#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# a and b are real numbers between 0 and 1, A(a,1) ,B(1,b) and C(0,0) are the vertices of a equilateral triangle ABC, then its area is (a) (7√3+12)/4 (b) (7√3-12)/4 (c) √3/4 (d) None of these....Explain pls. 9 years ago

The answer to is question is (d)(none of these)

Solution:

Reason for contradiction of option (a), the maximum area that a triangle can enclose where 0<a,b<1 is <1 (definitely its area  will be smaller than a square of side 1 and coordinates (0,0),(0,1),(1,0),(1,1) ).

However (7root(3)+12)/4 is >1 so it is rejected.

Reason for contradiction of option (b), If such a equilateral triangle ever exists with such coordinates  its sides should definitely be >1 because the distance between (0,0) and (a,1) where 0<a<1 is always >1.(Use distance formula)

In such a case the area would be > √3(1)2/4 which means area >  √3/4 but (7root(3)-12)/4 is less than √3/4

Reason for contradiction of option (c),

For the area to be √3/4 the sides must be 1, for this a,b can be = 0 but still it will form a right angled triangle with third side root(2) which will not be a equilateral triangle.

So option (c) is incorrect

So the only option left is (d)