Question

In the given figure, if AC is bisector of angle BAD such that AD = 3 cm, AC = 3 cm and AC = 5 cm, then the value of CD + BC is

With explanation please​

Answer 1

Answer:

(c) 8cm

Step-by-step explanation:

In triangle ACD

AD=base(b)=3cm

CD=perpendicular(p)=?

AC=hypotenuse(h)=5cm

according to Pythagoras theorem

$h^{2} = {p^{2}+b^{2} } \\p^{2} =h^{2} -b^{2} \\p= \sqrt{h^{2}-b^{2} } \\p=\sqrt{5^{2}-3^{2} } \\p=\sqrt{25-9}\\p=\sqrt{16}\\p=4$

∴CD=4cm

from the given figure

AD=AB

again

In triangle ABC

AC=hypotenuse(h)=5cm

AB=base(b)=3cm

BC=perpendicular(p)=?

according Pythagoras theorem

$h^{2} = {p^{2}+b^{2} } \\p^{2} =h^{2} -b^{2} \\p= \sqrt{h^{2}-b^{2} } \\p=\sqrt{5^{2}-3^{2} } \\p=\sqrt{25-9}\\p=\sqrt{16}\\p=4$

∴BC=4cm

now

CD+BC

=4cm+4cm

=8cm   answer.