Another useful property of Trapezium:
“Sum of the squares of the length of the diagonals = Sum of squares of lateral sides + 2 x product of bases.”
Diagonals AC^2 = BD^2 = 8^2 – 2^2 = 60

Using above property: AC^2 + BD^2 = AB^2 + CD^2 + 2 x AC x BD
60 + 60 = 4 + 4 + (2 x 8 x BC)
l(BC) = 7

Very handy for both questions.
Source: HandaKaFunda

for second question ..ans 7 cm

Right you are!

-J

Answer is coming to be 7.5 not 7. Dearest J are u sure it’s 7 ?

Kartik

I’m sorry it’s 7. I wish I could delete that comment.

Very sure ðŸ™‚ 0.5 cut off from each side….

regards

J

hi can you explain how you expand J = (p+q+r)^2+ q^2+r^2 with h in the first question

Use Pythagoras’ Theorem for s and t.

regards

J

Can you let me know the solution , as i am able to solve using area of

triangle and taking radius. is there any fast way to solve this

I did it differently – noted that ABD is a right angle and hence dropped a perpendicular BX to AD. Then used ADB ~ ABX to get AX = 0.5.

regards

J

Another useful property of Trapezium:

“Sum of the squares of the length of the diagonals = Sum of squares of lateral sides + 2 x product of bases.”

Diagonals AC^2 = BD^2 = 8^2 – 2^2 = 60

Using above property: AC^2 + BD^2 = AB^2 + CD^2 + 2 x AC x BD

60 + 60 = 4 + 4 + (2 x 8 x BC)

l(BC) = 7

Very handy for both questions.

Source: HandaKaFunda

7 cm