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Difficulty:
35%
(medium)
Question Stats:
72% (03:16) correct
28%
(03:01)
wrong
based on 25
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In the rectangle, the length is 14, and the width is 6. If the three triangles besides the rectangle are all regular isosceles triangle, then the area of the entire figure is?
A. 169
B. 151
C. 144
D. 121
E. 100
Attachment:
2.png [ 15.8 KiB | Viewed 1218 times ]
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Bunuel
We are given that the triangles are regular isosceles.
Hence -
In \(\triangle QCS\)
- QC = CS
- \(\angle CQS\) = \(\angle CSQ\) = \(45^{\circ}\)
In \(\triangle PAQ\)
- PA = AQ
- \(\angle APQ\) = \(\angle AQP\) = \(45^{\circ}\)
In \(\triangle PBR\)
- PB = BR
- \(\angle BPR\) = \(\angle BRP\) = \(45^{\circ}\)
In \(\triangle QCS\) we drop a perpendicular from C onto QS at point Y, the perpendicular will bisect \(\angle QCS\) and also bisect QS, resulting QY = QS.
The formed \(\triangle QYC\) is a 45 - 45 - 90 triangle. Hence, QY = CY = 3 units
Area of \(\triangle QYC\) = 1/2 * 3 * 3
Area of \(\triangle QCS\) = Area of \(\triangle QYC\) * 2 = 9 units
Similarly, Area of \(\triangle BPR\) = 9 units
Area of \(\triangle PAQ\) using similar approach will be 49 units.
Total Area = Area of rectangle + Area of the three traingles
= (14 * 6 ) + (9 + 49 + 9)
= 84 + 67
= 151
Option B
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Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
Power of Tiny Gains
1.01^(365) = 37.78
0.99^(365) = 0.03
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