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28 Dec 2011, 19:03
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Difficulty:
55%
(hard)
Question Stats:
60% (02:46) correct
40%
(02:27)
wrong
based on 48
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Not Attempted Yet
ABCD is a trapezium, in which AD and BC are parallel. If the four sides Ab,BC,CD,DA are 9 ,12, 15, 20. Find the magnitude of the sum of the squares of the two diagonals?.
a) 638
b) 786
c) 838
d) 648
e) 726
a) 638
b) 786
c) 838
d) 648
e) 726
28 Dec 2011, 22:46
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h -> height of the trapezium or the distance between the parallel sides
x -> distance between C` and D
y -> distance between B` and A
C` and B` are projections of B and C on AD .
BD^2 = h^2 + (12+x)^2
AC^2 = h^2 + (12+y)^2
also 9^2 = h^2 + y^2 and 15^2 = h^2 + x^2
Using these four equations I get :
BD^2 + AC^2 = 594 + 24*x*y
So the (ans-594) should be a multiple of 24.
Plugged in each answer and found out only B satisfies this.
There should be a shorter way to solve this, using properties of trapeziums . Most likely SL Loney should have a ready made answer to this.
x -> distance between C` and D
y -> distance between B` and A
C` and B` are projections of B and C on AD .
BD^2 = h^2 + (12+x)^2
AC^2 = h^2 + (12+y)^2
also 9^2 = h^2 + y^2 and 15^2 = h^2 + x^2
Using these four equations I get :
BD^2 + AC^2 = 594 + 24*x*y
So the (ans-594) should be a multiple of 24.
Plugged in each answer and found out only B satisfies this.
There should be a shorter way to solve this, using properties of trapeziums . Most likely SL Loney should have a ready made answer to this.
29 Dec 2011, 10:27
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I'm happy to show you this problem -- it has a very elegant solution. 
I'm going to start the way imdpro started:
h -> height of the trapezium or the distance between the parallel sides
x -> distance between C` and D
y -> distance between B` and A
C` and B` are projections of B and C on AD .
(In fact, if you construct this trapezium to scale, angle A is an obtuse angle, so really what project onto what is a little different, but that doesn't affect the calculations that follow.)
Clearly, 20 = 12 + x + y, so x + y = 8. The blue equation. That's important -- hold on to that for a moment.
Also, by Pythagorean Theorem on the side triangles,
x^2 + h^2 = 9^2 = 81
y^2 + h^2 = 15^2 = 225
The red & green equations: hold on to those for moment.
Now, Pythagorean Theorem for the diagonals.
(AC)^2 = h^2 + (20 -x)^2 = h^2 + 400 - 40x + x^2 = (h^2 + x^2) + 400 - 40x = 81 + 400 - 40x = 481 - 40x
(BD)^2 = h^2 + (20 -y)^2 = h^2 + 400 - 40y + x^2 = (h^2 + y^2) + 400 - 40y = 225 + 400 - 40y = 625 - 40y
sum of squares of diagonals = (AC)^2 + (BD)^2 = 481 - 40x + 625 - 40y = 1106 - 40(x + y) = 1106 - 40(8)
= 1106 - 320 = 786 ==> answer choice B
Does that make sense? Let me know if you have any questions.
Mike
I'm going to start the way imdpro started:
h -> height of the trapezium or the distance between the parallel sides
x -> distance between C` and D
y -> distance between B` and A
C` and B` are projections of B and C on AD .
(In fact, if you construct this trapezium to scale, angle A is an obtuse angle, so really what project onto what is a little different, but that doesn't affect the calculations that follow.)
Clearly, 20 = 12 + x + y, so x + y = 8. The blue equation. That's important -- hold on to that for a moment.
Also, by Pythagorean Theorem on the side triangles,
x^2 + h^2 = 9^2 = 81
y^2 + h^2 = 15^2 = 225
The red & green equations: hold on to those for moment.
Now, Pythagorean Theorem for the diagonals.
(AC)^2 = h^2 + (20 -x)^2 = h^2 + 400 - 40x + x^2 = (h^2 + x^2) + 400 - 40x = 81 + 400 - 40x = 481 - 40x
(BD)^2 = h^2 + (20 -y)^2 = h^2 + 400 - 40y + x^2 = (h^2 + y^2) + 400 - 40y = 225 + 400 - 40y = 625 - 40y
sum of squares of diagonals = (AC)^2 + (BD)^2 = 481 - 40x + 625 - 40y = 1106 - 40(x + y) = 1106 - 40(8)
= 1106 - 320 = 786 ==> answer choice B
Does that make sense? Let me know if you have any questions.
Mike
