Question

In ΔABC, In ΔABC, <ABC=90° and BO is parallel to AC. If AB=5.7cm, BO=3.8cm, CD=5.4cm. What is the length of BC?



​

Answer 1

Answer:

Given:

∠BDC=90

∘

∠ABC=90

∘

Let ∠BCD=x

∘

Using triangle sum property in △BDC, ∠DBC=90−x

∘

Also ∠ABD=x

∘

Using triangle sum property in △ADB, ∠BAD=90−x

∘

Now considering △BDC and △ADB

∠BDC=∠BDA

∠BCD=∠ABD

∠DBC=∠DAB

So by AAA

△BDC∼△ADB

Hence  

CD

BC

​

=

BD

AB

​

5.4

BC

​

=

3.8

5.7

​

BC=8.1

Hence BC=8.1cm

Solve any question of Triangles with:-

Patterns of problems

Patterns of problems

5 min read

hope this answer helps you!!!

Answer 2

Answer:-

mark as brainlist and like

Step-by-step explanation:

Given:

Given:∠BDC=90°

∠ABC=90°

Let ∠BCD=x°

Using triangle sum property in △BDC, ∠DBC=90−x

Using triangle sum property in △BDC, ∠DBC=90−x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DAB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CD

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC =

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC = BD

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC = BDAB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC = BDAB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC = BDAB

Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB∠BDC=∠BDA∠BCD=∠ABD∠DBC=∠DABSo by AAA △BDC∼△ADBHence CDBC = BDAB 5.4

BC

=

= 3.8

= 3.85.7

= 3.85.7

= 3.85.7

= 3.85.7 BC=8.1

= 3.85.7 BC=8.1Hence BC=8.1cm