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in Triangles by (32.1k points)
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Match the following columns:

Column I Column II
(a) A man goes 10m due east and then 20m due north. His distance from the starting point is ……m. (p) 25√3
(b) In an equilateral triangle with each side 10cm, the altitude is …..cm. (q) 5√3
(c) The area of an equilateral triangle having each side 10cm is …..cm2 . (r) 10√5
(d) The length of a diagonal of a rectangle having length 8m and breadth 6m is ….m (s) 10

The correct answer is: 

(a) - ……, (b)-……, (c)-……, (d)-…..,

1 Answer

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Best answer

(a) −(r) 

Let the man starts from A and goes 10 m due east at B and then 20 m due north at C. 

Then, in right-angled triangle ABC, we have: 

AB2 + BC2 = AC2

⟹ = \(\sqrt{10^2+20^2}\) = \(\sqrt{100+200}\) = 10√3 

Hence, the man is 10 √3m away from the starting points

(b) −(q) 

Let the triangle be ABC with altitude AD. 

In right-angled triangle ABC, we have:

(c) – (p) 

Area of an equilateral triangle with side a

(d) – (s) 

Let the rectangle be ABCD with diagonals AC and BD. 

In right-angled triangle ABC, we have: 

AC2 = AB2 + BC2 = 82 + 62 = 64 + 36 

⇒ AC = \(\sqrt{100} \) = 10 m

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