Heron's Formula
Exercise 12.2 Part 3
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?
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.
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
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`