PQ // BC
then
Angle P = Angle B, A is common angle the two triangles are similar
AP/AB = AQ/AC
8/18 = 12/ AC
AC = 12 * 18/8 = 27 (your ans: D)
AQ + QC = AC
QC = AC - AQ = 27 - 12 = 15 (if it was required)
tan60 = y/8
y = s tan60 = 8sqrt(3) (B)
Next time, post each question separately.
Answer:
<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>
Here angle B is 90°
So
and
Are right angled triangle
So we use Pythagoras thereon for solution
<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
perpendicular=p=8cm
Hypontenuse =h =10cm
According to Pythagoras thereon

















- Now in

Perpendicular=p=8cm
Base =b=15cm
- We need to find Hypontenuse =AD(x)
According to Pythagoras thereon













A is the 4
B is the 5
and C is also 4.
A is always the number in front of x^2. B is always the number i front of the normal x, and C is the number that is on the other side of the equal sign!
3 miles = 5,280 yards
6(250) = 1,500 yards
5,280 - 1,500 = 3,780 yards
(8/9-3/9)*2/3=5/9*2/3=10/27