# Find the Area of the Kite When the Sides are Tangents

## Geometry Math Problem: Find the area of the kite when the sides of the kite are tangents and radii of the circle

From figure AB = 4 cm and arc AP = PQ = QR = RB. PQ and QR are radii of the circle and PK and RK are tangents of the circle. Then find the area of the kite PQRK

## Solution

We select the semicircle and connect PO, QO and RO

Here

arc AP, PQ, QR and RB equal so

arc AP = arc PQ = arc QR = arc RB = 180/4 = 45°

⇒ ∠POQ = ∠QOR = 45°

Now connect QR then we get the triangle OQR

From triangle OQR

∠OQR = ∠ORQ {QO = RO}

⇒ ∠OQR = ∠ORQ = (180 – 45)/2

⇒ ∠OQR = ∠ORQ = 67.5°

Let QR = x

## Apply cosine rule in the triangle in OQR

QR² = QO² + RO² – 2 × QO × RO × cos 45°

⇒ x² = 2² + 2² – 2 × 2 × 2 × 1/√2

⇒ x² = 4 + 4 – 4√2

Thus, x² = 8 – 4√2

From figure area of the kite = 2 × Area of the triangle KQR {since ΔKQR and ΔKPQ are congruent triangles}

From triangle KQR

Let KR = y, then

tan ∠KQR = KR/QR

⇒ sin 67.5° = y/x

⇒ y = x tan 67.5°

thus, y = x(√2+1)

so, the area of the triangle KQR = ½ × QR × KR

⇒ Area of the triangle KQR = ½ × x × x(√2+1)

⇒ Area of the triangle KQR = ½ × x² × (√2+1)

so, Area of the triangle KQR = ½ × (8 – 4√2) × (√2+1)

⇒ Area of the triangle KQR = (4 – 2√2) × (√2+1)

Thus, the Area of the triangle KQR = 2√2 cm²

Now the area of the kite = 2 × 2√2

⇒ **Area of the kite = 4√2 cm²**