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Standard X

Mathematics

Relation between Area and Sides of Similar Triangles

A man walked ...

Question

A man walked $60$ m to the East from point $A$ , then turned right and walked $50$ m. At last, he turned left and went for $60$ m and reached point $B .$ Then the distance between $A$ and $B$ is

170

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130

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145

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150

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Solution

The correct option is B $130$ m

In right triangle $A E B,$
$A B = \sqrt{A E^{2} + E B^{2}} = \sqrt{50^{2} + 120^{2}}$
$\Rightarrow A B = 130$ m

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A man walked 60 m to the East from point A, then turned right and walked 50 m. At last, he turned left and went for 60 m and reached point B. Then the distance between A and B is

The correct option is B 130 m In right triangle AEB, AB=√AE2+EB2=√502+1202 ⇒AB=130 m

A man walked $60$ m to the East from point $A$ , then turned right and walked $50$ m. At last, he turned left and went for $60$ m and reached point $B .$ Then the distance between $A$ and $B$ is

The correct option is B $130$ m

In right triangle $A E B,$
$A B = \sqrt{A E^{2} + E B^{2}} = \sqrt{50^{2} + 120^{2}}$
$\Rightarrow A B = 130$ m