Question

AB = 9 cm. BC = 40 cm and
in the given figure, angle ACB = 90° = angle ACD.
If AB = 10 cm. BC = 6 cm and AD = 17 cm, find:
(1) АС
(ii) CD
please discribe​

Answer 1

Answer:

∆ABD

∠ACB = ∠ACD = 90°

and AB = 10 cm, BC = 6 cm and AD = 17 cm

To find:

(i) Length of AC

(ii) Length of CD

Proof:

(i) In right-angled triangle ABC

BC = 6 cm, AB = 110 cm

According to Pythagoras Theorem,

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

2

=AC

2

+BC

2

10^2=AC^2+6^210

2

=AC

2

+6

2

100=AC^2+36100=AC

2

+36

AC^2=100-36=64\ cmAC

2

=100−36=64 cm

AC^2=64\ cmAC

2

=64 cm

\therefore AC=\ \sqrt{8\times8}=8\ cm∴AC=

8×8

=8 cm

(ii) In right-angle triangle ACD

AD = 17 cm, AC = 8 cm

According to Pythagoras Theorem,

AD^2=AC^2+CD^2AD

2

=AC

2

+CD

2

17^2=8^2+CD^217

2

=8

2

+CD

2

289-64=CD^2289−64=CD

2

225=CD^2225=CD

2

CD=\sqrt{15\times15}=15\ cmCD=

15×15

=15 cm

Answer 2

Step-by-step explanation:

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