Question

In the figure. DE BCTAC = 23 cm. FC = 5 cm and DE = 12 cm. then BCS qual to #1 12​

Answer 1

Answer:

Given AD= 1.5 cm

∴BD = 3 cm and AB= AD+BD

= 1.5+3

= 4.5 cm

proof : Given , in triangle ADE and ABC

∠A is common and DE||BC.

∠ADE=∠ABC

⇒∠AED=∠ACB (corresponding angles)

∴ΔADE≅ΔABC (AA similarity)

ar(ΔABC)

ar(ΔADE)

=

AB

2

AD

2

=

(4.5)

2

(1.5)

2

=

9

1

ar(ΔABC)−ar(ΔADE)

ar(ΔADE)

=

9−1

1

∴

ar(trapeziumBCED)

ar(ΔADE)

=

8

1

Answer 2

yes because I don't know the answer bro