FINDING THE MISSING DIAGONAL OF A KITE

Find the length of the missing diagonal in each kite.

Problem 1 :

Find TR if QS = 24 ft.

Solution:

Given, Area = 384 ft² and QS = 24 ft

Area of kite = 1/2 × d₁ × d₂

Here d₁ = QS and d₂ = TR

Area = 1/2 × QS × TR

384 ft² = 1/2 × 24 ft × d₂

(384 ft²) = 12 ft × d₂

(384 ft²) / 12 ft = d₂

TR = d₂ = 32 ft

So, the length of other diagonal of a kite TR is 32 ft.

Problem 2 :

Find VX if WU = 28 in.

Solution :

Given, Area = 154 in² and WU = 28 in

Area of kite = 1/2 × d₁ × d₂

Here d₁ = WU and d₂ = VX

Area = 1/2 × WU × VX

154 in² = 1/2 × 28 in × d₂

(154 in²) = 14 in × d₂

(154 in²) / 14 in = d₂

VX = d₂ = 11 in

So, the length of other diagonal of a kite VX is 11 in.

Problem 3 :

Find HF if GE = 43 yd.

Solution :

Given, Area = 258 yd² and GE = 43 yd

Area of kite = 1/2 × d₁ × d₂

Here d₁ = GE and d₂ = HF

Area = 1/2 × GE × HF

258 yd² = 1/2 × 43 yd × d₂

(258 yd² × 2) = 43 yd × d₂

(516 yd²) / (43 yd) = d₂

HF = d₂ = 12 yd

So, the length of other diagonal of a kite HF is 12 yd.

Problem 4 :

Find NL if MK = 18 in.

Solution :

Given, Area = 324 in² and MK = 18 in

Area of kite = 1/2 × d₁ × d₂

Here d₁ = MK and d₂ = NL

Area = 1/2 × MK × NL

324 in² = 1/2 × 18 in × d₂

(324 in²) = 9 in × d₂

(324 in²) / 9 in = d₂

NL = d₂ = 36 in

So, the length of other diagonal of a kite NL is 36 in.

Problem 5 :

Find BD if AC = 47 yd.

Solution :

Given, Area = 493.5 yd² and AC = 47 yd

Area of kite = 1/2 × d₁ × d₂

Here d₁ = AC and d₂ = BD

Area = 1/2 × AC × BD

493.5 yd² = 1/2 × 47 yd × d₂

(493.5 yd² × 2) = 47 yd × d₂

(987 yd²) / (47 yd) = d₂

BD = d₂ = 21 yd

So, the length of other diagonal of a kite BD is 21 yd.

Problem 6 :

Find TV if SU = 30 ft.

Solution :

Given, Area = 153 ft² and SU = 30 ft

Area of kite = 1/2 × d₁ × d₂

Here d₁ = SU and d₂ = TV

Area = 1/2 × SU × TV

153 ft² = 30 ft × d₂ / 2

(153 ft²) = 15 ft × d₂

(153 ft²) / 15 ft = d₂

TV = d₂ = 10.2 ft

So, the length of other diagonal of a kite TV is 10.2 ft.

Problem 7 :

Find EG if DF = 15 ft.

Solution :

Given, Area = 360 ft² and DF = 15 ft

Area of kite = 1/2 × d₁ × d₂

Here d₁ = DF and d₂ = EG

Area = 1/2 × DF × EG

360 ft² = 15 ft × d₂ / 2

(360 ft² × 2) / 15 ft = d₂

(720 ft²) / 15 ft = d₂

EG = d₂ = 48 ft

So, the length of other diagonal of a kite EG is 48 ft.

Problem 8 :

Find XZ if WY = 32 in.

Solution :

Given, Area = 432 in² and WY = 32 in

Area of kite = 1/2 × d₁ × d₂

Here d₁ = WY and d₂ = XZ

Area = 1/2 × WY × XZ

432 in² = 1/2 × 32 in × d₂

(432 in²) = 16 in × d₂

(432 in²) / 16 in = d₂

XZ = d₂ = 27 in

So, the length of other diagonal of a kite XZ is 27 in.

Problem 9 :

Find QS if PR = 11 yd.

Solution :

Given, Area = 253 yd² and PR = 11 yd

Area of kite = 1/2 × d₁ × d₂

Here d₁ = PR and d₂ = QS

Area = 1/2 × PR × QS

253 yd² = 1/2 × 11 yd × d₂

(253 yd² × 2) = 11 yd × d₂

(506 yd²) / (11 yd) = d₂

QS = d₂ = 46 yd

So, the length of other diagonal of a kite QS is 46 yd.