Question

An equilateral triangle APB is constructed on side AB of the square ABCD having a side of 1 unit. what is the radius of the circlep assing through pointsC,P and D?

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Answer 1

Answer:

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Given :

Δ ABP is an equilateral triangle.

Then,

AP = BP (Same side of the triangle)

and AD = BC (Same side of square)

and, ∠ DAP = ∠ DAB - ∠ PAB = 90° - 60° = 30°

Similarly,

∠ BPC = ∠ ABC - ∠ ABP = 90° - 60° = 30°

∴ ∠ DAP = ∠ BPC

∴ Δ APD is congruent to Δ BPC ( SAS proved)

In Δ APD

AP = AD II as AP = AB (equilateral triangle)

We know that ∠ DAP = 30°

∴ ∠ APD = (180° - 30°)/2 (Δ APD is an isosceles triangle and ∠ APD is on of the base angles.)

= 150°/2 = 75°

= ∠ APD = 75°

Similarly, ∠ BPC = 75°

Therefore, ∠ DPC = 360° - (75°+75°+60°)

= ∠ DPC = 150°

Now in Δ PDC

PD = PC as Δ APD is congruent to Δ BPC

∴ Δ PDC is an isosceles triangle

And ∠ PDC = ∠ PCD = (180° - 150°)/2

or ∠ PDC = ∠ PCD = 15°

∠ DPC = 150°; ∠ PDC = 15° and ∠ PCD = 15°

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Answer 2

Answer: Radius of the circle passing through points C, P and D is 1 unit.

Step-by-step explanation:

Given : ABCD is square and side of square is 1 unit. Also ABP is a equilateral triangle constructed on side AB of sqaure.

From Given data-

• AB = AD = DC = BC = 1unit         (ABCD is SQUARE)

• AB = AP = PB = 1 unit.                  (ABP is equilateral triangle)

Now, Construct a line from point P going through centre O as shown in image attach so we will have Radius OP.

Now, we have 2 right angle triangle , Triangle OPB and Triangle OPA.

For Calculating OP, we will use Pythagorean therom.

According to Pythagorean therom,

(side of Right angle)² + (side of Right angle)² = (Hypotenus)²

(OP)² + (OB)² = (PB)²

OP = √(PB)²-(OB)² =√(1)² - (0.5)²  = √1 - 0.25 = √0.75 = 0.86 ≈ 0.9 ≈ 1 unit

Hence,

Radius of Circle is 1 unit.