Question

∆ABC, ∠ABC=90° and CD is angle bisector of ∠ACB which meets side AB at D. if AD=4cm, BD=3cm find the CD?

Answer 1

$CD=6\sqrt 2cm$

Step-by-step explanation:

In triangle ABC

Angle ABC=90 degree

CD is a angle bisector of angle ACB.

AD=4 cm BD=3 cm

By Angle bisector sector theorem

$\frac{AC}{BC}=\frac{AD}{BD}$

$\frac{AC}{BC}=\frac{4}{3}$

$AC=\frac{4}{3}BC$

AB=AD+DB=4+3=7 cm

In triangle ABC

$AC^2=AB^2+BC^2$

By using Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2$

Substitute the values

$(\frac{4}{3}BC)^2=(7)^2+BC^2$

$\frac{16}{9}BC^2-BC^2=49$

$\frac{16BC^2-9BC^2}{9BC^2}=49$

$\frac{7}{9}BC^2=49$

$BC^2=\frac{49\times 9}{7}=63$

$BC=\sqrt{63}=3\sqrt 7$cm

In triangle DBC

$DC^2=BC^2+BD^2$

$CD^2=63+(3)^2=63+9=72$

$CD=\sqrt{72}=6\sqrt 2cm$

$CD=6\sqrt 2cm$

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https://brainly.in/question/701435:Answered By Golda

Answer 2

Answer:

Step-by-step explanation:

Step-by-step explanation:

In triangle ABC

Angle ABC=90 degree

CD is a angle bisector of angle ACB.

AD=4 cm BD=3 cm

By Angle bisector sector theorem

AB=AD+DB=4+3=7 cm

In triangle ABC

By using Pythagoras theorem

Substitute the values

cm

In triangle DBC