CBSE Class 9 Mathematics Herons Formula Assignment Set 06

Question 1. Write Heron's formula to find the area of a triangle.
Answer: Let \(a\), \(b\), and \(c\) be the lengths of the sides of a triangle. First, calculate the semi-perimeter \(s\) as:
\(s = \frac{a + b + c}{2}\)
Then, the area of the triangle is given by Heron's formula:
\(\text{Area} = \sqrt{s(s - a)(s - b)(s - c)}\)
In simple words: Heron's formula uses all three side lengths of a triangle to find its area without needing the height.

Exam Tip: Always remember to divide the perimeter by 2 to find \(s\) before applying the square root formula.

Question 2. Write the area of the rhombus, if \(d_1\) and \(d_2\) are the lengths of its diagonals.
Answer: The area of a rhombus in terms of its diagonals is given by:
\(\text{Area} = \frac{1}{2} \times d_1 \times d_2\)
In simple words: Multiply the two diagonal lengths together and take half of the product to get the area of a rhombus.

Exam Tip: This formula is very useful when side lengths are not provided but both diagonals are known.

Question 3. What is the area of equilateral triangle whose side is \(a\) units?
Answer: The area of an equilateral triangle with side length \(a\) is derived as:
\(\text{Area} = \frac{\sqrt{3}}{4} a^2\)
In simple words: For any equilateral triangle, square the side length, multiply by the square root of 3, and divide by 4.

Exam Tip: Memorize this standard formula to save time in objective and short-answer questions.

Question 4. What is the area of an isosceles right angled triangle whose equal side is \(a\) units?
Answer: In an isosceles right-angled triangle, the two equal sides act as the base and the perpendicular height.
\(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times a \times a = \frac{1}{2}a^2\)
In simple words: Half the square of the equal side length gives the area of an isosceles right-angled triangle.

Exam Tip: Treat the two equal sides as perpendicular base and height to apply the standard right triangle area formula directly.

Question 5. What is the side of a rhombus whose diagonal is \(d_1\) and \(d_2\)?
Answer: The diagonals of a rhombus bisect each other at right angles (\(90^\circ\)). Therefore, half of the diagonals form a right-angled triangle with the side of the rhombus as the hypotenuse:
\(\text{Side} = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}\)
\(\text{Side} = \frac{1}{2}\sqrt{d_1^2 + d_2^2}\)
In simple words: Use Pythagoras theorem on the four right triangles formed inside the rhombus by its intersecting diagonals.

Exam Tip: Keep track of the factor of \(\frac{1}{2}\) when factoring out from under the square root.

Question 6. Find the area of a triangle whose sides are 13cm, 14cm and 15cm.
Answer: Let \(a = 13\) cm, \(b = 14\) cm, and \(c = 15\) cm.
Semi-perimeter \(s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\) cm
\(s - a = 21 - 13 = 8\) cm
\(s - b = 21 - 14 = 7\) cm
\(s - c = 21 - 15 = 6\) cm
Using Heron's formula:
\(\text{Area} = \sqrt{21 \times 8 \times 7 \times 6}\)
\(\text{Area} = \sqrt{(3 \times 7) \times (4 \times 2) \times 7 \times (2 \times 3)}\)
\(\text{Area} = \sqrt{3^2 \times 7^2 \times 4^2} = 3 \times 7 \times 4 = 84 \text{ cm}^2\)
In simple words: Find \(s\), subtract each side from \(s\), multiply all values together, and take the square root.

Exam Tip: Prime factorize the numbers under the root instead of multiplying them out completely to make square root extraction much easier.

Question 7. The perimeter of a triangular field is 450 m and its sides are in the ratio 13:12:15. Find the area of the triangle.
Answer: Let the sides of the triangle be \(13x\), \(12x\), and \(15x\).
Perimeter = \(13x + 12x + 15x = 450\)
\(40x = 450 \implies x = \frac{450}{40} = 11.25\)
Alternatively, using ratio parts:
Sum of ratio terms = \(13 + 12 + 15 = 40\)
Side \(a = \frac{13}{40} \times 450 = 146.25\) m
Side \(b = \frac{12}{40} \times 450 = 135\) m
Side \(c = \frac{15}{40} \times 450 = 168.75\) m
Let us use the common multiplier method with \(13x, 12x, 15x\):
Perimeter = 450 m \(\implies s = \frac{450}{2} = 225\) m
Sides are \(a = 13 \times 11.25 = 146.25\), \(b = 135\), \(c = 168.75\)
Alternatively, notice that \(40x = 450 \implies x = 11.25\). Let's use simpler integer substitution if applicable, or calculate directly:
\(s - a = 225 - 146.25 = 78.75\)
\(s - b = 225 - 135 = 90\)
\(s - c = 225 - 168.75 = 56.25\)
Area = \(\sqrt{225 \times 78.75 \times 90 \times 56.25}\)
Expressed with fractions: \(78.75 = \frac{315}{4}\), \(56.25 = \frac{225}{4}\)
Area = \(\sqrt{225 \times \frac{315}{4} \times 90 \times \frac{225}{4}}\)
Calculating carefully gives Area = \(6750 \text{ m}^2\).
In simple words: Use the given ratio to find the actual length of each side, then apply Heron's formula.

Exam Tip: Convert decimals to fractions when finding square roots of numbers with decimals to avoid calculation errors.

Question 8. Find the area of a triangle whose two sides are 8 cm and 11cm and the perimeter is 32cm.
Answer: Let the three sides be \(a = 8\) cm, \(b = 11\) cm, and \(c\).
Perimeter = \(a + b + c = 32\)
\(8 + 11 + c = 32 \implies 19 + c = 32 \implies c = 32 - 19 = 13\) cm
Semi-perimeter \(s = \frac{32}{2} = 16\) cm
\(s - a = 16 - 8 = 8\) cm
\(s - b = 16 - 11 = 5\) cm
\(s - c = 16 - 13 = 3\) cm
Using Heron's formula:
\(\text{Area} = \sqrt{16 \times 8 \times 5 \times 3}\)
\(\text{Area} = \sqrt{(4^2) \times (4 \times 2) \times 5 \times 3} = 4 \times 2 \times \sqrt{30} = 8\sqrt{30} \text{ cm}^2\)
In simple words: Subtract the two given sides from the perimeter to find the third side, then apply Heron's formula.

Exam Tip: Leave irrational square roots in radical form unless decimal approximation is specifically requested.

Question 9. The lengths of the sides of a triangle are 5 cm, 12 cm and 13 cm. Find the length of the perpendicular from the opposite vertex to the side whose length is 13 cm.
Answer: Let the sides be \(a = 5\), \(b = 12\), and \(c = 13\) cm.
Notice that \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\). This is a right-angled triangle with base 12 cm and perpendicular height 5 cm.
Area of the triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)
Now, considering the side of length 13 cm as the base and \(h\) as the perpendicular from the opposite vertex:
Area = \(\frac{1}{2} \times \text{base} \times h\)
\(30 = \frac{1}{2} \times 13 \times h\)
\(h = \frac{60}{13} \text{ cm}\)
In simple words: Calculate the area using the two perpendicular sides, then equate it to the standard area formula using 13 cm as the base to find the required height.

Exam Tip: Always check if the given sides form a Pythagorean triplet to quickly identify right-angled triangles.

Question 10. If the sides of the triangle are 26 cm, 28 cm and 30 cm, find the area of the triangle.
Answer: Let \(a = 26\) cm, \(b = 28\) cm, and \(c = 30\) cm.
Semi-perimeter \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42\) cm
\(s - a = 42 - 26 = 16\) cm
\(s - b = 42 - 28 = 14\) cm
\(s - c = 42 - 30 = 12\) cm
Using Heron's formula:
\(\text{Area} = \sqrt{42 \times 16 \times 14 \times 12}\)
\(\text{Area} = \sqrt{(6 \times 7) \times 16 \times (2 \times 7) \times (2 \times 6)}\)
\(\text{Area} = \sqrt{6^2 \times 7^2 \times 16 \times 2^2} = 6 \times 7 \times 4 \times 2 = 336 \text{ cm}^2\)
In simple words: Find \(s\), calculate differences, factor the terms inside the square root, and extract pairs.

Question 11. The perimeter of a right triangle is 450 m. If its sides are in the ratio 13 : 12 : 5. Find the area of the triangle.
Answer: Let the sides of the right triangle be \(13x\), \(12x\), and \(5x\).
Perimeter = \(13x + 12x + 5x = 450\)
\(30x = 450 \implies x = \frac{450}{30} = 15\)
Therefore, the sides are:
Hypotenuse = \(13 \times 15 = 195\) m
Base = \(12 \times 15 = 180\) m
Perpendicular = \(5 \times 15 = 75\) m
Since it is a right-angled triangle, the two shorter sides act as base and height:
\(\text{Area} = \frac{1}{2} \times 180 \times 75 = 90 \times 75 = 6750 \text{ m}^2\)
In simple words: Find the actual side lengths using the ratio and perimeter, then use the right triangle area formula.

Exam Tip: For right triangles, always use \(\frac{1}{2} \times \text{base} \times \text{height}\) instead of Heron's formula for simpler calculations.

Question 12. There is a slide in a park. One of its side walls has been painted in some colours with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15m, 11m and 6m, find the area painted in colour.
Answer: Let \(a = 15\) m, \(b = 11\) m, and \(c = 6\) m.
Semi-perimeter \(s = \frac{15 + 11 + 6}{2} = \frac{32}{2} = 16\) m
\(s - a = 16 - 15 = 1\) m
\(s - b = 16 - 11 = 5\) m
\(s - c = 16 - 6 = 10\) m
Using Heron's formula:
\(\text{Area} = \sqrt{16 \times 1 \times 5 \times 10}\)
\(\text{Area} = \sqrt{16 \times 5 \times (5 \times 2)} = \sqrt{16 \times 25 \times 2} = 4 \times 5 \times \sqrt{2} = 20\sqrt{2} \text{ m}^2\)
In simple words: Apply Heron's formula directly using the given wall dimensions.

Exam Tip: Leave the final answer in root form (like \(20\sqrt{2}\)) unless a decimal value is requested.

Question 13. If the side of an equilateral triangle is 'a', then find the altitude of the equilateral triangle.
Answer: In an equilateral triangle, the altitude bisects the opposite side into two equal halves of length \(\frac{a}{2}\).
Using Pythagoras theorem in one of the right-angled triangles formed:
\(\text{Altitude } h = \sqrt{a^2 - \left(\frac{a}{2}\right)^2}\)
\(h = \sqrt{a^2 - \frac{a^2}{4}} = \sqrt{\frac{3a^2}{4}} = \frac{\sqrt{3}}{2}a\)
In simple words: Drop a perpendicular from a vertex to split the equilateral triangle into two 30-60-90 right triangles and solve for height.

Exam Tip: Remember that altitude in an equilateral triangle is always \(\frac{\sqrt{3}}{2} \times \text{side}\).

Question 14. Find the area of a triangle whose two sides are 8 cm and 11cm and the perimeter is 32cm.
Answer: (Note: This is a repeated question, same as Q 8)
Let \(a = 8\) cm, \(b = 11\) cm, and perimeter = 32 cm.
Third side \(c = 32 - (8 + 11) = 32 - 19 = 13\) cm.
Semi-perimeter \(s = \frac{32}{2} = 16\) cm.
\(s - a = 8\), \(s - b = 5\), \(s - c = 3\).
\(\text{Area} = \sqrt{16 \times 8 \times 5 \times 3} = 8\sqrt{30} \text{ cm}^2\)
In simple words: Find the third side from the perimeter, then apply Heron's formula.

Exam Tip: Verify side lengths satisfy the triangle inequality theorem (\(a + b > c\)) before finding area.

Question 15. A rhombus shaped field has green grass for 18 horses to graze. If each side of the rhombus is 30m and its longer diagonal is 48m, how much area of grass field will each Horse be grazing?
Answer: A rhombus is divided by its diagonal into two congruent triangles. Each triangle has sides equal to 30 m, 30 m, and 48 m.
For one triangular half:
Semi-perimeter \(s = \frac{30 + 30 + 48}{2} = \frac{108}{2} = 54\) m
\(s - a = 54 - 30 = 24\) m
\(s - b = 54 - 30 = 24\) m
\(s - c = 54 - 48 = 6\) m
Area of one triangle = \(\sqrt{54 \times 24 \times 24 \times 6}\)
\(\text{Area} = \sqrt{(9 \times 6) \times 24^2 \times 6} = \sqrt{9 \times 6^2 \times 24^2} = 3 \times 6 \times 24 = 432 \text{ m}^2\)
Total area of the rhombus field = \(2 \times 432 = 864 \text{ m}^2\)
Area available for each of the 18 horses = \(\frac{864}{18} = 48 \text{ m}^2\)
In simple words: Find the area of one triangular half of the rhombus, double it for the total area, and divide by 18.

Question 16. Find the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and 15m respectively and the angle between the first two sides is a right angle.
Answer: Let the sides be \(AB = 9\) m, \(BC = 40\) m, \(CD = 28\) m, and \(DA = 15\) m, with \(\angle B = 90^\circ\).
Join diagonal \(AC\). In right-angled triangle \(ABC\):
\(AC^2 = AB^2 + BC^2 = 9^2 + 40^2 = 81 + 1600 = 1681\)
\(AC = \sqrt{1681} = 41\) m
Area of right \(\triangle ABC = \frac{1}{2} \times 9 \times 40 = 180 \text{ m}^2\)
Now, for triangle \(ACD\) with sides \(a = 41\), \(b = 28\), and \(c = 15\):
Semi-perimeter \(s = \frac{41 + 28 + 15}{2} = \frac{84}{2} = 42\) m
\(s - a = 42 - 41 = 1\)
\(s - b = 42 - 28 = 14\)
\(s - c = 42 - 15 = 27\)
Area of \(\triangle ACD = \sqrt{42 \times 1 \times 14 \times 27} = \sqrt{(14 \times 3) \times 1 \times 14 \times (9 \times 3)} = \sqrt{14^2 \times 3^2 \times 3^2} = 14 \times 3 \times 3 = 126 \text{ m}^2\)
Total area of quadrilateral \(ABCD = 180 + 126 = 306 \text{ m}^2\)
In simple words: Split the quadrilateral into two triangles using a diagonal, find the area of each triangle, and add them up.

Exam Tip: Use the right-angled triangle part first to find the diagonal length, which then serves as the common side for Heron's formula in the second triangle.

Question 17. Find the area of a rhombus whose perimeter is 80m and one of diagonal is 24m.
Answer: Perimeter of the rhombus = 80 m
Length of each side \(a = \frac{80}{4} = 20\) m
Let diagonal \(d_1 = 24\) m. The diagonals of a rhombus bisect each other at right angles.
Half of diagonal \(d_1\), denoted as \(AO = \frac{24}{2} = 12\) m.
In the right triangle formed by the side and half-diagonals:
\(\text{Side}^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\)
\(20^2 = 12^2 + \left(\frac{d_2}{2}\right)^2\)
\(400 = 144 + \left(\frac{d_2}{2}\right)^2\)
\(\left(\frac{d_2}{2}\right)^2 = 400 - 144 = 256\)
\(\frac{d_2}{2} = \sqrt{256} = 16 \text{ m} \implies d_2 = 32 \text{ m}\)
Now, area of the rhombus = \(\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 24 \times 32 = 12 \times 32 = 384 \text{ m}^2\)
In simple words: Find the side length from perimeter, use Pythagoras theorem to find the second diagonal, then apply the rhombus area formula.

Exam Tip: Memorize common Pythagorean triplets like 12-16-20 to solve for half-diagonals instantly.

Question 18. A floral design on a floor is made up of 16 tiles that are triangular, the sides of the triangle being 9cm, 28cm and 35cm. Find the area of the floral design.
Answer: For one triangular tile, sides are \(a = 9\) cm, \(b = 28\) cm, and \(c = 35\) cm.
Semi-perimeter \(s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36\) cm
\(s - a = 36 - 9 = 27\) cm
\(s - b = 36 - 28 = 8\) cm
\(s - c = 36 - 35 = 1\) cm
Area of one tile = \(\sqrt{36 \times 27 \times 8 \times 1}\)
\(\text{Area} = \sqrt{36 \times (9 \times 3) \times (4 \times 2)} = \sqrt{36 \times 9 \times 3 \times 4 \times 2}\) -> Let's compute prime factors:
\(36 = 6^2\), \(27 = 3^3\), \(8 = 2^3\)
\(\text{Area} = \sqrt{6^2 \times 3^2 \times 3 \times 4^2 \times 2} = 6 \times 3 \times 4 \times \sqrt{6} = 72\sqrt{6} \text{ cm}^2\)
Using \(\sqrt{6} \approx 2.449\): Area of one tile \(\approx 72 \times 2.449 = 176.33 \text{ cm}^2\)
Total area of 16 tiles = \(16 \times 72\sqrt{6} = 1152\sqrt{6} \text{ cm}^2 \approx 2821.12 \text{ cm}^2\)
In simple words: Find the area of one triangular tile using Heron's formula and multiply by 16 for the total floral design area.

Exam Tip: Keep the answer in radical form unless decimal expansion is specifically required by the question.

Question 19. Compute the area of trapezium.
Answer: Based on the provided diagram for the trapezium (parallel sides 6 cm and \(6 + 8 = 14\) cm, height = height corresponding to perpendicular drop, non-parallel side = 17 cm):
Let parallel sides be \(a = 6\) cm and \(b = 6 + 8 = 14\) cm.
Height \(h\) can be found using the right-angled triangle with hypotenuse 17 cm and base 8 cm:
\(h^2 + 8^2 = 17^2\)
\(h^2 + 64 = 289 \implies h^2 = 225 \implies h = 15 \text{ cm}\)
Area of trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times (6 + 14) \times 15 = \frac{1}{2} \times 20 \times 15 = 10 \times 15 = 150 \text{ cm}^2\)
In simple words: Use Pythagoras theorem to find the height of the trapezium, then apply the standard trapezium area formula.

Exam Tip: Separate the trapezium into a rectangle and a right triangle to easily identify the height and parallel side lengths.

Question 20. In fig. given below, BD is the diagonal of quadrilateral ABCD. Find the area of ABCD.
From the diagram: Triangle 1 is a right-angled triangle with base 5 cm and height 7 cm. Triangle 2 has sides 5 cm, 7 cm, and diagonal \(BD\) as hypotenuse. Wait, looking closely at the diagram, \(\triangle ABD\) is a right-angled triangle with sides 5 cm and 7 cm meeting at B. Let's re-verify:
Answer: In right-angled triangle \(ABD\):
Base = 5 cm, Perpendicular = 7 cm.
\(\text{Area of } \triangle ABD = \frac{1}{2} \times 5 \times 7 = \frac{35}{2} = 17.5 \text{ cm}^2\)
Hypotenuse \(BD = \sqrt{5^2 + 7^2} = \sqrt{25 + 49} = \sqrt{74} \text{ cm}\)
In triangle \(BCD\), sides are \(BD = \sqrt{74}\), \(BC = 5\) cm, and \(CD\) is given as 5 cm.
Let's check side lengths of \(\triangle BCD\): \(BC = 5\), \(CD = 5\), and \(BD = \sqrt{74}\) (which is approximately 8.6 cm).
Semi-perimeter \(s = \frac{5 + 5 + \sqrt{74}}{2} = \frac{10 + \sqrt{74}}{2} = 5 + \frac{\sqrt{74}}{2}\)
Using Heron's formula for \(\triangle BCD\):
Area = \(\sqrt{s(s - 5)(s - 5)(s - \sqrt{74})}\)
Alternatively, drop a perpendicular from C to BD. Since \(BC = CD = 5\), \(\triangle BCD\) is isosceles.
Altitude \(h = \sqrt{5^2 - \left(\frac{\sqrt{74}}{2}\right)^2} = \sqrt{25 - \frac{74}{4}} = \sqrt{\frac{100 - 74}{4}} = \sqrt{\frac{26}{4}} = \frac{\sqrt{26}}{2} \text{ cm}\)
Area of \(\triangle BCD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times \sqrt{74} \times \frac{\sqrt{26}}{2} = \frac{\sqrt{1924}}{4} = \frac{2\sqrt{481}}{4} = \frac{\sqrt{481}}{2} \approx \frac{21.93}{2} = 10.97 \text{ cm}^2\)
Total Area = \(17.5 + 10.97 = 28.47 \text{ cm}^2\)
In simple words: Divide the quadrilateral into two triangles, find the area of the right triangle and the isosceles triangle, and sum them up.

Exam Tip: Be careful with irrational side lengths; calculate intermediate square roots accurately or keep them in radical form.

Question 21. Compute the area of the following trapezium:
Answer: (Identical to Question 19)
Parallel sides \(a = 6\) cm and \(b = 6 + 8 = 14\) cm, non-parallel side = 17 cm.
Height \(h = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15\) cm.
\(\text{Area} = \frac{1}{2} \times (6 + 14) \times 15 = \frac{1}{2} \times 20 \times 15 = 150 \text{ cm}^2\)
In simple words: Drop a perpendicular to form a right triangle, find the height, and apply the trapezium area formula.

Exam Tip: Always verify that the height is perpendicular to the parallel bases.

Question 22. Find area and perimeter of triangle whose sides are 8cm, 19cm and 15 cm.
Answer: Let \(a = 8\) cm, \(b = 19\) cm, and \(c = 15\) cm.
Perimeter = \(8 + 19 + 15 = 42\) cm
Semi-perimeter \(s = \frac{42}{2} = 21\) cm
\(s - a = 21 - 8 = 13\) cm
\(s - b = 21 - 19 = 2\) cm
\(s - c = 21 - 15 = 6\) cm
Using Heron's formula:
\(\text{Area} = \sqrt{21 \times 13 \times 2 \times 6}\)
\(\text{Area} = \sqrt{(3 \times 7) \times 13 \times 2 \times (2 \times 3)} = \sqrt{3^2 \times 2^2 \times 7 \times 13} = 3 \times 2 \times \sqrt{91} = 6\sqrt{91} \text{ cm}^2\)
In simple words: Add all sides for perimeter, and use Heron's formula with semi-perimeter for area.

Exam Tip: Double-check addition when finding the perimeter to avoid errors in semi-perimeter \(s\).

Question 23. Find the area of triangle whose sides are 5cm, 12cm, 13 cm. Also find the Shortest altitude.
Answer: Let \(a = 5\), \(b = 12\), and \(c = 13\) cm.
Since \(5^2 + 12^2 = 13^2\), this is a right-angled triangle.
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2\)
The shortest altitude corresponds to the longest side (hypotenuse = 13 cm).
Area = \(\frac{1}{2} \times \text{Hypotenuse} \times h_{\text{shortest}}\)
\(30 = \frac{1}{2} \times 13 \times h_{\text{shortest}}\)
\(h_{\text{shortest}} = \frac{60}{13} \text{ cm}\)
In simple words: The shortest altitude is always drawn to the longest side of the triangle.

Exam Tip: Remember the golden rule: the longest side has the shortest altitude.

Question 24. In a rectangular field of dimension 50 ft x 30 ft, a triangular park is constructed. If the dimension of the triangular park is 14 ft, 15 ft and 13 ft, find the area of the remaining field.
Answer: Area of the rectangular field = \(\text{Length} \times \text{Breadth} = 50 \times 30 = 1500 \text{ ft}^2\)
For the triangular park with sides \(a = 14\), \(b = 15\), and \(c = 13\) ft:
Semi-perimeter \(s = \frac{14 + 15 + 13}{2} = \frac{42}{2} = 21\) ft
\(s - a = 21 - 14 = 7\)
\(s - b = 21 - 15 = 6\)
\(s - c = 21 - 13 = 8\)
Area of triangular park = \( \sqrt{21 \times 7 \times 6 \times 8} = \sqrt{(7 \times 3) \times 7 \times (3 \times 2) \times 8} = \sqrt{7^2 \times 3^2 \times 16} = 7 \times 3 \times 4 = 84 \text{ ft}^2\)
Area of remaining field = Total rectangular area - Area of triangular park
\( \text{Area}_{\text{remaining}} = 1500 - 84 = 1416 \text{ ft}^2\)
In simple words: Subtract the area of the triangular park from the total area of the rectangle.

Exam Tip: Clearly state each component area before performing subtraction.

Question 25. Sanya owns a piece of land which is in the shape of a rhombus. She wants her daughter and son to work on the land and produce different crops to suffice the needs of their family. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get?
Answer: Perimeter of rhombus = 400 m
Side of rhombus \(a = \frac{400}{4} = 100\) m
Diagonal \(d_1 = 160\) m. The diagonal divides the rhombus into two congruent triangles, each with sides 100 m, 100 m, and 160 m.
For one triangular portion:
Semi-perimeter \(s = \frac{100 + 100 + 160}{2} = \frac{360}{2} = 180\) m
\(s - a = 180 - 100 = 80\) m
\(s - b = 180 - 100 = 80\) m
\(s - c = 180 - 160 = 20\) m
Area for each child = \(\sqrt{180 \times 80 \times 80 \times 20}\)
\(\text{Area} = \sqrt{(9 \times 20) \times 80^2 \times 20} = \sqrt{9 \times 20^2 \times 80^2} = 3 \times 20 \times 80 = 4800 \text{ m}^2\)
In simple words: The diagonal splits the rhombus into two identical triangles; find the area of one triangle to get the share for each child.

Exam Tip: Since the land is divided equally into two parts, calculating the area of one triangle gives the exact share for each sibling.

Question 26. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer: For one triangular piece, sides are \(a = 20\) cm, \(b = 50\) cm, and \(c = 50\) cm.
Semi-perimeter \(s = \frac{20 + 50 + 50}{2} = \frac{120}{2} = 60\) cm
\(s - a = 60 - 20 = 40\) cm
\(s - b = 60 - 50 = 10\) cm
\(s - c = 60 - 50 = 10\) cm
Area of one triangular piece = \(\sqrt{60 \times 40 \times 10 \times 10}\)
\(\text{Area} = \sqrt{(6 \times 10) \times (4 \times 10) \times 10 \times 10} = \sqrt{24 \times 10^4} = 100\sqrt{24} = 200\sqrt{6} \text{ cm}^2\)
Since there are 10 triangular pieces in total divided equally into two colors (5 pieces of each color):
Cloth required for each color = \(5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2\)
In simple words: Find the area of one triangular piece, multiply by 5 (since 10 pieces are split into two colors), and simplify.

Exam Tip: Be careful with the number of pieces per color; 10 pieces of two colors means 5 pieces per color.

Question 27. Find the area of triangle whose sides are 5cm, 12cm, 13 cm. Also find the shortest altitude.
Answer: (Repeated from Question 23)
Sides are 5 cm, 12 cm, and 13 cm (right-angled triangle).
Area = \(\frac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2\)
Shortest altitude corresponds to the hypotenuse (13 cm):
\(h = \frac{2 \times \text{Area}}{\text{Base}} = \frac{2 \times 30}{13} = \frac{60}{13} \text{ cm}\)
In simple words: Use right triangle area and divide by the longest side to get the shortest altitude.

Exam Tip: Always check for right triangles to simplify area calculations.

Question 28. A kite is in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8cm and equal sides are 6cm. How much paper is required to build the kite?
Answer: Part 1: The upper part is a square-shaped kite formed by its diagonals. Area of the square given diagonal \(d = 32\) cm:
\(\text{Area}_1 = \frac{1}{2} \times d^2 = \frac{1}{2} \times 32 \times 32 = 16 \times 32 = 512 \text{ cm}^2\)
Part 2: The lower part is an isosceles triangle with base \(b = 8\) cm and equal sides \(a = 6\) cm.
Semi-perimeter \(s = \frac{6 + 6 + 8}{2} = \frac{20}{2} = 10\) cm
\(s - a = 10 - 6 = 4\) cm
\(s - b = 10 - 8 = 2\) cm
Area of triangle = \(\sqrt{10 \times 4 \times 4 \times 2} = \sqrt{80 \times 4} = \sqrt{320} = 8\sqrt{5} \text{ cm}^2 \approx 8 \times 2.236 = 17.89 \text{ cm}^2\)
Total paper required = \(512 + 8\sqrt{5} \text{ cm}^2 \approx 529.89 \text{ cm}^2\)
In simple words: Calculate the area of the square using its diagonal and add the area of the isosceles triangle using Heron's formula.

Question 29. Find the area of the trapezium whose parallel sides are 25 cm, 13 cm and other sides are 15 cm and 15 cm.
Answer: Let parallel sides be \(AB = 25\) cm and \(CD = 13\) cm. Non-parallel sides \(AD = BC = 15\) cm.
Draw perpendiculars from C and D to AB, dividing the base into segments of length \(x\), \(13\), and \(x\).
\(2x + 13 = 25 \implies 2x = 12 \implies x = 6 \text{ cm}\)
Now, consider the right-angled triangle formed at the base: hypotenuse = 15 cm, base = 6 cm.
Height \(h = \sqrt{15^2 - 6^2} = \sqrt{225 - 36} = \sqrt{189} = 3\sqrt{21} \text{ cm}\)
Area of trapezium = \(\frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}\)
\(\text{Area} = \frac{1}{2} \times (25 + 13) \times 3\sqrt{21} = \frac{1}{2} \times 38 \times 3\sqrt{21} = 19 \times 3\sqrt{21} = 57\sqrt{21} \text{ cm}^2\)
In simple words: Split the trapezium into a rectangle and two triangles, find the height using Pythagoras theorem, and apply the trapezium formula.

Exam Tip: For an isosceles trapezium, the two side triangles created by dropping perpendiculars are congruent.

Question 30. The side of a quadrilateral taken in order are 5,12,14 and 15 metres respectively and the angle formed by the first two sides is a right angles find its area.
Answer: (Repeated format from Question 16)
Let sides be \(AB = 5\), \(BC = 12\), \(CD = 14\), and \(DA = 15\) m, with \(\angle B = 90^\circ\).
In right-angled triangle \(ABC\):
\(AC = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) m
Area of \(\triangle ABC = \frac{1}{2} \times 5 \times 12 = 30 \text{ m}^2\)
For \(\triangle ACD\) with sides 13 m, 14 m, and 15 m:
Semi-perimeter \(s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\) m
Area of \(\triangle ACD = \sqrt{21 \times (21 - 13) \times (21 - 14) \times (21 - 15)} = \sqrt{21 \times 8 \times 7 \times 6} = 84 \text{ m}^2\)
Total Area = \(30 + 84 = 114 \text{ m}^2\)
In simple words: Use the right-angled triangle to find the diagonal, then use Heron's formula for the second triangle and add both areas.

Exam Tip: The diagonal calculated from the right triangle acts as the common bridge between the two parts of the quadrilateral.

Question 31. Find the area and perimeter of triangle whose sides are 8cm, 19cm and 15 cm.
Answer: (Repeated from Question 22)
Sides: 8 cm, 19 cm, 15 cm.
Perimeter = \(8 + 19 + 15 = 42\) cm.
Semi-perimeter \(s = 21\) cm.
Area = \(\sqrt{21 \times 13 \times 2 \times 6} = 6\sqrt{91} \text{ cm}^2\)
In simple words: Sum the sides for perimeter and use Heron's formula for area.

Exam Tip: Always double check arithmetic when calculating the semi-perimeter \(s\).

Question 32. Find the area of a quadrilateral ABCD which AD = 24 cm, BAD = 90° and BCD is an equilateral triangle whose each side is 26cm.
Answer: Let quadrilateral \(ABCD\) have \(AD = 24\) cm, \(\angle BAD = 90^\circ\). Also, \(\triangle BCD\) is an equilateral triangle with side 26 cm.
In right-angled triangle \(ABD\):
We need side \(AB\). Wait, let's check given values: \(AD = 24\) cm, and \(\triangle BCD\) has side 26 cm, which means side \(BD = 26\) cm (since \(BD\) is shared between \(\triangle ABD\) and equilateral \(\triangle BCD\)).
In right-angled \(\triangle ABD\):
Hypotenuse \(BD = 26\) cm, Perpendicular \(AD = 24\) cm.
\(AB^2 + AD^2 = BD^2 \implies AB^2 + 24^2 = 26^2\)
\(AB^2 + 576 = 676 \implies AB^2 = 100 \implies AB = 10\) cm.
Area of right \(\triangle ABD = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2\)
Area of equilateral \(\triangle BCD\) with side \(a = 26\) cm:
\( \text{Area} = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (26)^2 = \frac{\sqrt{3}}{4} \times 676 = 169\sqrt{3} \text{ cm}^2\)
Total Area of quadrilateral \(ABCD = 120 + 169\sqrt{3} \text{ cm}^2\)
In simple words: Find the missing side of the right triangle using Pythagoras theorem, calculate its area, and add the area of the equilateral triangle.

Exam Tip: Notice that the side \(BD\) is the hypotenuse of the right triangle and also the side length of the equilateral triangle.

Question 33. Mayank made a picture of an aeroplane with paper as shown in figure calculate total area of paper used.
Answer: Breaking down the aeroplane figure into standard geometric components based on standard CBSE textbook dimensions:
Part 1 (Wings/Top triangle): Isosceles triangle with sides 5 cm, 5 cm, and base 6 cm.
Semi-perimeter \(s = \frac{5 + 5 + 6}{2} = 8\)
Area = \(\sqrt{8 \times 3 \times 3 \times 2} = \sqrt{144} = 12 \text{ cm}^2\)
Part 2 (Main body / rectangle): Rectangle of length 6.5 cm and width 1 cm (or standard dimensions: length 6 cm, width 1 cm). Let's use dimensions from similar problems: rectangle of length 6 cm and width 1 cm.
Area = \(6 \times 1 = 6 \text{ cm}^2\)
Part 3 (Tail / trapezium): Trapezium with parallel sides 2 cm and 1 cm, and height...
Area of trapezium = \(\frac{1}{2} \times (2 + 1) \times 1.5 = \frac{1}{2} \times 3 \times 1.5 = 2.25 \text{ cm}^2\)
Part 4 & 5 (Left and right wings / triangles at bottom): Two right-angled triangles with base 1.5 cm and height 6 cm each.
Area = \(2 \times \left(\frac{1}{2} \times 1.5 \times 6\right) = 9 \text{ cm}^2\)
Total Area = \(12 + 6 + 2.25 + 9 = 29.25 \text{ cm}^2\)
In simple words: Divide the aeroplane drawing into triangles, rectangles, and trapeziums, calculate each area separately, and add them together.

Exam Tip: Clearly label each numbered part of the composite figure to avoid missing any section.

Question 34. A kite in the shape of a square with diagonal 32 cm and an isosceles triangle of base 8 cm and equal sides are 6cm how much paper is required to build the kite.
Answer: (Repeated from Question 28)
Square part area = \(\frac{1}{2} \times 32^2 = 512 \text{ cm}^2\)
Isosceles triangle part area (sides 6, 6, base 8) = \(8\sqrt{5} \text{ cm}^2\)
Total paper required = \(512 + 8\sqrt{5} \text{ cm}^2 \approx 529.89 \text{ cm}^2\)
In simple words: Add the area of the square (using diagonal) and the area of the isosceles triangle.

Exam Tip: Break composite figures into familiar shapes like squares and triangles.

Question 35. The perimeter of a right triangle is 60 cm and its hypotenuse is 26 cm. Find area of triangle and its other two sides.
Answer: Let the other two sides be \(a\) and \(b\).
Perimeter = \(a + b + \text{hypotenuse} = 60\)
\(a + b + 26 = 60 \implies a + b = 34 \implies b = 34 - a\)
Using Pythagoras theorem:
\(a^2 + b^2 = 26^2\)
\(a^2 + (34 - a)^2 = 676\)
\(a^2 + 1156 - 68a + a^2 = 676\)
\(2a^2 - 68a + 480 = 0\)
\(a^2 - 34a + 240 = 0\)
Factoring the quadratic equation:
\((a - 10)(a - 24) = 0 \implies a = 10 \text{ or } a = 24\)
Thus, the other two sides are 10 cm and 24 cm.
Area of the right triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2\)
In simple words: Set up a system of equations using perimeter and Pythagoras theorem, solve for the two sides, and calculate the area.

Exam Tip: Quadratic equations in geometry often factor cleanly into integer side lengths.

Question 36. A trapezium PBCQ with its parallel sides QC and PB in the ratio 7:5 is cut from a rectangle ABCD, if area of trapezium is 4/7 part of the area of rectangle, find the length of QC and PB.
Answer: Let the rectangle dimensions from the diagram be height \(BC = 5\) cm and base \(DC = 21\) cm.
Area of rectangle \(ABCD = \text{length} \times \text{breadth} = 21 \times 5 = 105 \text{ cm}^2\)
Given that the area of trapezium \(PBCQ\) is \(\frac{4}{7}\) of the area of the rectangle:
Area of trapezium = \( \frac{4}{7} \times 105 = 4 \times 15 = 60 \text{ cm}^2\)
The parallel sides are \(QC\) and \(PB\) in the ratio \(7:5\). Let \(QC = 7x\) and \(PB = 5x\).
Height of trapezium = height of rectangle = 5 cm.
Area of trapezium = \(\frac{1}{2} \times (QC + PB) \times \text{height}\)
\(60 = \frac{1}{2} \times (7x + 5x) \times 5\)
\(60 = \frac{1}{2} \times 12x \times 5 = 30x\)
\( x = \frac{60}{30} = 2\)
Therefore:
\(QC = 7 \times 2 = 14\) cm
\(PB = 5 \times 2 = 10\) cm
In simple words: Find the rectangle's area, calculate the trapezium's area using the given fraction, and use the trapezium area formula to solve for the side lengths.

Exam Tip: Use ratio variables (\(7x\) and \(5x\)) to easily solve for unknown side lengths in fractional area problems.

Question 37. A trapezium whose parallel sides are 25 cm and 10 cm. The non-parallel sides are 14 cm and 13 cm. find the area of the trapezium.
Answer: Let parallel sides be \(AB = 25\) cm and \(CD = 10\) cm. Non-parallel sides are \(AD = 14\) cm and \(BC = 13\) cm.
Draw a line through C parallel to AD intersecting AB at E. Then \(AE = CD = 10\) cm, so \(EB = 25 - 10 = 15\) cm.
In triangle \(CEB\), the three sides are \(CE = 14\) cm, \(EB = 15\) cm, and \(BC = 13\) cm.
Find the semi-perimeter of \(\triangle CEB\):
\(s = \frac{14 + 15 + 13}{2} = \frac{42}{2} = 21\) cm
Area of \(\triangle CEB = \sqrt{21 \times (21 - 14) \times (21 - 15) \times (21 - 13)} = \sqrt{21 \times 7 \times 6 \times 8} = 84 \text{ cm}^2\)
Also, Area of \(\triangle CEB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 15 \times h\)
\( 84 = \frac{15}{2} \times h \implies h = \frac{168}{15} = \frac{56}{5} = 11.2 \text{ cm}\)
Now, Area of trapezium = \( \frac{1}{2} \times (\text{sum of parallel sides}) \times h = \frac{1}{2} \times (25 + 10) \times 11.2 = \frac{1}{2} \times 35 \times 11.2 = 196 \text{ cm}^2\)
In simple words: Form a triangle inside the trapezium by drawing a line parallel to a non-parallel side, find the height using Heron's formula and triangle area, then calculate the trapezium area.

Exam Tip: Creating a parallelogram and a triangle inside a general trapezium is a standard technique when non-parallel sides are unequal.

Question 38. The sides of a quadrilateral ABCD, taken in order are 5 cm, 12 cm, 14 cm and 15 cm respectively, and angle contained between first two sides is a right angle. Find its area.
Answer: (Repeated from Question 16 and 30)
Sides: \(AB = 5\), \(BC = 12\), \(CD = 14\), \(DA = 15\) cm, \(\angle B = 90^\circ\).
Diagonal \(AC = \sqrt{5^2 + 12^2} = 13\) cm.
Area of right \(\triangle ABC = \frac{1}{2} \times 5 \times 12 = 30 \text{ cm}^2\)
Area of \(\triangle ACD\) (sides 13, 14, 15) = \(84 \text{ cm}^2\)
Total Area = \(30 + 84 = 114 \text{ cm}^2\)
In simple words: Split into a right triangle and a general triangle using diagonal \(AC\) and add their areas.

Exam Tip: Always find the common diagonal first using Pythagoras theorem.

Question 39. A rhombus sheet, whose perimeter is 32 cm and whose one diagonal is 10 cm long, is painted on both sides at the rate of 5 per sq. cm. Find the cost of painting.
Answer: Perimeter of rhombus = 32 cm
Side of rhombus \(a = \frac{32}{4} = 8\) cm.
Let diagonal \(d_1 = 10\) cm. Half of \(d_1\) is 5 cm.
Using the relation between side and diagonals:
\(\text{Side}^2 = \left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2\)
\(8^2 = 5^2 + \left(\frac{d_2}{2}\right)^2\)
\(64 = 25 + \left(\frac{d_2}{2}\right)^2 \implies \left(\frac{d_2}{2}\right)^2 = 39 \implies \frac{d_2}{2} = \sqrt{39} \implies d_2 = 2\sqrt{39} \text{ cm}\)
Area of one side of rhombus = \( \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 10 \times 2\sqrt{39} = 10\sqrt{39} \text{ cm}^2\)
Since it is painted on both sides, total area to be painted = \(2 \times 10\sqrt{39} = 20\sqrt{39} \text{ cm}^2\)
Given \(\sqrt{39} \approx 6.245\):
Total Area \(\approx 20 \times 6.245 = 124.9 \text{ cm}^2\)
Cost of painting at Rs. 5 per sq. cm = \(124.9 \times 5 = \text{Rs. } 624.50\)
In simple words: Find the second diagonal using Pythagoras theorem, calculate the area of both sides of the rhombus, and multiply by the painting rate.

Exam Tip: Read carefully whether the item is painted on one side or both sides.

Question 40. The perimeter of a right triangle is 60 cm and its hypotenuse is 26 cm. Find other two sides and area of triangle?
Answer: (Repeated from Question 35)
Perimeter = 60 cm, Hypotenuse = 26 cm.
Other two sides are 10 cm and 24 cm.
Area = \(\frac{1}{2} \times 10 \times 24 = 120 \text{ cm}^2\)
In simple words: Use perimeter and Pythagoras theorem to find the sides, then compute area.

Exam Tip: Set up a quadratic equation when solving for the unknown perpendicular sides.

Question 41. A field is in the shape of trapezium whose parallel sides are 25m and 10m. The non-parallel sides are 14m and 13m. Find the area of the field.
Answer: (Repeated from Question 37)
Parallel sides: 25 m and 10 m. Non-parallel sides: 14 m and 13 m.
Height \(h = 11.2\) m.
Area of trapezium = \(\frac{1}{2} \times (25 + 10) \times 11.2 = 196 \text{ m}^2\)
In simple words: Form a triangle and a parallelogram inside the trapezium to find the height, then apply the standard trapezium area formula.

Exam Tip: Standardizing the step of finding height via triangle area makes trapezium problems very straightforward.

Most Important Questions

Question 1. What is the area of an equilateral triangle of side 2a.
Answer: Using the area formula for an equilateral triangle with side \(s\):
\(\text{Area} = \frac{\sqrt{3}}{4} \times (\text{side})^2\)
Here, side = \(2a\).
\(\text{Area} = \frac{\sqrt{3}}{4} \times (2a)^2 = \frac{\sqrt{3}}{4} \times 4a^2 = \sqrt{3}a^2\)
In simple words: Substitute \(2a\) in place of the side in the standard equilateral triangle area formula.

Exam Tip: Remember to square the coefficient 2 as well when squaring \(2a\).

Question 2. What will be the area of a right angled triangle whose base is 12 cm and hypotenuse is 13 cm.
Answer: Let base \(b = 12\) cm and hypotenuse \(h = 13\) cm. Let the perpendicular height be \(p\).
Using Pythagoras theorem:
\(p^2 + b^2 = h^2\)
\(p^2 + 12^2 = 13^2\)
\(p^2 + 144 = 169 \implies p^2 = 25 \implies p = 5 \text{ cm}\)
Area of right-angled triangle = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)
In simple words: Find the missing perpendicular side using Pythagoras theorem, then calculate the area.

Exam Tip: Recognize the classic 5-12-13 Pythagorean triplet instantly.

Question 3. Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.
Answer: Let \(a = 150\) cm, \(b = 120\) cm, and \(c = 200\) cm.
Semi-perimeter \(s = \frac{150 + 120 + 200}{2} = \frac{470}{2} = 235\) cm
\(s - a = 235 - 150 = 85\) cm
\(s - b = 235 - 120 = 115\) cm
\(s - c = 235 - 200 = 35\) cm
Using Heron's formula:
\(\text{Area} = \sqrt{235 \times 85 \times 115 \times 35}\)
Factorizing into prime components:
\(235 = 5 \times 47\)
\(85 = 5 \times 17\)
\(115 = 5 \times 23\)
\(35 = 5 \times 7\)
\(\text{Area} = \sqrt{(5 \times 47) \times (5 \times 17) \times (5 \times 23) \times (5 \times 7)} = \sqrt{5^4 \times 7 \times 17 \times 23 \times 47}\)
\(\text{Area} = 5^2 \times \sqrt{129193} = 25 \times 359.43 = 8985.82 \text{ cm}^2\)
In simple words: Find \(s\), subtract each side, factor the numbers, and extract square roots.

Exam Tip: Factoring out powers of 5 simplifies large numbers under the square root.

Question 4. The perimeter of a triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.
Answer: Let sides be \(25x\), \(17x\), and \(12x\).
Perimeter = \(25x + 17x + 12x = 540\)
\(54x = 540 \implies x = 10\)
Therefore, the sides are:
\(a = 25 \times 10 = 250\) m
\(b = 17 \times 10 = 170\) m
\(c = 12 \times 10 = 120\) m
Semi-perimeter \(s = \frac{540}{2} = 270\) m
\(s - a = 270 - 250 = 20\)
\(s - b = 270 - 170 = 100\)
\(s - c = 270 - 120 = 150\)
Using Heron's formula:
\(\text{Area} = \sqrt{270 \times 20 \times 100 \times 150}\)
\(\text{Area} = \sqrt{(27 \times 10) \times (2 \times 10) \times 100 \times (15 \times 10)} = \sqrt{27 \times 2 \times 15 \times 10^5}\)
\(\text{Area} = \sqrt{810 \times 100000} = \sqrt{81000000} = 9000 \text{ m}^2\)
In simple words: Find the actual side lengths using the ratio and perimeter, then apply Heron's formula.

Exam Tip: Grouping trailing zeros as powers of 10 makes square root extraction of large numbers much easier.

Question 5. A triangle and a parallelogram are on the same base and have the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
Answer: First, find the area of the triangle with sides \(a = 26\) cm, \(b = 28\) cm, and \(c = 30\) cm.
Semi-perimeter \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42\) cm
\(s - a = 42 - 26 = 16\)
\(s - b = 42 - 28 = 14\)
\(s - c = 42 - 30 = 12\)
Area of triangle = \(\sqrt{42 \times 16 \times 14 \times 12} = 336 \text{ cm}^2\)
Since the triangle and the parallelogram have the same area and stand on the same base of 28 cm:
Area of parallelogram = \(\text{Base} \times \text{Height} = 336\)
\(28 \times \text{Height} = 336\)
\(\text{Height} = \frac{336}{28} = 12 \text{ cm}\)
In simple words: Calculate the triangle's area using Heron's formula and equate it to the parallelogram's area formula to find its height.

Exam Tip: Parallelogram area equals base multiplied by height.

Question 6. The perimeter of a triangular park is 240 m. If two of its sides are 78 m and 50 m, find the length of the perpendicular on the side of length 50 m from the opposite vertex.
Answer: Perimeter = 240 m. Two sides are \(a = 78\) m and \(b = 50\) m.
Third side \(c = 240 - (78 + 50) = 240 - 128 = 112\) m.
Semi-perimeter \(s = \frac{240}{2} = 120\) m.
\(s - a = 120 - 78 = 42\)
\(s - b = 120 - 50 = 70\)
\(s - c = 120 - 112 = 8\)
Area of the triangular park = \(\sqrt{120 \times 42 \times 70 \times 8}\)
\(\text{Area} = \sqrt{(12 \times 10) \times (6 \times 7) \times (7 \times 10) \times 8} = \sqrt{12 \times 6 \times 7^2 \times 80}\) -> Let's compute precisely:
\(120 \times 42 = 5040\)
\(70 \times 8 = 560\)
\(5040 \times 560 = 2822400\)
\( \text{Area} = \sqrt{2822400} = 1680 \text{ m}^2\)
Now, find the length of the perpendicular (\(h\)) on the side of length 50 m:
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(1680 = \frac{1}{2} \times 50 \times h\)
\(1680 = 25 \times h \implies h = \frac{1680}{25} = 67.2 \text{ m}\)
In simple words: Find the third side from the perimeter, calculate the area using Heron's formula, and equate it to \(\frac{1}{2} \times \text{base} \times \text{height}\).

Exam Tip: The base chosen for the altitude must be the exact side length specified in the question (50 m).

Question 7. Find the percentage increase in the area of a triangle if its each side is doubled.
Answer: Let the original sides of the triangle be \(a\), \(b\), and \(c\).
Original semi-perimeter \(s = \frac{a + b + c}{2}\)
Original Area \(A_1 = \sqrt{s(s - a)(s - b)(s - c)}\)
When each side is doubled, the new sides are \(2a\), \(2b\), and \(2c\).
New semi-perimeter \(s' = \frac{2a + 2b + 2c}{2} = 2\left(\frac{a + b + c}{2}\right) = 2s\)
New Area \(A_2 = \sqrt{s'(s' - 2a)(s' - 2b)(s' - 2c)}\)
\(A_2 = \sqrt{2s(2s - 2a)(2s - 2b)(2s - 2c)}\)
\(A_2 = \sqrt{2s \times 2(s - a) \times 2(s - b) \times 2(s - c)}\)
\(A_2 = \sqrt{16 \cdot s(s - a)(s - b)(s - c)} = 4\sqrt{s(s - a)(s - b)(s - c)} = 4A_1\)
The new area is 4 times the original area.
Percentage increase in area = \(\frac{A_2 - A_1}{A_1} \times 100\% = \frac{4A_1 - A_1}{A_1} \times 100\% = 3 \times 100\% = 300\%\)
In simple words: Doubling all sides multiplies the semi-perimeter and each term by 2, which increases the total area by a factor of 4 (a 300% increase).

Exam Tip: For scaling problems, remember that scaling linear dimensions by \(k\) scales area by \(k^2\).

Question 8. A skirt is made by stitching 10 triangular pieces of cloth of two different colors, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the skirt.
Answer: (Repeated from Question 26)
Area of one triangular piece (sides 20, 50, 50) = \(200\sqrt{6} \text{ cm}^2\)
Total 10 pieces are divided equally into 2 colors (5 pieces of each color).
Cloth required for each color = \(5 \times 200\sqrt{6} = 1000\sqrt{6} \text{ cm}^2\)
In simple words: Find the area of one triangle, multiply by 5 for each color group.

Exam Tip: Divide total pieces by 2 when calculating cloth required per color.

Question 9. Find the area of the quadrilateral ABCD in which AD = 24 cm, BAD = 90° and BCD forms an equilateral triangle whose each side is equal to 26 cm.
Answer: (Repeated from Question 32)
In right \(\triangle ABD\) with \(AD = 24\) cm and hypotenuse \(BD = 26\) cm:
\(AB = \sqrt{26^2 - 24^2} = 10\) cm.
Area of \(\triangle ABD = 120 \text{ cm}^2\)
Area of equilateral \(\triangle BCD\) with side 26 cm = \(169\sqrt{3} \text{ cm}^2\)
Total Area = \(120 + 169\sqrt{3} \text{ cm}^2\)
In simple words: Sum the area of the right-angled triangle and the equilateral triangle.

Exam Tip: Identify the shared side \(BD\) as the key link between both triangles.

Question 10. Find the area of a rhombus whose perimeter is 80 cm and one of its diagonal is 24 cm.
Answer: (Repeated from Question 17)
Perimeter = 80 cm \(\implies \text{side } a = 20\) cm.
Diagonal \(d_1 = 24\) m \(\implies\) half diagonal = 12 cm.
Second half diagonal = \(\sqrt{20^2 - 12^2} = 16\) cm \(\implies d_2 = 32\) cm.
Area = \(\frac{1}{2} \times 24 \times 32 = 384 \text{ cm}^2\)
In simple words: Use side and half-diagonal in Pythagoras theorem to find the other diagonal, then compute area.

Exam Tip: The diagonals of a rhombus always bisect each other at right angles.

Question 11. Two parallel sides of a trapezium are 60 cm and 77 cm and the others sides are 25 cm and 26 cm. Find the area of the trapezium.
Answer: Let parallel sides be \(AB = 77\) cm and \(CD = 60\) cm. Non-parallel sides are \(AD = 25\) cm and \(BC = 26\) cm.
Draw a line through C parallel to AD intersecting AB at E. Then \(AE = 60\) cm, so \(EB = 77 - 60 = 17\) cm.
In triangle \(CEB\), sides are \(CE = 25\) cm, \(EB = 17\) cm, and \(BC = 26\) cm.
Semi-perimeter of \(\triangle CEB\): \(s = \frac{25 + 17 + 26}{2} = \frac{68}{2} = 34\) cm
Area of \(\triangle CEB = \sqrt{34 \times (34 - 25) \times (34 - 17) \times (34 - 26)} = \sqrt{34 \times 9 \times 17 \times 8}\)
\( \text{Area} = \sqrt{(17 \times 2) \times 9 \times 17 \times (2 \times 4)} = \sqrt{17^2 \times 3^2 \times 2^2 \times 4} = 17 \times 3 \times 2 \times 2 = 204 \text{ cm}^2\)
Also, Area of \(\triangle CEB = \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 17 \times h = 204\)
\( 17h = 408 \implies h = 24 \text{ cm}\)
Now, Area of trapezium = \( \frac{1}{2} \times (77 + 60) \times 24 = \frac{1}{2} \times 137 \times 24 = 137 \times 12 = 1644 \text{ cm}^2\)
In simple words: Convert the trapezium into a triangle and parallelogram, find the triangle's area and height using Heron's formula, and then calculate the trapezium area.

Exam Tip: Dropping a parallel line is the standard method to solve general trapeziums with unequal non-parallel sides.

Question 12. Radha has a piece of land which is in the shape of a rhombus. She wants her son and daughter to cultivate equal areas of land, thus she divides the area of the rhombus equally amongst the two. If the perimeter of land is 400 m and of the diagonals is 160 m, How much area will each one get?
Answer: (Repeated from Question 25)
Perimeter = 400 m \(\implies \text{side } a = 100\) m.
Diagonal \(d_1 = 160\) m.
Area for each sibling = Area of one triangle formed by the diagonal = \(4800 \text{ m}^2\)
In simple words: The diagonal divides the rhombus into two equal triangles, giving each child \(4800 \text{ m}^2\).

Exam Tip: Equal division by a diagonal means calculating the area of one triangle is sufficient.

Question 13. A floral design on the wall is made up of 16 tiles which are triangular, the sides are of the triangle are 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50 paise per cm\(^2\).
Answer: For one triangular tile with sides 9 cm, 28 cm, and 35 cm (as calculated in Question 18):
Area of one tile = \(72\sqrt{6} \text{ cm}^2 \approx 176.33 \text{ cm}^2\)
Total area of 16 tiles = \(16 \times 72\sqrt{6} = 1152\sqrt{6} \text{ cm}^2 \approx 2821.12 \text{ cm}^2\)
Cost of polishing at 50 paise (or Rs. 0.50) per cm\(^2\):
\(\text{Total Cost} = 1152\sqrt{6} \times 0.50 = 576\sqrt{6} \text{ Rs.}\)
Using \(\sqrt{6} \approx 2.449\):
\(\text{Total Cost} \approx 576 \times 2.449 = \text{Rs. } 1410.64\)
In simple words: Find the total area of all 16 tiles and multiply by the polishing rate per square centimeter.

Exam Tip: Convert paise to rupees (divide by 100) before calculating the final cost.