APPLICATION OF HERON’S FORMULA IN FINDING AREAS OF QUADRILATERALS

Heron’s formula can be applied to find the area of a quadrilateral by dividing the quadrilateral into two triangular parts.

Q1. 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? 

Q2.  A rhombus has perimeter 100 m and one of its diagonal is 40 m. Find the area of the rhombus.

Solution: ABCD is the rhombus having perimeter = 100 m and AC = 40 m.

Now, we have m

 .

Solution: Let s be the semi-perimeter of a tile. Then, 

Q4. Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA (figure). Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠B = 90°, which group cleaned more area and by how much? Find the total area cleaned by the students.

Solution: In ΔABC = 9 m and BC = 40 m, ∠B = 90°, we have

Therefore, the first group has to clean the area of triangle ABC, 

Area of ΔABC

m2 = 180 m²

The second group has to clean the area of triangle ACD, 

Here, m = 42 m

Therefore, area of Δ

So first group cleaned 180 m2, which is (180 − 126) m2, i.e., 54 m2 more than the area cleaned by the second group.

Total area cleaned by all the students = (180 + 126) m2 = 306 m2.

Q5.  A field is in the shape of a trapezium whose parallel sides are 50 m and 15 m. The non-parallel sides are 20 m and 25 m. Prove that area of the trapezium is m2.

Solution: ABCD is a trapezium in which m, BC=25 m and  AD= 20 m.

BCandADare non-parallel sides of trapezium. Through the vertex C, we draw and CE meeting AB at E. ABED becomes a parallelogram. Then AE = 15 m,  BE=35  m and CE=20 m. 

AE = DC = 15 m [opposite sides of ||gm]

and AD = CE = 20 m [opposite sides of ||gm]

Area of the trapezium

Q6. A farmer has a triangular field with sides 240 m, 200 m, 360 m, where she grew wheat. In another triangular field with sides 240 m, 320 m, 400 m adjacent to the previous field, she wanted to grow potatoes and onions (figure). She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew potatoes by joining the mid-point of the longest side to the opposite vertex and grew potatoes in one part and onions in the other part. How much area (in hectares) has been used for wheat, potatoes and onions? (1 hectare = 10,000 m2).

Solution: Let ABC be the field where wheat is grown. Also let ACD be the field which has been divided in two parts by joining C to the mid-point E of AD. In triangle ABC, we have

a=200 m,  b = 240 m, c=360 m

Therefore,

Now Ar(ΔAEC) = Ar(ΔDEC) [ Median divides Δ into two parts of equal area]

In ΔACD.

Q7.  A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?

Solution: Since diagonals of a square bisect each other at right angle. Therefore,

cm

Now, Area of part I = 2 × area of ΔAOB