Question

In the given figure, ABCD is a square and APAB is an equilateral triangle. (i) Prove that AAPD ≈ ABPC. (ii) Show that ZDPC = 15°​

Answer 1

Step-by-step explanation:

Information provided with us:

➡ The side of a square is 4 m.

The same square is divided into four equal squares.

What we have to find out :

➡ The required area of the each square

Formula used :

➡ Area of square = (side)²

Given that :

➡ The Side of a square = 4 m.

Now ,

➡ By using formula of area of square and putting the given values in it we get ,

⬛Area of Square

\rm \implies \: (4 {)}^{2} =4 \times 4 = 16 \: {m}^{2}⟹(4)

2

=4×4=16m

2

We know that :

➡ The same square is divided into four equal squares.

Henceforth :

⬛ The area of each square

\rm \implies \: \dfrac{16}{4}⟹

4

16

\bf\implies \: 4 \: {m}^{2}⟹4m

2

➡ Therefore 4 m² is the required area of the each square

Answer 2

Step-by-step explanation:

To prove:- ∆APD ≈ ∆BPC

prove-- In ∆ APD and in ∆BPC

AD = BC [ Side of square ]

Ap = PB [ side of equilateral ∆ ]

<DAB + <PAB = 90 + 60 = 150° [Each angle of square is 90 and each angle of equilateral ∆ is 60 ].

<DAP = <PBC [ By ABOVE LINE]

SO ∆APD AND ∆BPC IS CONGRUENT

BY SAS CRITERIA