Question

∆ABC ~ ∆DEF such that AB =9.1 cm and DE= 6.5cm .If the perimeter of∆DEF is 25 cm, what is the perimeter of ∆ABC?
Pls explain with diagram

Answer 1

Answer:

Given:- Triangle ABC congruent Triangle DEF

AB=9.1, DE=6.5,Perimeter of Triangle DEF = 25cm

To prove:- Find perimeter of Triangle ABC

Proof:- Triangle ABC ~ DEF (Given)

:• Angle A = Angle F

Angle B = Angle E

Angle C = Angle D

Perimeter of Triangle DEF = 25 cm

EF+DE+FD=25cm

9.1+6.5+FD=25cm

15.6+FD=25cm

FD=25-15.6cm

FD=9.4cm

:• EF=AB=9.1cm

DE=BC=6.5cm

FD=AC=9.4cm

Hence,Perimeter of Triangle ABC

AB+BC+AC

9.1+6.5+9.4=25cm

Proved•

Answer 2

QUESTION :-

∆ABC ~ ∆DEF such that AB =9.1 cm and DE= 6.5cm .If the perimeter of∆DEF is 25 cm, what is the perimeter of ∆ABC?

SOLUTION:-

➠ ΔABC ~ ΔDEF such that, AB=9.1cm and DE=6.5cm.

➠ ΔABC ~ ΔDEF (given)

➠ΔABC is congruent to ΔDEF ,since all sides & angle are congruent.

➠.°. AB = DE, BC=EF ,AC=DF & A =D ,B=E , C = F

➠in ΔDEF,

➠ DE + EF + DF =25

➠ 9.1+6.5 +DF =25

➠ .°.DF = 25-15.6

➠ .°. DF = 9.4cm.

[NOTE:- AC = DF , .°. AC = 9.4cm]

➠ in ΔABC,

➠ AB+BC+ AC

➠ 9.1+6.5+9.4

➠ 25cm.

➠ .°. the perimeter of ΔABC is 25cm.

➠ VERIFICATION :-

➠AB+BC+AC =25

➠.°. 9.1+6.5+9.4 =25

➠ .°. 25=25