Question

Rahim and Nadeem are two friends whose plots are adjacent to each other. Rahim’s son made a drawing of the plots with necessary details. It is decided that Rahim will fence the triangular plot ABC and Nadeem will fence along the sides AF, FE and BE.

Observe the diagram carefully and answer the following questions:

(Use `sqrt(2)` = 1.41 and `sqrt(3)` = 1.73)

1. Find the length of BC.   (1)
2. Find the length of AG.   (1)
3. 1. Calculate perimeter of ΔАВС.   (2)
      OR
   2. Calculate the length of (AF + FE + EB).   (2)

Answer 1

i. In right angle ΔBCD

cos B = `(BD)/(BC)`

⇒ cos 45° = `(BH + HD)/(BC)`

⇒ `1/sqrt(2) = (40 + 10)/(BC)`   ...(∵ BH = 40 m and HD = 10 m)

⇒ BC = `50sqrt(2)`

⇒ BC = 50 × 1.41   ...`(∵ sqrt(2) = 1.41)`

⇒ BC = 70.5 m

ii. In right angle ΔAGF

tan A = `(GF)/(AG)`

⇒ tan 45° = `38/(AG)`

⇒ 1 = `38/(AG)`   ...(∵ tan 45° = 1)

⇒ AG = 38 m

iii. a. AB = AG + DG + BD

= 38 + 40 + 50

AB = 128 m

BC = 70.5 m

For length AC,

In right angle ΔACD

sin C = `(AD)/(AC)`

⇒ sin 60° = `(AG + DG)/(AC)`

⇒ `sqrt(3)/2 = (38 + 40)/(AC)`

⇒ `sqrt(3)/2 = 78/(AC)`

⇒ AC = `156/sqrt(3) xx sqrt(3)/sqrt(3)`

⇒ AC = `(156sqrt(3))/3`

⇒ AC = `52sqrt(3)` 

⇒ AC = 89.96 m

∴ Perimeter of ΔАBC = AB + BC + AC

= 128 + 70.5 + 89.96

= 288.46 m

OR

b. Length of (AF + FE + EB)

AF = 53.58 m

FE = 53 m

For length EB,

In right angle ΔBHE

cos B = `(BH)/(EB)`

⇒ cos 30° = `40/(EB)`

⇒ `sqrt(3)/2 = 40/(EB)`

⇒ EB = `(40 xx 2)/sqrt(3)`

⇒ EB = `80/sqrt(3)`

⇒ EB = 46.24 m

Now, AF + FE + EB

= 53.58 + 53 + 46.24

= 152.82 m