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
3
Answer:
2.1) Altitude through A
2.2) 358 square metres
2.3) Rs 225
2.4) True
Answered by
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
====================
Question 2.3
(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
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