ABC is a triangle, D is the mid-point on side BC.

Given, AB=2 cm, AC=4 cm and BD=3 cm
Find the value of AD.

1 cm

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2 cm

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3 cm

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4 cm

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Solution

The correct option is A
1 cm

From Apollonius's theorem, we can say

$A B^{2} + A C^{2} = 2 (B D^{2} + A D^{2})$
Substituting the values we get,
$2^{2} + 4^{2} = 2 (3^{2} + A D^{2}) 4 + 16 = 2 (9 + A D^{2}) A D^{2} + 9 = 10 \Rightarrow A D^{2} = 1 \Rightarrow A D = 1 c m$

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ABC is a triangle, D is the mid-point on side BC. Given, AB=2 cm, AC=4 cm and BD=3 cm Find the value of AD.

The correct option is A 1 cm From Apollonius's theorem, we can say AB2+AC2=2(BD2+AD2) Substituting the values we get, 22+42=2(32+AD2)4+16=2(9+AD2)AD2+9=10⇒AD2=1⇒AD=1 cm

ABC is a triangle, D is the mid-point on side BC.

Given, AB=2 cm, AC=4 cm and BD=3 cm
Find the value of AD.

The correct option is A
1 cm

From Apollonius's theorem, we can say

$A B^{2} + A C^{2} = 2 (B D^{2} + A D^{2})$
Substituting the values we get,
$2^{2} + 4^{2} = 2 (3^{2} + A D^{2}) 4 + 16 = 2 (9 + A D^{2}) A D^{2} + 9 = 10 \Rightarrow A D^{2} = 1 \Rightarrow A D = 1 c m$