In the following figure- 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-i - MBC - ABDii ar BYXD -2 ar MBC iii arBYXD -2 ar ABMN iv - FCB - ACEv ar CYXE -2 ar FCB vi ar CYXE -ar ACFG vii 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-

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Triangle Inequality

In the follow...

Question

In the following figure, 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:

(i) ΔMBC ≅ ΔABD

(ii)

(iii)

(iv) ΔFCB ≅ ΔACE

(v)

(vi)

(vii)

Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.

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Solution

(i) We know that each angle of a square is 90°.

Hence, ∠ABM = ∠DBC = 90º

⇒ ∠ABM + ∠ABC = ∠DBC + ∠ABC

⇒ ∠MBC = ∠ABD

In ΔMBC and ΔABD,

∠MBC = ∠ABD (Proved above)

MB = AB (Sides of square ABMN)

BC = BD (Sides of square BCED)

∴ ΔMBC ≅ ΔABD (SAS congruence rule)

(ii) We have

ΔMBC ≅ ΔABD

⇒ ar (ΔMBC) = ar (ΔABD) ... (1)

It is given that AX ⊥ DE and BD ⊥ DE (Adjacent sides of square

BDEC)

⇒ BD || AX (Two lines perpendicular to same line are parallel to each other)

ΔABD and parallelogram BYXD are on the same base BD and between the same parallels BD and AX.

Area (BYXD) = 2 area (ΔMBC) [Using equation (1)] ... (2)

(iii) ΔMBC and parallelogram ABMN are lying on the same base MB and between same parallels MB and NC.

2 ar (ΔMBC) = ar (ABMN)

ar (BYXD) = ar (ABMN) [Using equation (2)] ... (3)

(iv) We know that each angle of a square is 90°.

∴ ∠FCA = ∠BCE = 90º

⇒ ∠FCA + ∠ACB = ∠BCE + ∠ACB

⇒ ∠FCB = ∠ACE

In ΔFCB and ΔACE,

∠FCB = ∠ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

ΔFCB ≅ ΔACE (SAS congruence rule)

(v) It is given that AX ⊥ DE and CE ⊥ DE (Adjacent sides of square BDEC)

Hence, CE || AX (Two lines perpendicular to the same line are parallel to each other)

Consider ΔACE and parallelogram CYXE

ΔACE and parallelogram CYXE are on the same base CE and between the same parallels CE and AX.

⇒ ar (CYXE) = 2 ar (ΔACE) ... (4)

We had proved that

∴ ΔFCB ≅ ΔACE

ar (ΔFCB) ≅ ar (ΔACE) ... (5)

On comparing equations (4) and (5), we obtain

ar (CYXE) = 2 ar (ΔFCB) ... (6)

(vi) Consider ΔFCB and parallelogram ACFG

ΔFCB and parallelogram ACFG are lying on the same base CF and between the same parallels CF and BG.

⇒ ar (ACFG) = 2 ar (ΔFCB)

⇒ ar (ACFG) = ar (CYXE) [Using equation (6)] ... (7)

(vii) From the figure, it is evident that

ar (BCED) = ar (BYXD) + ar (CYXE)

⇒ ar (BCED) = ar (ABMN) + ar (ACFG) [Using equations (3) and (7)]

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Similar questions

Q. In fig,

A B C

is a right triangle right angled at

A . B C E D, A C F G

and

A B M N

are square on the sides

B C, C A

and

A B

respectively. Line segment

A X ⊥ D E

meets

B C

Y

. Show that:
(i)

△ M B C ≅ △ A B D

(ii)

a r (B Y X D) = 2 a r (M B C)

(iii)

a r (B Y X D) = a r (A B M N)

(iv)

△ F C B ≅ △ A C E

(v)

a r (C Y X E) = 2 a r (F C B)

(vi)

a r (C Y X E) = a r (A C F G)

(vii)

a r (B C E D) = a r (A B M N) + a r (A C F G)

Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler
proof of this theorem in Class X.

Question 8
In the following figure, 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 $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Question 8
In the following figure, 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 $A X ⊥ D E$ meets BC at Y. Show that:

(i) $Δ M B C ≅ Δ A B D$
(ii) $a r e a (B Y X D) = 2 a r e a (Δ M B C)$
(iii) $a r (B Y X D) = a r (A B M N)$
(iv) $Δ F C B ≅ Δ A C E$
(v) $a r (C Y X E) = 2 a r (Δ F C B)$
(vi) $a r (A C F G) = a r (C Y X E)$
(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

In the figure, 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 $A X ⊥ D E$ meets BC at Y Show that
$(i) △ M B C ≅ △ A B D (i i) a r (B Y X D) = 2 a r (△ M B C) (i i i) a r (B Y X D) = a r (A B M N) (i v) △ F C B ≅ △ A C E (v) a r (C Y X E) = 2 a r (△ F C B) (v i) a r (C Y X E) = a r (A C F G) (v i i) a r (B C E D) = a r (A M B N) + a r (A C F G)$

Q. If the given figure, ABC is a right triangle right angled at A, BCED, ACFGand ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y.

Show that
(i) Δ MBC

≅

Δ ABD

(ii) ar (BYXD) = 2 ar (Δ MBC)

(iii) ar (BYXD) = ar (ABMN)

(iv) Δ FCB

≅

Δ ACE

(v) ar (CYXE) = 2 ar (ΔFCB)

(vi) ar (CYXE) = ar (ACFG)

(vii) ar (BCED) = ar (ABMN) + ar (ACFG)

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