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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, the point D divides the side BC of ∆ABC in the
ratio m: n. Prove that area of ∆ ABD: area of ∆ ADC = m: n.
(b) In the figure (2) given below, P is a point on the side BC of ∆ABC such that PC =
2BP, and Q is a point on AP such that QA = 5 PQ, find area of ∆AQC: area of
∆ABC.
(c) In figure (3) given below, AD is a median of ∆ABC and P is a point in AC
such that area of ∆ADP: area of ∆ABD = 2:3. Find
(i) AP: PC
(ii) area of ∆PDC: area of ∆ABC.
(a) Given:
From fig (1)
In ∆ABC, the point D divides the side BC in the ratio m: n.
BD: DC = m: n
To prove:
area of ∆ ABD: area of ∆ ADC = m: n
Proof:
area of ∆ ABD = ½ × base × height
ar (∆ ABD) = ½ × BD × AE ….. (1)
ar (∆ ACD) = ½ × DC × AE ….. (2)
let us divide (1) by (2)
[ar (∆ ABD) = ½ × BD × AE] / [ar (∆ ACD) = ½ × DC × AE]
[ar (∆ ABD)] / [ar (∆ ACD)] = BD/DC
= m/n [it is given that, BD: DC = m: n]
Hence proved.
(b) Given:
From fig (2)
In ∆ABC, P is a point on the side BC such that PC = 2BP, and Q is a point on AP such
that QA = 5 PQ.
To Find:
area of ∆AQC: area of ∆ABC
Now,
It is given that: PC = 2BP
PC/2 = BP
We know that, BC = BP + PC
Now substitute the values, we get
BC = BP + PC
= PC/2 + PC
= (PC + 2PC)/2
= 3PC/2
2BC/3 = PC
ar (∆APC) = 2/3 ar (∆ABC) …… (1)
It is given that, QA = 5PQ
QA/5 = PQ
We know that, QA= QA + PQ
So, QA = 5/6 AP
ar (∆AQC) = 5/6 ar (∆APC)
= 5/6 (2/3 ar (∆ABC)) [From (1)]
ar (∆AQC) = 5/9 ar (∆ABC)
ar (∆AQC)/ ar (∆AQC) = 5/9
Hence proved.
(c) Given:
From fig (3)
AD is a median of ∆ABC and P is a point in AC such that area of ∆ADP: area of ∆ABD
= 2:3
To Find:
(i) AP: PC
(ii) area of ∆PDC: area of ∆ABC
Now,
(i) we know that AD is the median of ∆ABC
ar (∆ABD) = ar (∆ADC) = ½ ar (∆ABC) ……. (1)
It is given that,
ar (∆ADP): ar (∆ABD) = 2: 3
AP: AC = 2: 3
AP/AC = 2/3
AP = 2/3 AC
Now,
PC = AC – AP
= AC – 2/3 AC
= (3AC-2AC)/3
= AC/3 …….. (2)
So,
AP/PC = (2/3 AC) / (AC/3)
= 2/1
AP: PC = 2:1
(ii) we know that from (2)
PC = AC/3
PC/AC = 1/3
So,
ar (∆PDC)/ar (∆ADC) = PC/AC
= 1/3
ar (∆PDC)/1/2 ar (∆ABC) = 1/3
ar (∆PDC)/ar (∆ABC) = 1/3 × ½
= 1/6
ar (∆PDC): ar (∆ABC) = 1: 6
Hence proved.
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