Question

please answer it........

Answer 1

in ∆ABD

using Pythagoras theorem

AB=√80=4√5 cm,

Now in ∆ABD & ∆ ABC,

angle B = angle D (each 90°),

angle A = angle A (common),

therefore by AA similarity criterion

∆ ABD similar to ∆ ABC,

hence

AC/AB = AB/AD,

(4+CD)/4√5 = 4√5/4,

(4+CD)=√5×4√5,

(4+CD)=20,

then

CD=20-4=16 cm

Answer 2

Hey mate,

Please refer to the figure above ⤴,

Given,

Angle(ABC) = 90°

BD = 8 cm

AD = 4 cm

Assumptions :

Let Angle(ACB) = x

Angle (CBD) = 90 - x

So,

Angle (ABD) = x

Angle (BAD) = 90 - x

By property of similar triangles,

We can say Triangle(ABC), triangle (ADB) and (BDC) are similar,

So,

BD/CD = AD/DB

8/CD = 4/8

CD = 16 cm

Hope this helps you out!