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ML Aggarwal Solutions Class 9 Mathematics Solutions for Theorems on Area Exercise 14 in Chapter 14 - Theorems on Area
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(a) In the figure (1) given below, AD is median of ∆ABC and P is any point on AD.
Prove that
(i) Area of ∆PBD = area of ∆PDC.
(ii) Area of ∆ABP = area of ∆ACP.
(b) In the figure (2) given below, DE || BC. Prove that
(i) area of ∆ACD = area of ∆ ABE.
(ii) Area of ∆OBD = area of ∆OCE.
(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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