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The area of rectangular region ABCD is equal to the area of the isosce
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15 Mar 2021, 03:22
15 Mar 2021, 03:22
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The area of the rectangular region ABCD is equal to the area of the isosceles trapezoid PQRS, with PS equal to QR. What is the length of segment \(\overline{QR}\)?
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Question part of the project GRE Quantitative Reasoning Daily Challenge - (2021) EDITION
GRE - Math Book
The area of rectangular region ABCD is equal to the area of the isosce
[#permalink]
15 Mar 2021, 05:22
15 Mar 2021, 05:22
2
Solution:
Area of rectangle ABCD=6*8=48
Area of trapezium PQRS=\(\frac{1}{2}\)×(15+9)×ht of trapezium
48=12×ht
Ht of trapezium=4
Now, we can construct a perpendicular line from Q to other end next to R, let's say to point M.
Thus,RM=3.
RM=3,consider the trapezium as a rectangle, thus both sides will be 15. As the length of SR=9 the additional length of 6 will be equally distributed to both side. Thus RM=3
The two side as 3-4 and a right angle. Therefore the hypotenuse will be 5.
(Hypotenuse)QR=5
Hope this helps!
Posted from my mobile device
Area of rectangle ABCD=6*8=48
Area of trapezium PQRS=\(\frac{1}{2}\)×(15+9)×ht of trapezium
48=12×ht
Ht of trapezium=4
Now, we can construct a perpendicular line from Q to other end next to R, let's say to point M.
Thus,RM=3.
RM=3,consider the trapezium as a rectangle, thus both sides will be 15. As the length of SR=9 the additional length of 6 will be equally distributed to both side. Thus RM=3
The two side as 3-4 and a right angle. Therefore the hypotenuse will be 5.
(Hypotenuse)QR=5
Hope this helps!
Posted from my mobile device
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The area of rectangular region ABCD is equal to the area of the isosce
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15 Mar 2021, 05:25
15 Mar 2021, 05:25
1
Area of Rectangle = Area of Trapezoid
\((8)(6) = \frac{1}{2}(15 + 9)(H)\)
\(H = \frac{96}{24} = 4\)
Now drop a perpendicular either from S or R to PQ
We will get triangle(s) with base as \(3\) and height as \(4\)
Apply Pythagoras Theorem;
\(QR^2 = 3^2 + 4^2 = 25\)
Hence, \(QR = 5\)
\((8)(6) = \frac{1}{2}(15 + 9)(H)\)
\(H = \frac{96}{24} = 4\)
Now drop a perpendicular either from S or R to PQ
We will get triangle(s) with base as \(3\) and height as \(4\)
Apply Pythagoras Theorem;
\(QR^2 = 3^2 + 4^2 = 25\)
Hence, \(QR = 5\)
