# Triangle PQR is equilateral. QA is perpendicullar to PR and B is the midpoint of QA.What is the length of PB ? QR=30

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The triangle PQR is equilateral . Then:

PQ = PR = QR = 30

Given.

QA perpend. to PR

B is mid point of QA

Since the triangle if equolateral:

==> QA divides PR into two equal parts.

==> PA = RA = 30/2 = 15

==> QA = sqrt(QP^2 - PA^2 )= sqrt(30^2 - 15^2) = 15sqrt3

Since B is mid point of QA

**==> QB = PB = RB= QA/2 = 15sqrt3/2**

Since the triangle is equilateral, then the lengths of it's sides are equal.

PQ = PR = QR = 30 units

Since QA is perpendicular to PR, then QA is the height of the triangle PQR.

We'll consider the right angled triangle QAR, whose right angle is A = 90 degrees, R = 60 degrees and QA is a cathetus and QR=30 is the hypothenuse.

First, we'll compute the length of QA:

sin R = QA/QR

sin 60 = QA/30

QA = 30*sin 60

QA = 30*sqrt3/2

QA = 15*sqrt3

QA = 25.98

Since B is the midpoint of QA, then QB = BA = 25.98/2

AB = 12.99

Now, we'll calculate PB from the right angled triangle PAB, A=90 degrees, PB is the hypothenuse.

We'll apply Pythagorean Theorem

PB^2 = PA^2 + BA^2

PA = 30/2

PA = 15

PB^2 = 15^2 + 12.99^2

**PB = 19.84**

Since PQR is an equilateral triangle, each side is equal to a, say.

Now triangle PQA is right angled triangle with right angle at A. Now we find QA by Pythagoras theorem:

Then QA = sqrt{PQ^2- QA^2) = sqrt(a^2-(a/2)^2) = (sqrt3)a/2.

Now consider the right angled triangle PBA with right angle at A.We use Pythagoras theorem to find PB:

PB = sqrt(PB^2+PA^2)

PB = sqrt{(1/2QA)^2 +(1/2PR^2)}

PB = (1/2)sqrt{3a^2/4 + a^2}

PB = (1/4)(sqrt7)a.

Therefore the length of PB in terms of the sides of the equolateral triangle is (1/4)(sqrt7)a.