Math, asked by atharv16218, 20 days ago

Boojho is a farmer. He has a triangular land ABC as shown in figure. AB = 32m, BC = 40m and AC = 24m.
He wants to divide his land in two parts by drawing an altitude and planting roses along this altitude.
The cost of planting the roses is Rs 20 per m.
Question 2.1
Which altitude should he choose to minimize the cost of planting the roses?
(A) Altitude through A
(B) Altitude through B
(C) Altitude through C
Question 2.2
Area of Boojho’s field is
(A) 864 square metres
(B) 384 square metres
(C) 225 square metres
(D) 358 square metres
Question 2.3
What is the minimum cost of planting the roses?
(A) Rs 450
(B) Rs 358
(C) Rs 225
(D) Rs 384
Question 2.4
Areas of two parts will be equal: True/ False

Answers

Answered by nawazish61
3

Answer:

2.1) Altitude through A

2.2) 358 square metres

2.3) Rs 225

2.4) True

Answered by rkcomp31
2

SOLUTION:

In the Δ ABC Given that

AB=32, BC=40 cm and Ac=24 m

AB²+AC²=32²+24²

=1024+576

=1600=40²

Thus AB²+AC²=BC²

Hence Δ ABC  is a right triangle with ∠A=90°

==========

Q2.1

BC =40 m is the longest side thus the altitude to BC

will be smallest

cORRECT OPTION: (A) Altitude through A

=========

Q 2.2

Area of Boojho’s field is

=ar( \triangle ABC) =\frac 12 AB \times AC \\\\=\frac 12 \times 32 \times 24\\\\\bf \therefore \ The \ area \ of \ Boojho’s \ field =384 \ cm^2\\\\\bf \therefore \ Option (B) 384  \ square \ metres \is \ correct\\\\

====================

Question 2.3

area \ of \ \triangle ABC =\frac 12 \times AD \times BC =384\\\\\Rightarrow  \frac 12 \times AD \times 40 =384\\\\\Rightarrow  AD =\frac{ 2 \times  384  }{40} =19.2 \ m\\\\The \ cost \ of\ planting \ the\ roses = \ Rs \ 20 \ per \ m.\\\\\bf \therefore Total \ cost \ of\ planting \ the\ roses:\\\\= 20 \times 19.2 =  \ Rs  \ 384

(B) 384 square metres

============

Question 2.4

In the Δ ABD

BD²=AB²- AD²

= 32²-19.2²

=1024-368.64

BD²=655.36

BD= 25.6 m

Ar(  \triangle ABD) =\frac 12 \times AD  \times BD\\\\=\frac 12 \times 19.2 \times 25.6=245.76 \ m^2\\\\Thus \ Ar(  \triangle ACD) =384-245.76=138.24 \ m^2\\\\\bf \therefore Ar(  \triangle ABD \neq Ar(  \triangle ACD\\\\So\ the \ statement \ is \ False

Attachments:
Similar questions