Following offsets were taken from a survey line to a hedge:

Distance (in m) 0 5 10 15 20 30 40

Offsets (in m) 3 4 5.5 5 6 4 4.5

The area between survey line and the hedge is (by trapezoidal method):

185 m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

187.5 m^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

189.5 m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

289.5 m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $187.5 m^{2}$
For area as given by Trapezoidal method is
$A = d [(\frac{O_{0} + O_{n}}{2}) + O_{1} + O_{2} . . . . . . O_{n - 1}]$
Now, since 'd' is not constant
$∴ A_{1} = 5 [\frac{3 + 6}{2} + 4 + 5.5 + 5] = 95 m^{2}$
$a_{2} = 10 [\frac{6 + 4.5}{2} + 4] = 92.5 m^{2}$
total area,
$A = A_{1} + A_{2}$
$= 95 + 95.5 = 187.5 m^{2}$

Suggest Corrections

Similar questions

A D

A B C D E F A

\begin{matrix} S i z e o f a g r i c u l t u r a l h o l d i n g s (i n h e c) & N u m b e r o f f a m i l i e s 0 - 5 & 10 5 - 10 & 15 10 - 15 & 30 15 - 20 & 80 20 - 25 & 40 25 - 30 & 20 30 - 35 & 5 \end{matrix}

50

k m

Commuted Distance $(k m)$	$50 - 59$	$40 - 49$	$30 - 39$	$20 - 29$	$10 - 19$
No. of Workers	$4$	$5$	$9$	$18$	$14$

Class:	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$	$35 - 40$	$40 - 45$
Frequency:	$5$	$6$	$15$	$10$	$5$	$4$	$2$	$2$

20

Number of plants	$0 - 2$	$2 - 4$	$4 - 6$	$6 - 8$	$8 - 10$	$10 - 12$	$12 - 14$
Number of houses	$1$	$2$	$1$	$5$	$6$	$2$	$3$

Join BYJU'S Learning Program