Question

Please solve question no. 4

Answer 1

Answer:

ABCD is a parallelogram.

∴ ​∠A = ​∠C and ∠B =​ ∠D (Opposite angles)

And ​∠A + ​∠B = 180o (Adjacent angles are supplementary)

∴ ​∠B = 180o − ∠A

⇒ 180o − 60o = 120o ( ∵∠A = 60o)

∴ ​∠A = ​∠C = 60o and ∠B =​ ∠D​ = 120o

(i) In ∆ APB, ∠​PAB = 60°2=30°60°2=30°and ∠PBA = 120°2=60°120°2=60°

∴​ ∠​APB​ = 180o − (30o + 60o) = 90o

(ii) In ∆ ADP, ∠​PAD = 30o and ∠ADP = 120o

∴ ∠APB = 180o − (30o + 120o) = 30o

Thus, ∠​PAD = ​∠APB = ​30o

Hence, ∆ADP is an isosceles triangle and AD = DP.

In ∆ PBC, ∠​ PBC = 60o, ∠​ BPC = 180o − (90o +30o) = 60oand ∠​ BCP = 60o (Opposite angle of ∠A)

∴ ∠ PBC = ∠​ BPC = ∠​ BCP

Hence, ∆PBC is an equilateral triangle and, therefore, PB = PC = BC.​

(iii) DC = DP + PC

From (ii), we have:

DC = AD + BC [AD = BC, opposite sides of a parallelogram]

⇒ DC = AD + AD

⇒ DC = 2 AD

Answer 2

Step-by-step explanation:

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