Question

Each side of triangle PQR is 16 units. S is the foot of the
perpendicular dropped from Q on PR, T is the mid-
point of QS. The length of PT is​

Answer 1

$\bold {Length\ of\ PT\ = 4\sqrt(7)\ units.}$

Step-by-step explanation:

Given,Each side of a triangle PQR are equal.

PQ=QR=RS=16 unit

⇒ΔPQR is an equilateral triangle.

S is the foot of the perpendicular drawn from point Q on PR.

QS is the height and the median both of ΔPQR ffrom Q on PR.

⇒PS=SR=$\frac{1}{2}PR$=8 units

In ΔPQS,

∠PSQ = 90°

By Pythagoras Theorem,

$\bold {(PQ)^2=(PS)^2+(QS)^2}$

$(16)^2=(8)^2+(QS)^2$

$256=64+(QS)^2$

$(QS)^2= 256-64$

          = 192

$QS= 8\sqrt(3) units$

since,T is the mid point of QS,

$QT=TS =4\sqrt(3)\ unit$

Now,In ΔPTS,∠PST = 90°

By Applying Pythagoras Theorem,

$\bold {(PT)^2=(PS)^2+(TS)^2}$

$(PT)^2=(8)^2+(4\sqrt(3))^2$

$(PT)^2=64+48$

$(PT)^2=112$

$PT= 4\sqrt(7)$

$\bold {Hence,\Length\ of\ PT\ = 4\sqrt(7)\ units.}$