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805+ (Hard)|   Geometry|      
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 24 Jul 2020, 01:42
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[GMAT math practice question]

The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendicular to line RV and AR = 10cm , DS = 16cm, CV = 9cm, ST = 3cm and RS = 8cm. What is the area of parallelogram ABCD?

Attachment:
7.24ps.png
7.24ps.png [ 7.66 KiB | Viewed 3272 times ]

A. \(110 cm^2\)

B. \(122 cm^2 \)

C. \(130 cm^2 \)

D. \(144 cm^2 \)

E. \(150 cm^2\)
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B
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 24 Jul 2020, 03:07
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Since the slope of AD and BC are equal, VT = 8 and BT = 9-6=3

We get,
Attachment:
7.24ps.png
7.24ps.png [ 17.11 KiB | Viewed 3206 times ]

Area of ABCD = Area of PQRS - Area of APB - Area of BQC - Area of CRD - Area of DSA

Area of ABCD = Area of PQRS - [Area of APB + Area of CRD ] - [Area of BQC+Area of DSA ]

Area of ABCD = 19*13- 7*11 - 8*6 = 122
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 25 Jul 2020, 02:15
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nick1816
Since the slope of AD and BC are equal, VT = 8 and BT = 9-6=3

We get,
Attachment:
7.24ps.png

Area of ABCD = Area of PQRS - Area of APB - Area of BQC - Area of CRD - Area of DSA

Area of ABCD = Area of PQRS - [Area of APB + Area of CRD ] - [Area of BQC+Area of DSA ]

Area of ABCD = 19*13- 7*11 - 8*6 = 122
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Can you please explain the highlighted portion ? I don't really follow.
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 25 Jul 2020, 05:10
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Attachment:
7.24ps (1).png
7.24ps (1).png [ 8.56 KiB | Viewed 3058 times ]
AD and BC are 2 opposite sides of parallelogram. So, they are parallel or their slopes are equal. Also, their length are equal.

Hence, DE = 16-10 = 6 = CF

and AE = 8 = BF or VT

CF = CV-BT

BT = 9-6=3

If you still have any doubt, you can ask.



Anuraagsaboo1
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Since the slope of AD and BC are equal, VT = 8 and BT = 9-6=3

We get,
Attachment:
The attachment 7.24ps.png is no longer available

Area of ABCD = Area of PQRS - Area of APB - Area of BQC - Area of CRD - Area of DSA

Area of ABCD = Area of PQRS - [Area of APB + Area of CRD ] - [Area of BQC+Area of DSA ]

Area of ABCD = 19*13- 7*11 - 8*6 = 122

Can you please explain the highlighted portion ? I don't really follow.
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 26 Jul 2020, 19:09
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=>

Attachment:
7.24ps(a).png
7.24ps(a).png [ 13.57 KiB | Viewed 2885 times ]

The area of the trapezoid \(ARSD\) is \((\frac{1}{2})(10 + 16)·8 = 104.\)

The area of the trapezoid \(CDSV\) is \((\frac{1}{2})(9 + 16)·11 = \frac{275}{2}.\)

The area of the trapezoid \(BART\) is \((\frac{1}{2})(3 + 10)·11 = \frac{143}{2}.\)

The area of the trapezoid \(BTVC\) is \((\frac{1}{2})(3 + 9)·8 = 48.\)

Then the area of \(ABCD\) is

\(□ARSD + □CDSV - □BARD - □BTVC\)

\(= 104 + \frac{275}{2} – \frac{143}{2} – 48\)

\(= \frac{208}{2} + \frac{275}{2} – \frac{143}{2} – \frac{96}{2}\)

\(= \frac{244}{2}\)

\(= 122 cm^2.\)

Therefore, B is the correct answer.
Answer: B
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The figure shows parallelogram ABCD. AR , DS , BT and CV are perpendi 02 Nov 2023, 00:18
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nick1816
Since the slope of AD and BC are equal, VT = 8 and BT = 9-6=3

We get,
Attachment:
7.24ps.png

Area of ABCD = Area of PQRS - Area of APB - Area of BQC - Area of CRD - Area of DSA

Area of ABCD = Area of PQRS - [Area of APB + Area of CRD ] - [Area of BQC+Area of DSA ]

Area of ABCD = 19*13- 7*11 - 8*6 = 122
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Hi, I have a doubt. Bunuel chetan2u nick1816 MathRevolution

If we have the values of AD (= 10) and AB (=\(\sqrt{ 7*7 + 11*11 }\)=\(\sqrt{ 170 }\)), then why can't we use base x height (10*\(\sqrt{ 170 }\) = 10*(~13) = 130) as the answer?
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