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10 Nov 2022, 08:04
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B
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Difficulty:
35%
(medium)
Question Stats:
72% (03:16) correct
28%
(03:01)
wrong
based on 25
sessions
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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 1152 times ]
10 Nov 2022, 13:17
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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
Attachments
Screenshot 2022-11-11 013408.png [ 70.42 KiB | Viewed 988 times ]
