(a) In the given figure, D is a point on BC such that angle BAD = angleC and AB = 7 cm,
BD = 4 cm.
(i) Prove that triangle ABD ~triangle CBA
(ii) Find the area of ABC : area of ADC.
Answers
Answer:
(a) In the given figure, D is a point on BC such that angle BAD = angleC and AB = 7 cm,
BD = 4 cm.
(i) Prove that triangle ABD ~triangle CBA
(ii) Find the area of ABC : area of ADC.
Answer:
49/33
Step-by-step explanation:
In Triangle ABD & Triangle ABC,
Angle BAD=Angle BCA(Given)
Angle B is common
Therefore, Triangle BAD=Triangle BCA(AA)
BA/BC=AD/CA=BD/BA
Area of Triangle ABD/Area of Triangle ABC=BD²/AB²
=4²/7²
=16/49---(i)
49(Area of Triangle ABD)=16(Area of Triangle ABC)
49(Area of Triangle ABD)=16(Area of Triangle ABD + Area of Triangle ADC)
49(Area of Triangle ABD)-16(Area of Triangle ABD)=16(Area of Triangle ADC)
33(Area of Triangle ABD)=16(Area of Triangle ADC)
Area of Triangle ABD/Area of Triangle ADC=16/33----(ii)
Dividing (i) by (ii),
Area of Triangle ABC/Area of Triangle ADC=49/33