Heron's Formula

Exercise 12.2 Part 2

Question5: 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 getting?

Answer: Given; AB = BC = CD = AD = 30 cm and AC = 48 m
So, AO = 24 m

exercise 12.2 answer 3 figure

In ΔAOB; `BO^2 = AB^2 - AO^2`
`= 30^2 - 24^2`
`= 900 – 576 = 324`
Or, `BO = 18  m`
So, second diagonal of rhombus `= 18 xx 2 = 36  m`

Area of Rhombus `= ½ xx d_1 xx d_2`
`= ½ xx 48 xx 36 = 864  sq  m`
So, area for each cow `= 864/18 = 48  sq  m`


Question 6: 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?

exercise 12.2 question figure 2

Answer: a = 20 cm, b = 50 cm, c= 50 cm
So, `s = (20 + 50 + 50)/2 = 60`


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

`=sqrt(60(60-50)(60-50)(60-20))`

`=sqrt(60xx10xx10xx40)`

`=sqrt(10^4xx2^2xx6)=200sqrt6 sq cm`

Cloth of each colour `= 10 xx 200sqrt6  sq  cm`


Question 7: 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 this figure. How much paper of each shade has been used in it?

exercise 12.2 question figure 3

Answer: Area of square `= ½ xx 32 xx 32 = 512  sq  cm`
(Same as area of rhombus when diagonals are given)
Hence, Area of shape I = shape II `= 512/2 = 256` sq cm

Now, for triangular shape; a = 6 cm, b = 6 cm and c = 8 cm
For this shape; `s = (6 + 6 + 8)/2 = 10`

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

`=sqrt(10(10-6)(10-6)(10-8))`

`=sqrt(10xx4xx4xx2)=8sqrt5 sq cm`

Question 8: A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50p per cm2.

exercise 12.2 question figure 4

Answer:Given; a = 9 cm, b = 28 cm and c = 35 cm
So, `s = (9 + 28 + 35)/2 = 36`


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

`=sqrt(36(36-35)(36-28)(36-9))`

`=sqrt(36xx1xx8xx27)`

`=sqrt(6^2xx2^2xx3^2xx6)=36sqrt6 sq cm`

Cost `= 16 xx 36sqrt6 xx 0.50 = Rs. 288sqrt6`

Question 9: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Answer: ABCD is the given trapezium in which AB = 10 cm, DC – 25 cm, BC = 14 cm and AD = 13 cm

exercise 12.2 answer 4 figure

Draw BE||AD
In ΔBEC; a = 13 cm, b = 14 cm and c = 15 cm
Hence, `s = (13 + 14 + 15)/2 = 21`

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

`=sqrt(21(21-13)(21-14)(21-15))`

`=sqrt(21xx8xx7xx6)`

`=sqrt(7xx2xx2xx2xx2xx7xx3)=84 sq cm`

Now, altitude BM can be calculated as follows:
Area of triangle `= ½ xx \Ba\se\ xx \He\ig\ht`
`= ½ xx EC xx BM`
Or, `84 = ½ xx 15 xx BM`
Or, `BM = (84 xx 2)/15 = 11.2  m`

Now, area of trapezium `= ½ xx \He\ig\ht\ xx` (sum of parallel sides)
`= ½ xx 11.2 xx (10 + 25) = 5.6 xx 35 = 196  sq  m`



Copyright © excellup 2014