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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, BC || AE and CD || BE. Prove that: area of
∆ABC= area of ∆EBD.
(b) In the figure (2) given below, ABC is a right-angled triangle at A. AGFB is a
square on the side AB and BCDE is a square on the hypotenuse BC. If AN ⊥ ED,
prove that:
(i) ∆BCF ≅ ∆ ABE.
(ii) area of square ABFG = area of rectangle BENM
(a) Given:
From fig (1)
BC || AE and CD || BE
To prove:
area of ∆ABC= area of ∆EBD
Proof:
By joining CE.
We know that, from ∆ABC and ∆EBC
ar (∆ABC) = ar (∆EBC) ….. (1)
From EBC and ∆EBD
ar (∆EBC) = ar (∆EBD) …… (2)
From (1) and (2), we get
ar (∆ABC) = ar (∆EBD)
Hence proved.
(b) Given:
ABC is a right-angled triangle at A. Squares AGFB and BCDE is drawn on the side AB
and hypotenuse BC of ∆ABC. AN ⊥ ED which meets BC at M.
To prove:
(i) ∆BCF ≅ ∆ ABE.
(ii) area of square ABFG = area of rectangle BENM
From the figure (2)
(i) ∠FBC = ∠FBA + ∠ABC
So,
\begin{array}{l} \angle \mathrm{FBC}=90^{\circ}+\angle \mathrm{ABC} \ldots . .(1) \\ \angle \mathrm{ABE}=\angle \mathrm{EAC}+\angle \mathrm{ABC} \\ \mathrm{So}, \\ \angle \mathrm{ABE}=90^{\circ}+\angle \mathrm{ABC} \ldots \ldots(2) \end{array}
From (1) and (2), we get
∠FBC = ∠ABE ….. (3)
So, BC = BE
Now, in ∆BCF and ∆ABE
BF = AB
By using SAS axiom rule of congruency,
∴ ∆BCF ≅ ∆ ABE
Hence proved.
(ii) we know that,
∆BCF ≅ ∆ ABE
So, ar (∆BCF) = ar (∆ABE) ….. (4)
\begin{aligned} \angle \mathrm{BAG}+\angle \mathrm{BAC} &=90^{\circ}+90^{\circ} \\ &=180^{\circ} \end{aligned}
So, GAC is a straight line.
Now, from ∆BCF and square AGFB
ar (∆BCF) = ½ ar (square AGFB) …. (5)
From ∆ABE and rectangle BENM
ar (∆ABE) = ½ ar (rectangle BENM) ….. (6)
From (4), (5), and (6)
½ ar (square AGFB) = ½ ar (rectangle BENM)
ar (square AGFB) = ar (rectangle BENM)
Hence proved.
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