Question

a field in the shape of a rhombus have the distance between pairs of opposite vertices are as 14 M and 48 M what is the cost of fencing the field at rupees 20 per metre

Answer 1

Answer:

Cost of fencing the field = 20,00 Rs

Solution:

Distance between pairs of opposite vertices are as 14 M and 48 M

I.e. diagonal of rhombus
$d_{1} = 14 \: m \\ \\ d_{2} = 48 \: m$

Side of Rhombus : since diagonal of Rhombus bisect each other at 90°,so it forms a right triangle ,thus apply Pythagoras theorem

$a = \sqrt{ { (\frac{14}{2}) }^{2} + ( { \frac{48}{2}) }^{2} } \\ \\ = \sqrt{( {7)}^{2} + ( {24)}^{2} } \\ \\ = \sqrt{49 + 576} \\ \\ a = \sqrt{625} \\ \\ a = 25 \: m$

Side of Rhombus is 25 M

Perimeter of Rhombus = 25×4

= 100 M

cost of fencing the field at rupees 20 per metre = 20× 100

= 2,000 Rs

Answer 2

Answer:

this is your answer

Step-by-step explanation:

hope it helps you

thanku