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Match the following columns:
CBSE | Class 10 | Exercise MCQ | Maths | RS Aggarwal | Triangles
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)-…..,

 

Correct 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,

????????2 + ????????2 = ????????2

\begin{array}{l} AC = \sqrt {{{10}^2} + {{20}^2}} \\ AC = \sqrt {100 + 200} \\ AC = 10\sqrt 3 \end{array}

Hence, the man is 10 √3???? ???????????????? ???????????????? ????ℎ???? ???????????????????????????? ????????????????????.

(b) −(q)

 

 

 

Let the triangle be ABC with altitude AD.

In right-angled triangle ABC,

????????2 = ????????2 + ????????2

????????2 = 102 − 52

\begin{array}{l} AD = \sqrt {100 - 25} \\ \end{array}

AD = √75 => 5√3 ????m

(c) – (p)

 

 

 

Area of an equilateral triangle with side,

\begin{array}{l} a = \frac{{\sqrt 3 }}{4}{a^2}\\ a = \frac{{\sqrt 3 }}{4}{10^2}\\ a = \sqrt 3 \times 5 \times 5\\ a = 25\sqrt 3 c{m^2} \end{array}

(d) – (s)

Let the rectangle be ABCD with diagonals AC and BD.

In right-angled triangle ABC,

????????2 = ????????2 + ????????2

=> 82 + 62

=> 64 + 36

???????? = √100

=> 10 m

i

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