An isosceles triangle has its base length 2a and its height is h. On each side of the triangle a square is drawn external to the triangle. What is the area of the figure thus formed?

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CDS Elementary Mathematics 3 Sep 2023 Official Paper
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  1. 6a+ 2h+ 2ah
  2. 6a2 + 2h2 + ah
  3. 4a+ 2h+ ah
  4. 6a2 + h2 + ah

Answer (Detailed Solution Below)

Option 2 : 6a2 + 2h2 + ah
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Detailed Solution

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Given:

An isosceles triangle has its base length 2a & its height is h.

Formula used:

Area of Triangle = 1/2 × base × height

Area of Square = side2

Calculation:

F1 Vinanti Defence 01.12.23 D8

According to the figure,

Area of isosceles triangle = 1/2 × b × h 

⇒ 1/2 × 2a × h  ( base = 2a)     .....(1)

Again, according to the figure,

By Pythagoras theorem

⇒ AC = √DC2 + AD2 

⇒ AC = √a2 + h2

Then, Area of square drawn on AB & AC 

Side2 = (√a2 + h2)2 = (a2 + h2)

Now, both side will be equal then, 

Area = 2(a2 + h2)     .....(2)

Again, according to the figure,

Area of square drawn on BC which is (2a)2

Area of square = (2a)2 = (4a2)    .....(3)

By adding above equation (1), (2) & (3) we get,

Area of the figure,

(1/2 × 2a × h) + 2(a2 + h2) + (4a2)

⇒ ah + 2a2 + 2h2 + 4a2

⇒(2a2 +4a2) + 2h2 + ah

⇒ 6a2 + 2h2 + ah

∴ The area of the figure is  6a2 + 2h2 + ah

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