in trapezium abcd ab is parallel to dc. diagonals intersect at o if ab=cd. prove that area of triangle aob=9 times area of triangle cod
Answers
Step-by-step explanation:
Let the breadth of the rectangle be x cm.
Then, the length of the rectangle is (x+9) cm.
So, area of rectangle = length x breadth =x(x+9)cm
2
Now, length of new rectangle =(x+9+3) cm =(x+12) cm and
breadth of new rectangle =(x+3) cm.
So, area of new rectangle = length × breadth =(x+12)(x+3)cm
2
According to the given condition,
(x+12)(x+3)=x(x+9)+84
⇒x
2
+12x+3x+36=x
2
+9x+84
⇒15x+36=9x+84
⇒15x−9x=84−36
⇒6x=48
⇒x=8
So, breadth of the rectangle is 8 cm and length
=8+9=17 cm.
Answer:
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Let's construct a diagram according to the given question.
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD
In trapezium ABCD,
AB is parallel to CD and AB = 2 CD ---------- (1)
Diagonals AC and BD intersect at ‘O’
In ΔAOB and ΔCOD,
∠AOB = ∠COD (vertically opposite angles)
∠ABO = ∠CDO (alternate interior angles)
⇒ ΔAOB ~ ΔCOD (AA criterion)
⇒ Area of ΔAOB / Areaof ΔCOD = (AB)2 / (CD)2 [Theorem 6.6]
(2CD)2 / (CD)2 = 4CD2 / CD2 = 4 / 1 [From equation (1)]
Thus, Area of ΔAOB : Area of ΔCOD = 4:1
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Step-by-step explanation:
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