Question
Download Solution PDFAn 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?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
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:
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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