Math, asked by Manasmk, 1 year ago

in the figure given below AD is a median of triangle ABC and P is a point in AC such that area of triangle DAP:area of triangle ABD.Find
(1)AP:PC
(2)area of triangle PDC : area of triangle ABC

Answers

Answered by Anonymous
1

(a) Given:

∆ABC in which AD is the median. P is any point on AD. Join PB and PC.

To prove:

(i) Area of ∆PBD = area of ∆PDC.

(ii) Area of ∆ABP = area of ∆ACP.

Proof:

From fig (1)

AD is a median of ∆ABC

So, ar (∆ABD) = ar (∆ADC) …. (1)

Also, PD is the median of ∆BPD

Similarly, ar (∆PBD) = ar (∆PDC) …. (2)

Now, let us subtract (2) from (1), we get

ar (∆ABD) – ar (∆PBD) = ar (∆ADC) – ar (∆PDC)

Or ar (∆ABP) = ar (∆ACP)

Hence proved.

(b) Given:

∆ABC in which DE || BC

To prove:

(i) area of ∆ACD = area of ∆ ABE.

(ii) Area of ∆OBD = area of ∆OCE.

Proof:

From fig (2)

∆DEC and ∆BDE are on the same base DE and between the same || line DE and BE.

ar (∆DEC) = ar (∆BDE)

Now add ar (ADE) on both sides, we get

ar (∆DEC) + ar (∆ADE) = ar (∆BDE) + ar (∆ADE)

ar (∆ACD) = ar (∆ABE)

Hence proved.

Similarly, ar (∆DEC) = ar (∆BDE)

Subtract ar (∆DOE) from both sides, we get

ar (∆DEC) – ar (∆DOE) = ar (∆BDE) – ar (∆DOE)

ar (∆OBD) = ar (∆OCE)

Hence proved.

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Answered by protyayroychoudhury
0

Answer:

Given: From fig ( 3 )

AD is a median of △ABC and P is a point in AC such that area of ΔADP : area of ΔABD

=2:3

To Find:

(i) AP: PC

(ii) area of ΔPDC : area of ΔABC

Now,

(i) we know that AD is the median of ΔABC

ar(ΔABD)=ar(ΔADC)=1/2

ar (ΔABC)…….(1)

It is given that, ar(ΔADP):ar(ΔABD)=2:3

AP:AC=2:3

AP/AC=2/3

AP=2/3AC

Now, PC=AC−AP

=AC−2/3AC

=(3AC−2AC)/3

=AC/3……..(2)

So AP/PC=(2/3AC)/(AC/3)

=2/1

AP:PC=2:1

(ii) we know that from ( 2) PC=AC/3

PC/AC=1/3

So,

ar(ΔPDC)/ar(ΔADC)

=PC/AC

=1/3

ar (ΔPDC)/1/2 ar (ΔABC)=1/3

ar (ΔPDC)/ar(ΔABC)=1/3×1/2

=1/6

ar(ΔPDC): ar (ΔABC)=1:6

Hence proved.

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