Lesson: Quadrilaterals

Comment on Quadrilaterals

Hi brent,

Could you please help me understand the approach to solving the following question:

https://gmatclub.com/forum/what-is-the-area-of-rectangular-region-r-166186.html

Is my assumption that diagonals of a rectangle are angle bisectors incorrect?

Is that valid only for square and rhombus?

Thanks in advance!
gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-area-of-rectangular-region-r-1661...

Unless the rectangle in question has 4 equal sides (i.e., the rectangle is a square), then the diagonals do not bisect the angles.

For example, in the following diagonal, you can see that the angles are not bisected:
https://www.math.nmsu.edu/~pmorandi/math112f00/graphics/RectangleWithDia...

As you suggest, the diagonals bisect the angles in squares and rhombuses only.

Cheers,
Brent

Hi, I can't understand the area of a circle: if the area is 4a²π, why the radius is 2a, not 4a? An area of a circle should be Pi (radius)^2? I think I have just missed something..

Image
In the figure above the square has two sides which are tangent to the circle. If the area of the circle is 4a²π, what is the area of the square?

A. 2a²
B. 4a
C. 4a²
D. 16a²
E. 64a²

Thanks in advance!
gmat-admin's picture

Question link: https://gmatclub.com/forum/in-the-figure-above-the-square-has-two-sides-...

We're told the area = 4a²π
NOTE: the a is squared but the 4 is not. That is, the area = 4(a²)π

Take 4(a²)π and make the following EQUIVALENCIES:
4(a²)π = (π)(4)(a²)
= (π)(2²)(a²)
= (π)(2a)²

Since the area of circle = (π)(radius)², we can see that the radius must have length 2a

Does that help?

Here's my full solution: https://gmatclub.com/forum/in-the-figure-above-the-square-has-two-sides-...

Cheers,
Brent

Hi Brent,

Could you please help me understand the approach to solving the following question:

In the figure above, what is the perimeter of rectangle ABPQ?

(1) The area of rectangular region ABCD is 3 times the area of rectangular region ABPQ.
(2) The perimeter of rectangle ABCD is 54.

https://gmatclub.com/forum/in-the-figure-above-what-is-the-perimeter-of-rectangle-abpq-2821.html
gmat-admin's picture

I'm happy to help.

Here's my step-by-step solution (with diagrams!): https://gmatclub.com/forum/in-the-figure-above-what-is-the-perimeter-of-...

Cheers,
Brent

Thanks Brent!

Hi Brent, please help me with the below question

https://gmatclub.com/forum/pqrs-is-a-parallelogram-and-st-tr-what-is-the-ratio-of-the-area-of-218625.html

Hi Brent, in your solution to the below question, I did not understand one part - why is x/y = y/(x/2)?
Shouldn't it be x/y - x/2/y?

----------------------------------------------------------

A project requires a rectangular sheet of cardboard satisfying the following requirement: When the sheet is cut into identical rectangular halves, each of the resulting rectangles has the same ratio of length to width as the original sheet. Which of the following sheets comes closest to satisfying the requirement?

(A) A sheet measuring 7 inches by 10 inches
(B) A sheet measuring 8 inches by 14 inches
(C) A sheet measuring 10 inches by 13 inches
(D) A sheet measuring 3 feet by 5 feet
(E) A sheet measuring 5 feet by 8 feet


Here's an algebraic solution:

Let x be length of the LONG side of the original rectangle
Let y be length of the SHORT side of the original rectangle
Then cut the rectangle into two pieces
Image

We want the resulting rectangles to have the same ratio of length to width as the original sheet.
In other words, we want x/y = y/(x/2)
Cross multiply to get: x²/2 = y²
Multiply both sides by 2 to get: x² = 2y²
Divide both sides by y² to get: x²/y² = 2
Take square root of both sides to get: x/y = √2

IMPORTANT: For the GMAT, everyone should know the following APPROXIMATIONS: √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2

So, we know that x/y ≈ 1.4
In other words, the ratio (LONG side)/(SHORT side) ≈ 1.4
----------------------------------------------------------
gmat-admin's picture

We need the ratio of the sides of the BIG rectangle to equal the ratio of the sides of a SMALL rectangle.
In other words, we want:
(longest length of BIG rectangle)/(shortest length of BIG rectangle) = (longest length of SMALL rectangle)/(shortest length of SMALL rectangle)

See my diagram at https://gmatclub.com/forum/a-project-requires-a-rectangular-sheet-of-car...

BIG RECTANGLE:
longest length = x
shortest length = y
So, ratio = x/y

SMALL RECTANGLE:
longest length = y
shortest length = x/2
So, ratio = y/(x/2)

The resulting EQUATION is: x/y = y/(x/2)

Does that help?

Cheers,
Brent

Hi Brent,

The question specifically mentions "the ratio of length to width", does it have any bearing on how we tackle the question? We used the longest length to shortest length ratio which essentially is length : width (big rectangle) = width : length (small rectangle), reversing the order for the shorter rectangle.

Thanks so much for all the help!
gmat-admin's picture

Good question!

Since the length to width ratios for BOTH rectangles need to be the same, the ONLY way to accomplish this is to "reverse the order"
That is, for both rectangles, we the equation MUST be one of the following:

1) (longer side of BIG rectangle)/(shorter side of BIG rectangle) = (longer side of SMALL rectangle)/(shorter side of SMALL rectangle)

2) (shorter side of BIG rectangle)/(longer side of BIG rectangle) = (shorter side of SMALL rectangle)/(longer side of SMALL rectangle)

Does that help?

Cheers,
Brent

Hi Brent, just a question in terms of Rhombus'.

1) You state that a Rhombus has 4 equal sides and that the diagonals of a Rhombus are perpendicular bisectors. Could you kindly explain as to what advantage we hold in terms of knowing that the diagonals of a rhombus are perpendicular bisectors? I.E. How would this concept be tested?

2) Also, later in the video (around 5:25) you demonstrate another equation to find the area of a rhombus where you set one diagonal to 7 and the other to 4. Given that a rhombus has 4 equal sides, wouldn't both diagonals always be the same length?

Thank you as always! P.S. Happy Happy Halloween.
gmat-admin's picture

Happy Halloween, brownpure!

We're taking our son out trick or treating shortly, but I thought I'd respond before leaving :-)

1) If you were told that quadrilateral ABCD has perpendicular bisectors, then you'd know that the quadrilateral is a rhombus.

2) Be careful. The term "perpendicular bisector" doesn't mean the diagonals are equal; it means each diagonal cuts the other diagonal into 2 equal length. If you check out the link that follows, you can see that the 2 diagonals of the given rhombus don't have equal lengths: https://goo.gl/images/N5dzNU

Cheers,
Brent

Awesome thank you! Hope you and your family enjoy :).
gmat-admin's picture

We had a great time - thanks!

Hi Brent,

Can you please explain your approach in solving the below-mentioned problem.
Thanks a lot
Fatima-Zahra

https://gmatclub.com/forum/what-is-the-area-of-the-shaded-region-in-the-figure-above-208853.html#p1604778
gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-area-of-the-shaded-region-in-the-...

I started by recognizing that:
Area of shaded region = (area of large 8 x 9 rectangle) - (area of unshaded trapezoid)

That is, I took the area of the large 8 x 9 rectangle, and then subtracted area of unshaded trapezoid.

The UNSHADED trapezoid has height 3, and the two parallel sides (the bases) have lengths 4 and 6.

At this point, I applied the formula for area of a trapezoid (see video for formula)

We get: Area of trapezoid = (4 + 6)(3)/2 = 15, which I then subtracted from the area of large 8 x 9 rectangle

Does that help?

Cheers,
Brent

https://gmatclub.com/forum/what-is-the-area-of-the-shaded-region-in-the-figure-shown-264465.html

Couldn't we have proved sufficiency for Statement 1 by this:

Area of unshaded region = 1/2 of area of rectangle as the width and length represent height and base?
gmat-admin's picture

Question link: https://gmatclub.com/forum/what-is-the-area-of-the-shaded-region-in-the-...

That makes total sense. In fact, it's the same reasoning that @Princ used above my solution.
My solution presents an alternative approach.

Cheers,
Brent

https://gmatclub.com/forum/the-figure-above-represents-an-l-shaped-garden-what-is-the-value-of-k-207705.html

I understand the algebraic method to prove statement 2 is insufficient. But visually, could the diagram have been drawn differently with more clues to make the statement sufficient? If so how?
gmat-admin's picture

Here's my full solution (with diagrams): https://gmatclub.com/forum/the-figure-above-represents-an-l-shaped-garde...

Please let me know if that helps.

Cheers,
Brent

https://gmatclub.com/forum/pqrs-is-a-parallelogram-and-st-tr-what-is-the-ratio-of-the-area-of-218625.html

I'm wondering if we could just eyeball this one? Since ST is equal to TR (i.e. QT is the midpoint), we get two equal triangles SQT and QRT). QS divides the parallelogram into half and if we cut QPS into two triangles equal to SQT and QRT then the parallelogram has got 4 triangles, the area which equals the parallelogram. So the proportion of one area of triangle to the entire parallelogram (i.e. 4 triangles) would 1:4?

Does that work or has my mind just dangerously conceived of an easy way out of this which really isn't valid?
gmat-admin's picture

Question link: https://gmatclub.com/forum/pqrs-is-a-parallelogram-and-st-tr-what-is-the...

That's a perfectly valid/fast (GMAT-style) solution - nice work!!

Cheers,
Brent

Hi Brent,

Would you please explain how did we calculate that Angle JQM to be 60 degrees?
https://gmatclub.com/forum/properties-of-polygons-questions-254633.html

Thanks!
gmat-admin's picture

I think you posted the wrong link.
https://gmatclub.com/forum/properties-of-polygons-questions-254633.html is just a list of linked questions pertaining to polygons.
Cheers,
Brent

Hey Brent,

considering this Q:

https://gmatclub.com/forum/in-the-figure-shown-above-line-segment-qr-has-length-12-and-rectangle-165552.html

isn´t there a faster way to solve it than with a quadratic equation?

Cheers,

Philipp
gmat-admin's picture

I can't think of a faster approach.
Here's my full solution: https://gmatclub.com/forum/in-the-figure-shown-above-line-segment-qr-has...

Cheers,
Brent

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