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
(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.
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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