Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are

Question

https://gmatclub.com/forum/download/file.php?id=53352 Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are 7 inches apart and 6 and 8 inches long, respectively. What is the radius of the circle in inches? (A) 4 (B) 5 (C) 4\sqrt{2} (D) 5\sqrt{2} (E) 7 2020-03-17_2213.png

GMATinsight · Accepted Answer

Answer: Option  B

Bunuel · Answer

MANHATTAN GMAT OFFICIAL SOLUTION: Redraw the figure as closely to scale as possible (use the grid on your scrapboard!), labeling the known dimensions and the radius in question. https://gmatclub.com/forum/download/file.php?id=26866 In order for the trapezoid vertices to lie on the circle, the trapezoid must be symmetrical about the dotted line, which passes through the center of the circle. By drawing this vertical and the radii to points B and C we have created two right triangles, allowing us to use the Pythagorean Theorem. In fact, we might play an educated hunch that the triangles are 3–4–5 common right triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3 and 4 side. The unknown vertical sides are thus 4 and 3, which sum to 7 as they must. Algebraically, we can set up the following equations from the picture: x^2 + 3^2 = r^2 y^2 + 4^2 = r^2 x + y = 7 Setting the two equations for r^2 equal: x^2 + 3^2 = y^2 + 4^2 x^2 – y^2 = 4^2 – 3^2 (x + y)(x – y) = 7 Since (x + y)(x – y) = 7, (x – y) = 1. Solve for x and y: (x + y) = 7 (x – y) = 1 2x = 8 x = 4 y = 7 – x = 3 The radius of the circle is 5, because r^2 = 3^2 + 4^2 = 25. The correct answer is B. 2015-06-08_1627.png

Guilio · Answer

Is this a general rule: if trapezoid is inscribed in the circle, then it  MUST BE  isosceles trapezoid? Something like the triangle inscribed in a "half circle" is always right triangle?

GMATinsight · Answer

Hi,

Yes any trapezium in a circle will be an Isosceles Trapezium...

Since two sides are parallel so the line joining the centre and Mid points of those two parallel line will be a straight line and will serve the purpose of line of symmetry.

arnavjb · Answer

Alternate Solution (Approximating Method):

Area of Square inside Trapezoid = 6x6= 36

Area of remaining part of Trapezoid = Sum of Area of 2 remaining triangles inside the Trapezoid = 1/2 .b .h = 1/2 x (8-6) x7 = 1/2 x 2 x 7 = 7

Total Area of Trapezoid = 36 + 7 = 43

To find radius of the circle use the answer choices to estimate which of the respective answer choices leads to an approximately correct corresponding circle area:

A) 4. Area= π. 4^2 = Aprox. 48 (TOO SMALL)
B) 5. Area= π. 5^2 = Aprox. 75 (CORRECT)
C) 7. Area= π. 7^2 = Aprox. 150 (TOO LARGE)
D) 4√2. Area= π. (4√2) ^2 = Aprox. 96 (TOO LARGE)
E) 5√2. Area= π. (5√2) ^2 = Aprox. 150 (TOO LARGE)

Answer is (B)

anairamitch1804 · Answer

In order for the trapezoid vertices to lie on the circle, the trapezoid must be symmetrical about the dotted line, which passes through the center of the circle. By drawing this vertical and the radii to points B and C we have created two right triangles, allowing us to use the Pythagorean Theorem. In fact, we might play an educated hunch that the triangles are 3–4–5 common right triangles. This checks out: If hypotenuse r is 5, then each triangle has a 3 and 4 side. The unknown vertical sides are
thus 4 and 3, which sum to 7 as they must. Algebraically, we can set up the following equations from the picture:

x^2 + 3^2 = r^2
y^2 + 4^2 = r^2
x + y = 7
Setting the two equations for r2 equal:
x^2 + 3^2 = y^2 + 4^2
x^2 – y^2 = 4^2 – 3^2
(x + y)(x – y) = 7
Since (x + y)(x – y) = 7, (x – y) = 1.
Solve for x and y:
(x + y) = 7
(x – y) = 7
2x = 8
x = 4
y = 7 – x = 3
The radius of the circle is 5, because r^2 = 3^2 + 4^2 = 25.
The correct answer is B.

Parasmadan · Answer

Is this a realistic gmat question?

Karpov92 · Answer

The area of the trapezoid that you calculated here is wrong. There's no square inside the trapezoid shown in the picture, rather it's a rectangle.
Direct formula : Area of a trapezoid = 1/2(sum of parallel sides) * height = 1/2(6+8)* 7 = 49. 
Alternative Method :
Area of Recatngle inside Trapezoid = 6x7= 42
Area of remaining part of Trapezoid =Sum of Area of 2 remaining triangles inside the Trapezoid = 1/2 .b .h = 1/2 x (8-6) x7 = 1/2 x 2 x 7 = 7
Total Area of Trapezoid = 42 + 7 = 49. The rest of the solution makes sense.

To find the radius of the circle, use the answer choices to estimate which of the respective answer choice leads to an approximately correct corresponding circle area:
A) 4. Area= π. 4^2 = Aprox. 48 (TOO SMALL)
B) 5. Area= π. 5^2 = Aprox. 75 (CORRECT)
C) 7. Area= π. 7^2 = Aprox. 150 (TOO LARGE)
D) 4√2. Area= π. (4√2) ^2 = Aprox. 96 (TOO LARGE)
E) 5√2. Area= π. (5√2) ^2 = Aprox. 150 (TOO LARGE)

baraa900 · Answer

can you explain it step by step?

the second answer can anyone explain it from this part? 
 (Setting the two equations for r^2 equal= )

Karpov92 · Answer

First of all, remember one rule, any trapezium in a circle will be an Isosceles Trapezium.

Now draw a straight line shown in the picture that passes through the center of the circle; it will divide the trapezoid into two congruent quadrilaterals.
 We can set up the following equations from the picture:

x^2 + 3^2 = r^2--------1.
y^2 + 4^2 = r^2-------2.
x + y = 7------3
Setting the two equations for r2 equal 1=2 :
x^2 + 3^2 = y^2 + 4^2
x^2 – y^2 = 4^2 – 3^2
(x + y)(x – y) = 7
Since (x + y)(x – y) = 7, (x – y) = 1.
Solve for x and y:
(x + y) = 7
(x – y) = 7
2x = 8
x = 4
y = 7 – x = 3
The radius of the circle is 5, because from eq. 1 : r^2 = 3^2 + 4^2 = 25.
The correct answer is B.

karantambat · Answer

Can somebody tell me whether this is a realistic gmat question?

Karpov92 · Answer

No, if you're targeting a score below 750. Most of the problems in MGMAT Advanced Quant book are extremely tough and highly unlikely to come to GMAT unless you're doing exceptionally well in the quant section(Q50 or Q51).
So relax. Just remember the key Takeaway from this qs - Any trapezium inscribed in a circle must be an isosceles trapezium.

Hit Kudos if you like my relpy.

Fdambro294 · Answer

Since the Trapezoid is Inscribed inside the Circle, the Center of the Circle will be Common with the center of the Trapezoid.

If we draw in this Center and Call it = O

Rule: Any Perpendicular Line drawn from the Center O to a chord inside the Circle will BISECT that Chord

(1st) We can draw 2 Line Segments:

OX ----> where Point X is the Mid-Point of Parallel Side AB and OX is Perpendicular to Side AB

and

OY ------> where Point Y is the Mid-Point of Parallel Side DC and OY is Perpendicular to Side DC

(2nd) fro the Center 0, we can draw 2 Radii connecting Center O to the Inscribed Vertices:

Vertex A and Vertex D

we know have 2 Right Triangles: Triangle XOA and Triangle YOD -----> each have the Radius = r as the Hypotenuse

(Side OX) + (Side OY) = Vertical Distance between the 2 Parallel Lines = 7

Let Side OX = h

Let Side OY = (7 - h)

then using Pythagoras:

Triangle XOA - (3)^2 + (h)^2 = (r)^2

Triangle YOD - (4)^2 + (7 - h)^2 = (r)^2

we can set the 2 Equations equal to each other:

9 + (h)^2 = 16 + (7 - h)^2

9 + (h)^2 = 16 + 49 - 14h + (h)^2

9 = 65 - 14h

14h = 56

h = 4

which means: (7 - h) = 3

Thus: Right Triangle XOA has Legs: 3 and h = 4

and

Right Triangle YOD has Legs: 4 and (7 - h) = 3

the Hypotenuse = Radius = 5

(B)

CEdward · Answer

I don't quite understand the implication of this.

Why does it matter whether the trapezium is isosceles ?

Fdambro294 · Answer

He is just summarizing a takeaway that may apply to (and may be used to answer) an all together different question:

Anytime a Trapezoid (a quadrilateral with only one pair of parallel sides) is inscribed Inside a Circle, the 2 non parallel sides must be equal.

Posted from my mobile device

SonamRa · Answer

Hey Folks, 
Although I am not a master but tried a simple solution for this question, so sharing the same. Many of you might find it inappropriate but I found it useful.

Let's draw a straight line from point A to line CD. This should intersect the line CD at point E from a distance of 1 cm from point D as AB & CD are parallel and all points lie of the circle.
So now AC which is 2R (radius) becomes hypotenuse with side AE and EC, as given the distance between lines is 7 cm so AE=7 cm, And EC = CD-1 = 8-1 = 7 cm.
So, AC^2 = AE^2 + EC^2 = 49 + 49
Gives AC = 7\sqrt{2}
2R = 7\sqrt{2}
R= 7/\sqrt{2}
R= 5

CEdward · Answer

This is exactly what I did, but how did you end up with those last two steps?

R = 7\sqrt{2} / 2 is it not?

KabxW · Answer

Hey everyone!
Again this is super-super tough and while going through the Advance MGMAT Book, I too took sometime too solve this. But, after wasting 3-4 minutes applying equations., drawing on grid and even Algebra (Not to say they are bad, they're just too time-consuming and often it might be better to skip or give a guess on such a question), I resorted to Logic and Estimation:

Just look at the trapezoid. Roughly speaking, the radius will be the hypotenuse (BD) divided by 2. Just bear with me, I know I am telling Maths with some tweak but just try this out. 
Hypotenuse = 7^2 + 6^2 = 49+36 = Under root(85)
Approximate this too. It will be a bit greater than 9 so take 9.2 (Just saying 9.2 keep the 9 in your mind)

We divide by 2. Somwhere it's between 4.5-5. So essentially, on very rough estimates if we just apply direct Pythagoras Theorem, we get radius to be above 4.5 (This is for sure)
Check out the answers. Only 5 is closest, rest all are too large, except 4, which ofcourse cannot be the answer.

It'll take a maximum of 75 seconds, if you trust your guts and be on the edge. (Thanks MGMAT, your attitude coming in haha)
Thus, answer is (B) 5 and a hard question estimated as one step solution is done.

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Trapezoid ABCD is inscribed in a circle. Parallel sides AB and CD are

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