Question

Construct a quadrilateral ABCD where AB =4.5cm ,BC = 5.5 cm ,CD = 4 cm ,

AD = 6cm , AC = 7 CM stap dy stap​

Answer 1

Answer:

RA = \frac{1}{3} ACRA=

3

1

AC

Construction :-

Draw PS parallel to BR to meet AC at S.

Proof :-

In Δ BCR, P is the mid-point of BC and PS is parallel to BR.

Where, S is the mid-point of CR

So, CS = SRCS=SR ----- (1)

Again, In Δ APS, Q is the mid-point of AP and QR is parallel to PS.

Where, R is the mid-point of AS.

So, AR = RSAR=RS ----- (2)

From equations (1) and (2),

We get, AR = RS = SC

⇒ AC = AR + RS + SCAC=AR+RS+SC

⇒ AC = AR + AR + ARAC=AR+AR+AR

⇒ AC = 3ARAC=3AR

∴ AR= \frac{1}{3} ACAR=

3

1

AC

--Please refer to the attached image for clarification of diagram and points--

--Please take into consideration another method attached in the form of an image--

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Regards