Question

Triangle PQR is an equilateral triangle. Point M is on the base QR such that MR =1/3 QR. If PQ = 12cm. Find PM.​

Answer 1

Given :-

• An equilateral ∆ PQR with PQ = QR = RM = 12 cm.

• M is any point on QR such that MR = 1/3 QR.

To Find :-

• Length of PM.

Construction :-

• Through P, Construct PN perpendicular to QR meeting QR at N.

$\large\underline{\bold{Solution-}}$

In ∆ PNQ and ∆ PNR

• PQ = PR [Sides of equilateral triangle]

• PN = PN [ Common ]

• ∠ PNQ = ∠ PNR [ Each 90° ]

• ⇛ ∆ PNQ ≅ ∆ PNR [ RHS Congruency Rule ]

• ⇛ QN = NR = 6 cm [ as QR = 12 cm]

Now,

• MR = 1/3 QR

• ⇛ MR = 1/3 × 12 = 4 cm.

Now,

• MN = RN - MR = 6 - 4 = 2 cm.

Now,

In ∆ PNR

Using Pythagoras Theorem,

• ⇛ PR² = RN² + PN²

• ⇛ PR² = (RM + MN)² + PN²

• ⇛ PR² = RM² + MN² + 2 × RM × MN + PN²

• ⇛ PR² = (MN² + PN²) + RM² + 2 × RM × MN

• ⇛ PR² = PM² + RM² + 2 × RM × MN

On Substituting the values of PR, RM and MN, we get

$\sf \: {(12)}^{2} = {(PM)}^{2} + {(4)}^{2} + 2 \times 4 \times 2$

$\sf \: 144 = {PM}^{2} + 16 + 16$

$\sf \: 144 = {PM}^{2} + 32$

$\sf \: {PM}^{2} = 144 - 32$

$\sf \: {PM}^{2} = 112$

$\sf \: PM \: = \: \sqrt{112}$

$\sf \: PM \: = \: \sqrt{2 \times 2 \times 2 \times 2 \times 7}$

$\bf\implies \:PM \: = \: 4 \sqrt{7} \: cm$