Question

Four cows are tied with a rope of length 7 cm at four corners of a quadrilateral field of unequal sides. Find the total area grazed.

Answer 1

Won't it be equal to a full circle of radius 7 cm????

Answer 2

This question seems a little ambiguous, but I'll give it a try. Cut the quad in half which will result in 2 triangles. The hypotenuse of both triangles is going to be 14. It doesn't matter how long each side is since the area will be the same for each. Remember pathagorean's therom? A^2 + B^2 = C^2. C is the hypotenuse and we can assume that A = B. We can transform this equation into squrt(14) = 2A^2. We divide both sides by 2 and get squrt(14)/2= A^2. Now square root that again which becomes 14^(1/4)/2^(1/2) = A. Now all you do to find the are is square A which becomes squrt(14)/2. I'm pretty sure it's right, but I wouldn't count on stranger's advice from the internet.

Answer 3

@dunn0053
thanks for the try..............but it is not correct
Now, i am quite sure about my solution that the area is equal to the area of a full circle of radius 7 cm....☺

Answer 4

Well then the area would be pi * r^2 which is 49pi, but I dunno that the answer would be a circle if they talk about quadrilaterals in the question.

Answer 5

My logic is simple and easy to understand......

the sum of the four interior angles of a quadrilateral is always 360 degree
so all the four angles will total up to 360 but they will be in sectors of different angles but all will have radius 7 cm
so when you put them all together, they will form a full circle because a circle = 360 degree
so the answer is πr^2 = 3.14*49cm^2 = 153.86 cm^2

Answer 6

\[\theta 1 / 360 \times \pi r ^{2} + \theta2 / 360 \times \pi r ^{2} +\]\[\theta3/ 360 \times \pi r ^{2} +\theta4 / 360 \times \pi r ^{2}\]
\[= (\theta1 /360+ \theta2 /360 + \theta3/360 + \theta 4/360) \times \pi r ^{2}\]

\[=( \theta1 +\theta2 +\theta3 +\theta4)/360 \times \pi r ^{2}\]
\[= 360/360 \times \pi r ^{2}\]
\[= \pi r ^{2}\]
= (22/7) *7 *7
= 22 * 7
= 154 unit sq.