krit.club logo

Heron's Formula - Application of Heron's Formula in finding Areas of Quadrilaterals

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Heron's Formula for Triangles: The foundation for finding quadrilateral areas involves calculating the area of a triangle when all three sides are known. If a,b,a, b, and cc are the sides, we first find the semi-perimeter ss. Visually, ss represents half the boundary length of the triangular region.

•

Quadrilateral Decomposition: To find the area of any general quadrilateral, it is divided into two triangles by drawing one of its diagonals. Visually, imagine drawing a straight line between two opposite corners (vertices) of a four-sided figure, creating two separate triangular shapes that share a common side (the diagonal).

•

Sum of Triangular Areas: The total area of a quadrilateral is the sum of the areas of the two triangles formed by the diagonal. Visually, if a quadrilateral ABCDABCD is split by diagonal ACAC, the total shaded region is the combination of the shaded regions of △ABC\triangle ABC and △ADC\triangle ADC.

•

Application of Pythagoras Theorem: In many problems, a quadrilateral might have one right angle. If the diagonal's length is unknown, the Pythagoras theorem (h2=p2+b2h^2 = p^2 + b^2) is applied to the right-angled triangle to find the length of the common diagonal. Visually, this diagonal acts as the hypotenuse of the right triangle.

•

Properties of Special Quadrilaterals: For specific shapes like a rhombus or a parallelogram, a diagonal divides the figure into two congruent triangles. Visually, this means both triangles have the same shape and size, so you can calculate the area of one triangle and simply multiply it by 2 to get the total area.

•

Consistent Units and Precision: When applying Heron's formula, ensure all side lengths are in the same units (e.g., all in cmcm or all in mm). When calculating square roots that are not perfect squares, it is common practice to round to two decimal places for the final area calculation.

•

Perimeter vs Semi-perimeter: Distinguish between the perimeter (P=a+b+cP = a+b+c) and the semi-perimeter (s=P2s = \frac{P}{2}). Visually, the perimeter is the total distance around the triangle, while ss is the value used within the square root of the area formula.

📐Formulae

Semi-perimeter of a triangle: s=a+b+c2s = \frac{a + b + c}{2}

Heron's Formula for Area of a Triangle: Area=s(s−a)(s−b)(s−c)Area = \sqrt{s(s-a)(s-b)(s-c)}

Area of Quadrilateral ABCDABCD: Area(△ABC)+Area(△ADC)Area(\triangle ABC) + Area(\triangle ADC)

Pythagoras Theorem (to find diagonal dd): d=side12+side22d = \sqrt{side_1^2 + side_2^2} (if ∠=90∘\angle = 90^\circ)

Area of a Right-angled Triangle: Area=12×base×heightArea = \frac{1}{2} \times base \times height

💡Examples

Problem 1:

Find the area of a quadrilateral ABCDABCD in which AB=3 cm,BC=4 cm,CD=4 cm,DA=5 cmAB = 3\text{ cm}, BC = 4\text{ cm}, CD = 4\text{ cm}, DA = 5\text{ cm} and AC=5 cmAC = 5\text{ cm}.

Solution:

Step 1: The quadrilateral ABCDABCD is divided into two triangles, △ABC\triangle ABC and △ADC\triangle ADC by diagonal ACAC. Step 2: For △ABC\triangle ABC, the sides are a=3,b=4,c=5a=3, b=4, c=5. Since 32+42=523^2 + 4^2 = 5^2 (9+16=259+16=25), it is a right-angled triangle. Area(△ABC)=12×3×4=6 cm2Area(\triangle ABC) = \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2. Step 3: For △ADC\triangle ADC, the sides are a=5,b=4,c=5a=5, b=4, c=5. s=5+4+52=142=7 cms = \frac{5+4+5}{2} = \frac{14}{2} = 7\text{ cm}. Area(△ADC)=7(7−5)(7−4)(7−5)=7×2×3×2=84≈9.17 cm2Area(\triangle ADC) = \sqrt{7(7-5)(7-4)(7-5)} = \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} \approx 9.17\text{ cm}^2. Step 4: Total Area =Area(△ABC)+Area(△ADC)=6+9.17=15.17 cm2= Area(\triangle ABC) + Area(\triangle ADC) = 6 + 9.17 = 15.17\text{ cm}^2.

Explanation:

We divide the quadrilateral using the given diagonal ACAC. We use the simple right-angle area formula for one triangle and Heron's formula for the other, then add them together.

Problem 2:

A park is in the shape of a quadrilateral ABCDABCD has ∠C=90∘,AB=9 m,BC=12 m,CD=5 m\angle C = 90^\circ, AB = 9\text{ m}, BC = 12\text{ m}, CD = 5\text{ m} and AD=8 mAD = 8\text{ m}. How much area does it occupy?

Solution:

Step 1: Join BDBD to form two triangles. In right △BCD\triangle BCD, using Pythagoras: BD=BC2+CD2=122+52=144+25=169=13 mBD = \sqrt{BC^2 + CD^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13\text{ m}. Step 2: Area(△BCD)=12×12×5=30 m2Area(\triangle BCD) = \frac{1}{2} \times 12 \times 5 = 30\text{ m}^2. Step 3: For △ABD\triangle ABD, sides are a=9,b=8,c=13a=9, b=8, c=13. s=9+8+132=15 ms = \frac{9+8+13}{2} = 15\text{ m}. Area(△ABD)=15(15−9)(15−8)(15−13)=15×6×7×2=1260=635≈35.5 m2Area(\triangle ABD) = \sqrt{15(15-9)(15-8)(15-13)} = \sqrt{15 \times 6 \times 7 \times 2} = \sqrt{1260} = 6\sqrt{35} \approx 35.5\text{ m}^2. Step 4: Total Area =30+35.5=65.5 m2= 30 + 35.5 = 65.5\text{ m}^2.

Explanation:

Since one angle is 90∘90^\circ, we use Pythagoras to find the diagonal length BDBD. This diagonal allows us to split the quadrilateral into a right triangle and a general triangle, calculating their areas separately.