NCERT Solutions
Class 9 Maths
Areas of Parallelograms and Triangles
Ex.9.4 Q.8
In Fig. 9.34, ABC is a right triangle right angled at A.
BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively.
Line segment AX ⊥ DE meets BC at Y. Show that:
1. ∆ MBC ≅ ∆ ABD
2. ar (BYXD) = 2 ar (MBC)
3. ar (BYXD) = ar (ABMN)
4. ∆ FCB ≅ ∆ ACE
5. ar (CYXE) = 2 ar (FCB)
6. ar (CYXE) = ar (ACFG)
7. ar (BCED) = ar (ABMN) + ar (ACFG)
Note: Result (vii) is the famous Theorem of Pythagoras.
You shall learn a simpler proof of this theorem in Class X.
1. In ΔMBC and ΔABD, we have
BC = BD
[Sides of the square BCED]
MB = AB
[Sides of the square ABMN]
∠MBC = ∠ABD
[since each 900 + ∠ABC]
Therefore, BY SAS criterion of congruence, we have
∆ MBC ≅ ∆ ABD
2. ΔABD and square BYXD have the same base BD and are
between the same parallels BD and AX.
Therefore, ΔABD = 
× ar (BYXD)
But ∆ MBC ≅ ∆ ABD
[Proved in part (1)]
So, ar (∆ MBC) ≅ ar (∆ ABD)
Therefore,
ar (∆ MBC) ≅ ar (∆ ABD) = × ar (BYXD)
=> ar (BYXD) = 2 × ar (∆ MBC)
3. Square ABMN and ∆ MBC have the same base MB and
are between the same parallels MB and NAC.
ar (∆ MBC) = 12 × ar (ABMN)
=> ar (ABMN) = 2 × ar (∆ MBC) = ar (BYXD)
[Using part (2)]
4. In ΔACE and ΔBCF, we have
CE = BC
[Sides of the square BCED]
AC = CF
[Sides of the square ACFG]
∠ACE = ∠BCF
[since each 900 + ∠BCA]
Therefore, BY SAS criterion of congruence, we have
∆ ACE ≅ ∆ BCF
5. ΔACE and square CYXE have the same base CE
and are between same parallels CF and AYX.
Therefore, ar (∆ ACE) = × ar (CYXE)
=> ar (∆ FCB) = × ar (CYXE)
[Since ∆ ACE ≅ ∆ BCF in part (iv)]
=> ar (CYXE) = 2 × ar (∆ FCB)
6. Square ACFG and ∆ BCF have the same base CF
and are between the same parallels CF and BAG.
ar (∆ BCF) = × ar (ACFG)
=> × ar (CYXE) = × ar (ACFG)
[Using part (v)]
=> ar (CYXE) = ar (ACFG)
7. From part (3) and (4), we have
ar (BYXD) = ar (ABMN)
and ar (CYXE) = ar (ACFG)
On adding, we get
ar (BYXD) + ar (CYXE) = ar (ABMN) + ar (ACFG)
=> ar (BCED) = ar (ABMN) + ar (ACFG)