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PAVAN JEET SINGH

· started a discussion

· 1 Months ago

no figure (Question:646794.0)

isme figure koun banayega Guru ji

Question:
A surveyor in his field book, has drawn the plot as shown in the figure. The area of the polt is:

Options:
A) \(\cfrac{1}{2}(az+by+ct+dx)\)
B) \(\cfrac{1}{2}(bt+cx+ay+az)\)
C) \(\cfrac{1}{2}(cx+bt+bt+az)\)
D) \(\cfrac{1}{2}(d+t)(c+x)+\cfrac{1}{2}(a+b)(y+z)\)
Solution:
Ans: (b)

Area of \(\triangle\)ABC = \(\cfrac{1}{2}\)az  .......... (i) 

Area of \(\triangle\)BCF = \(\cfrac{1}{2}\) ay .......... (ii) Area of \(\triangle\)CFD = \(\cfrac{1}{2}\) bt .......... (iii) Area of \(\triangle\)DEF = \(\cfrac{1}{2}\) cx .......... (iv) 

form equation (i), (ii) and (iv), we get area of field

= \(\cfrac{1}{2}\)az + \(\cfrac{1}{2}\)ay + \(\cfrac{1}{2}\)bt + \(\cfrac{1}{2}\)cx

= \(\cfrac{1}{2}\)(az + ay + bt + cx)

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