Heron's Formula

Exercise 12.2 Part 1

Question 1: A park, in the shape of a quadrilateral ABCD, has ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer: The following figure shows AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm. ∠C = 90°

exercise 12.2 answer 1 figure

In ΔBDC; BD2 = BC2 + DC2
= 122 + 52
= 144 + 25 = 169
Or, BD = 13 cm

Area of ΔBCD `= ½ xx BC xx CD`

`= ½ xx 12 xx 5 = 30  sq  cm`

For ΔABD; a = 9 cm, b = 13 cm and c = 8 cm
Now, `s = (9 + 13 + 8)/2 = 15`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(15(15-9)(15-13)(15-8))`

`=sqrt(15xx6xx2xx7)=6sqrt35 sq cm`

Total Area = 30 + 36 = 66 sq cm (approx)


Question 2: Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Answer: For ΔABC; `s = (3 + 4 + 5)/2 = 6`

exercise 12.2 answer 2 figure

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(6(6-3)(6-4)(6-5))`

`=sqrt(6xx3xx2xx1)=6 sq cm`

For Δ ADC; `s = (5 + 5 + 4)/2 = 7`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(7(7-5)(7-5)(7-4))`

`=sqrt(7xx2xx2xx3)=2sqrt21=9 sq cm`(approx)

Total Area = 6 + 9 = 15 sq cm (approx)


Question 3: Radha made a picture of an aeroplane with coloured paper as shown in the figure. Find the total area of the paper used.

Answer: Part I (Triangle): a = 5 cm, b = 5 cm and c = 1 cm
So, `s = (5 + 5 + 1)/2 = 5.5`

exercise 12.2 question figure 1

Area`=sqrt(s(s-a)(s-b)(s-c) )`

`=sqrt(5.5(5.5-5)(5.5-5)(5.5-1))`

`=sqrt(5.5xx0.5xx0.5xx4.5)`

`=0.75sqrt11=2.488  sq  cm` (approx)

Part II (Rectangle): l = 6.5 cm and b = 1 cm
Area `= 6.5 xx 1 = 6.5  sq  cm`

Part III (Trapezium): It is composed of three equilateral triangles with sides = 1 cm
Area of equilateral triangle `= (sqrt3)/(4) xx (si\de)^2`

`= (sqrt3)/(4) xx 1^2`

Hence, area of part III `= (3sqrt3)/(4) = 1.3  sq  cm `(approx)

Part IV and IV(Triangles): b = 6 cm and h = 1.5 cm
Area of triangle `= ½ xx h xx b`

`= ½ xx 1.5 xx 6 = 4.5  sq  cm`

Area of Part IV + Part V = 9 sq cm
Total area `= 2.488 + 6.5 + 1.3 + 9 = 19.3  sq  cm`

Question 4: A triangle and a parallelogram have the same base and 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: Given; a = 26 cm, b = 28 cm and c = 30 cm, and based of parallelogram = 28 cm
For triangle: s = (26 + 28 + 30)/2 = 84/2 = 42

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(42(42-26)(42-28)(42-30))`

`=sqrt(42xx16xx14xx12)`

`=sqrt(7xx2xx3xx4xx4xx2xx7xx2xx2xx3)`

`=sqrt(7^2xx3^2xx16^2)`

`=7xx3xx16=336 sq cm`

Area of parallelogram = base X height
Or, 336 = 28 X height
Or, Height = 336/28 = 12 cm



Copyright © excellup 2014