1680}{25} = \frac{3360}{50} = 67.2\text{ cm}\).
In simple words: Find the third side, calculate the total area using Heron's formula, and then equate the area to \(\frac{1}{2} \times \text{base} \times \text{height}\) using the 50 cm side as the base.

Exam Tip: The base chosen for calculating height must be the specific side mentioned in the question.

Question 15. An Isosceles triangle has perimeter 45cm and each of equal side be 16cm. Find area of triangle and length of Altitudes in form of square root
Answer: Given Perimeter \(P = 45\text{ cm}\) and each equal side \(a = 16\text{ cm}\).
Third side \(c = \text{Perimeter} - (16 + 16) = 45 - 32 = 13\text{ cm}\).
Semi-perimeter \(s = \frac{45}{2}\text{ cm}\).
Area of triangle = \(\sqrt{\frac{45}{2}\left(\frac{45}{2}-16\right)\left(\frac{45}{2}-16\right)\left(\frac{45}{2}-13\right)}\)
Area = \(\sqrt{\frac{45}{2} \times \frac{13}{2} \times \frac{13}{2} \times \frac{19}{2}}\)
Area = \(\sqrt{\frac{45 \times 13^2 \times 19}{16}} = \frac{13}{4}\sqrt{45 \times 19} = \frac{13}{4}\sqrt{855} = \frac{13}{4}\sqrt{9 \times 95} = \frac{39\sqrt{95}}{4}\text{ cm}^2\).
Now, let us find the length of the altitudes:
1. Altitude on the base \(c = 13\text{ cm}\):
Area = \(\frac{1}{2} \times \text{base} \times h_1\)
\(\frac{39\sqrt{95}}{4} = \frac{1}{2} \times 13 \times h_1\)
\(h_1 = \frac{39\sqrt{95}}{4} \times \frac{2}{13} = \frac{3\sqrt{95}}{2}\text{ cm}\).
2. Altitude on one of the equal sides (\(16\text{ cm}\)):
\(\frac{39\sqrt{95}}{4} = \frac{1}{2} \times 16 \times h_2\)
\(\frac{39\sqrt{95}}{4} = 8 \times h_2\)
\(h_2 = \frac{39\sqrt{95}}{32}\text{ cm}\).
In simple words: Find the third side, compute the area using Heron's formula in fractional form, and then use the area formula with respective bases to find both altitudes.

Exam Tip: Keep fractions inside square roots until the end to simplify surds accurately.