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firas92
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03 Jul 2019, 07:46
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Difficulty:
65%
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Question Stats:
57% (02:32) correct
43%
(03:07)
wrong
based on 70
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On a semi circle with diameter AB, chord CD is drawn parallel to the diameter. The ends of the diameter and chord are joined to form a trapezium. Given that the diameter of the semi circle is 8cm and the length of the non parallel sides is 2cm, find the length of the chord.
A. 3.5cm
B. 4cm
C. 7cm
D. 7.5cm
E. 9cm
A. 3.5cm
B. 4cm
C. 7cm
D. 7.5cm
E. 9cm
03 Jul 2019, 15:33
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Very hard to explain this question without a diagram, so I'd suggest anyone reading this draw one (and even then I'm not sure it's easy to follow, sorry!) : if you draw the diameter AB and the line we want, CD, so they're horizontal, and then draw vertical lines from C and D down to the line AB, you'll make a big rectangle inside the trapezoid. We just want to know the horizontal length of this rectangle. Let's call the two vertices of this rectangle that lie on the diameter X and Y, so X is beneath C, and Y is beneath D. Notice we've made two small right triangles on either side of the trapezoid (ACX and BDY), each with a hypotenuse of 2.
Now if we draw a radius from the center O of the circle to point D, we'll make another right triangle, ODY. The hypotenuse of this triangle is a radius, so is 4. ODY and BDY share a vertical side DY, which we can call "h". If the horizontal length OY is "c", then the length BY is 4-c, because those two lengths form a radius. So we have two right triangles, and we've labeled their sides, and we can use Pythagoras for each:
h^2 + c^2 = 4^2
h^2 + (4 - c)^2 = 2^2
and subtracting the second equation from the first, and using the difference of squares:
c^2 - (4 - c)^2 = 16 - 4
(c + 4 - c)(c - (4-c)) = 12
(4)(2c - 4) = 12
2c = 7
and 2c is the length of the rectangle that we're looking for.
Now if we draw a radius from the center O of the circle to point D, we'll make another right triangle, ODY. The hypotenuse of this triangle is a radius, so is 4. ODY and BDY share a vertical side DY, which we can call "h". If the horizontal length OY is "c", then the length BY is 4-c, because those two lengths form a radius. So we have two right triangles, and we've labeled their sides, and we can use Pythagoras for each:
h^2 + c^2 = 4^2
h^2 + (4 - c)^2 = 2^2
and subtracting the second equation from the first, and using the difference of squares:
c^2 - (4 - c)^2 = 16 - 4
(c + 4 - c)(c - (4-c)) = 12
(4)(2c - 4) = 12
2c = 7
and 2c is the length of the rectangle that we're looking for.
05 Jul 2019, 06:29
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1. You schould draw trapezium ACDB
2. Draw a high from D (let it be DH)
3. Draw a radii to D (let it be OD=4)
4. Then OH= 4-x and HB = x
5. sqrt (4^2-(4-x)^2) = sqrt (2^2-x^2)
6. After solving equation, you will get x=0.5
7. CD= 8-(2*0.5)=7
8. You are awesome!!!
2. Draw a high from D (let it be DH)
3. Draw a radii to D (let it be OD=4)
4. Then OH= 4-x and HB = x
5. sqrt (4^2-(4-x)^2) = sqrt (2^2-x^2)
6. After solving equation, you will get x=0.5
7. CD= 8-(2*0.5)=7
8. You are awesome!!!
\((a-b)^2 = a^2 - 2ab +b^2\) 
