Selina solutions

Selina solutions

Grade 7

Pythagoras Theorem | Exercise 16

Question 11

Q11) In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.

Solution:

∆ACD = ∠ABC = 90°

and AD = 13 cm, BC = 12 cm, AB = 3 cm

To find : Length of DC.

(i) In right angled ∆ABC

AB = 3 cm, BC = 12 cm

According to Pythagoras Theorem,

AC^2=AB^2+BC^2

AC^2=3^2+12^2

AC=\sqrt{9+144}=\sqrt{153}cm

(ii) In right angled triangle ACD

AD = 13 cm, AC=\sqrt{153}

According to Pythagoras Theorem,

DC^2=AB^2-AC^2

DC^2=169-153

DC=\sqrt{16}=4\ cm

∴ Length of DC is 4 cm

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Question 11

Q11) In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.

Solution:

∆ACD = ∠ABC = 90°

and AD = 13 cm, BC = 12 cm, AB = 3 cm

To find : Length of DC.

(i) In right angled ∆ABC

AB = 3 cm, BC = 12 cm

According to Pythagoras Theorem,

AC^2=AB^2+BC^2

AC^2=3^2+12^2

AC=\sqrt{9+144}=\sqrt{153}cm

(ii) In right angled triangle ACD

AD = 13 cm, AC=\sqrt{153}

According to Pythagoras Theorem,

DC^2=AB^2-AC^2

DC^2=169-153

DC=\sqrt{16}=4\ cm

∴ Length of DC is 4 cm

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Book a free class now

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