We are give the radius =10
Let the base of the semicircle is AOB.

Let points X and Y be on edges of the traffic lane (XY=12)
OX=6

Now we have a right triangle with one side as 6 and the hypotenuse as 10, we can easily calculate the height using the Pythagorus theorem
= sqrt (6^2+10^2) =8

we need 1/2 ft clearance.

so height must be (8-1/2) =71/2 ft


B.

The figure above shows the dimensions of a semicircular cross section of a one-way tunnel. The single traffic lane is 12 feet wide and is equidistant from the sides of the tunnel. If vehicles must clear the top of the tunnel by at least 1/2 foot when they are inside the traffic lane, what should be the limit on the height of vehicles that are allowed to use the tunnel?

A. 5½ ft
B. 7½ ft
C. 8 ½ ft
D. 9½ ft
E. 10 ft

To GUARANTEE that ANY vehicle inside the 12 foot lane will clear the tunnel by 1/2 foot, we must consider the WORST-CASE SCENARIO:
The WIDEST possible vehicle -- a truck that is 12-feet wide -- traveling down the 12 foot lane.



In the figure above, O is the center of the circle and rectangle ABCD represents a 12-foot wide truck.
CD is the height of a truck that would TOUCH the top of the tunnel.
Since AD = 12, OD = 6.
Since the diameter of the semi-circle is 20, radius OC = 10
Thus, ∆OCD is a 6-8-10 triangle, implying that CD=8.
Since a truck with a height of 8 feet would touch the top of the tunnel, the maximum height that guarantees CLEARING the tunnel by 1/2 foot = 8 - 1/2 = 7.5 feet.

The correct answer is B.

Guys ,

From where do i get the gprep software or questions to practice..

Hi Rich ,

That i have already downloaded the software besides that?

Shreyans