CBSE Class 9 Mathematics Herons Formula Assignment Set 12

Fill the gap:

Question 1. The length of diagonal of a square of side 5 cm is _______
Answer: \(5\sqrt{2}\text{ cm}\)
Side of the square = \(5\text{ cm}\)
Length of diagonal = \(a\sqrt{2} = 5\sqrt{2}\text{ cm}\)
In simple words: Multiply the side length of the square by \(\sqrt{2}\) to find the diagonal.

Exam Tip: Remember the standard formula for the diagonal of a square with side \(a\) is \(a\sqrt{2}\).

Question 2. The height of an equilateral \(\Delta\) whose each side is 6 cm is _______
Answer: \(3\sqrt{3}\text{ cm}\)
Side of the equilateral triangle (\(a\)) = \(6\text{ cm}\)
Height (\(h\)) = \(\frac{\sqrt{3}}{2}a = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3}\text{ cm}\)
In simple words: The height of an equilateral triangle is found using the formula \(\frac{\sqrt{3}}{2}a\).

Exam Tip: Memorize the height formula for an equilateral triangle to save time during exams.

Question 3. The length of diagonal of a rectangle whose adjacent sides are 8 cm and 6 cm respectively is _______
Answer: \(10\text{ cm}\)
Length (\(l\)) = \(8\text{ cm}\)
Breadth (\(b\)) = \(6\text{ cm}\)
Diagonal = \(\sqrt{l^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\)
In simple words: Use Pythagoras theorem on the rectangle's sides to find the diagonal length.

Exam Tip: Recognize Pythagorean triples like \(6, 8, 10\) to write the answer instantly.

Question 4. The side of an equilateral \(\Delta\) whose area is \(16\sqrt{3}\text{ cm}^2\) is _______
Answer: \(8\text{ cm}\)
Area = \(\frac{\sqrt{3}}{4}a^2 = 16\sqrt{3}\)
\(a^2 = \frac{16\sqrt{3} \times 4}{\sqrt{3}} = 64\)
\(a = \sqrt{64} = 8\text{ cm}\)
In simple words: Equate the given area to the equilateral triangle area formula and solve for side \(a\).

Exam Tip: Cancel out common radical terms like \(\sqrt{3}\) on both sides to simplify calculations quickly.

Question 5. The side of a square whose area is 4 hectare is _______
Answer: \(200\text{ m}\)
\(1\text{ hectare} = 10,000\text{ m}^2\)
Area = \(4 \times 10,000 = 40,000\text{ m}^2\)
Side = \(\sqrt{\text{Area}} = \sqrt{40,000} = 200\text{ m}\)
In simple words: Convert hectares to square meters first, then take the square root to get the side length.

Exam Tip: Always check units carefully; convert hectares to square meters (\(1\text{ ha} = 10^4\text{ m}^2\)) before proceeding.

Question 6. The area of a square field whose side is 10 cm each is _______
Answer: \(100\text{ cm}^2\)
Side (\(a\)) = \(10\text{ cm}\)
Area = \(a^2 = (10)^2 = 100\text{ cm}^2\)
In simple words: Square the side length to find the area of a square.

Exam Tip: Ensure units are written in square units (\(\text{cm}^2\)) for area.

Question 7. The area of a square field whose perimeter is 20 cm is _______
Answer: \(25\text{ cm}^2\)
Perimeter = \(4a = 20\text{ cm}\)
Side (\(a\)) = \(\frac{20}{4} = 5\text{ cm}\)
Area = \(a^2 = (5)^2 = 25\text{ cm}^2\)
In simple words: Divide the perimeter by 4 to get the side, then square it to find the area.

Question 8. The cost of white washing a room measuring 5m, 4m, 3m respectively at the rate of Rs 50/sq m is _______
Answer: Rs 3700
Dimensions of the room: length \(l = 5\text{ m}\), breadth \(b = 4\text{ m}\), height \(h = 3\text{ m}\)
Total surface area to be whitewashed (four walls + ceiling) = \(lh + 2(lh + bh) = 2(lh + bh) + lb\) or calculating lateral surface area plus ceiling area:
Lateral surface area = \(2(l + b) \times h = 2(5 + 4) \times 3 = 2 \times 9 \times 3 = 54\text{ m}^2\)
Area of ceiling = \(l \times b = 5 \times 4 = 20\text{ m}^2\)
Total area = \(54 + 20 = 74\text{ m}^2\)
Cost = \(74 \times \text{Rs } 50 = \text{Rs } 3700\)
In simple words: Calculate the area of the four walls and the ceiling, then multiply by the rate per square meter.

Exam Tip: For whitewashing rooms, unless the floor is explicitly mentioned, include the four walls and the ceiling.

Question 9. The area of a \(\Delta\) whose sides are 5 cm, 12 cm, and 13 cm is _______
Answer: \(30\text{ cm}^2\)
Since \(5^2 + 12^2 = 25 + 144 = 169 = 13^2\), the triangle is a right-angled triangle.
Base = \(5\text{ cm}\), Height = \(12\text{ cm}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 12 = 30\text{ cm}^2\)
In simple words: Check if the sides form a Pythagorean triplet to easily use the right triangle area formula.

Exam Tip: Spotting right-angled triangles saves time compared to using Heron's formula.

Question 10. Area of an equilateral \(\Delta\) whose each side is \(x\) is _______
Answer: \(\frac{\sqrt{3}}{4}x^2\)
Standard formula for the area of an equilateral triangle with side \(x\).
In simple words: Use the standard area formula for equilateral triangles with variable \(x\).

Exam Tip: Memorize standard algebraic area formulas for regular polygons.

Question 11. The area of a \(\Delta\) whose sides are \(a, b, c\) is _______
Answer: \(\sqrt{s(s-a)(s-b)(s-c)} \text{ where } s = \frac{a+b+c}{2}\)
This is Heron's formula for the area of a triangle.
In simple words: Heron's formula uses the semi-perimeter \(s\) and all three side lengths.

Question 12. The area of trapezium is _______
Answer: \(\frac{1}{2}(a+b)h \text{ where } a \parallel b\)
Area = Half the sum of parallel sides multiplied by the distance between them.
In simple words: Average the two parallel sides and multiply by the height.

Exam Tip: Clearly state that \(a\) and \(b\) represent the lengths of the parallel sides.

Question 13. The area of a rhombus is _______
Answer: \(\frac{1}{2} d_1 d_2 \text{ where } d_1, d_2 \text{ are diagonals}\)
Area is half the product of its diagonals.
In simple words: Multiply the two diagonals together and divide by 2.

Exam Tip: This formula is specific to rhombuses and kites where diagonals intersect at right angles.

Question 14. The height of a \(\Delta\) whose base is 9 cm and area is \(72\text{ cm}^2\) is _______
Answer: \(16\text{ cm}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = 72\)
\(\frac{1}{2} \times 9 \times h = 72\)
\(h = \frac{72 \times 2}{9} = 8 \times 2 = 16\text{ cm}\)
In simple words: Rearrange the triangle area formula to solve for height: \(h = \frac{2 \times \text{Area}}{\text{base}}\). Note: The printed answer key value of 8 is a typo; correct calculation gives 16 cm.

Exam Tip: Double-check arithmetic when solving for dimensions from a given area.

Question 15. The area of an isosceles \(\Delta\) whose equal side is 'a' and base is 'b' is _______
Answer: \(\frac{b}{4}\sqrt{4a^2 - b^2}\)
Height of isosceles triangle = \(\sqrt{a^2 - \left(\frac{b}{2}\right)^2} = \frac{1}{2}\sqrt{4a^2 - b^2}\)
Area = \(\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times b \times \frac{1}{2}\sqrt{4a^2 - b^2} = \frac{b}{4}\sqrt{4a^2 - b^2}\)
In simple words: Drop a perpendicular to split the base in half, use Pythagoras theorem for height, then calculate area.

Exam Tip: Remembering this direct formula saves considerable time in objective questions.