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In the figure, what is the area of Triangle ABC?

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Bunuel
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In the figure, what is the area of Triangle ABC? [#permalink] New post   21 Jun 2017, 01:00
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In the figure, what is the area of Triangle ABC ?

(A) 25

(B) 50

(C) \(\frac{100}{\sqrt{2}}\)

(D) 100

(E) \(100\sqrt{2}\)

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2017-06-21_1155.png
2017-06-21_1155.png [ 9.28 KiB | Viewed 4425 times ]

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Difficulty Level: 600
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sashiim20
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Re: In the figure, what is the area of Triangle ABC? [#permalink] New post   21 Jun 2017, 01:34
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Bunuel wrote:

In the figure, what is the area of Triangle ABC ?

(A) 25

(B) 50

(C) \(\frac{100}{\sqrt{2}}\)

(D) 100

(E) \(100\sqrt{2}\)

Show Spoiler
Attachment:
2017-06-21_1155.png


Given from the diagram \(\triangle\) ABC is isosceles triangle. Therefore \(\angle ABC = \angle ACB\).
AD is the height and median on BC. Hence, BD = DC. Given side \(AC = 10.\)

Given from the diagram \(\triangle\) ADC is isosceles right triangle (45:45:90 triangle). Therefore \(\angle DAC = \angle ACD\) \(= 45^{\circ}\) .
Sides of isosceles right triangle (45:45:90 triangle) are in ratio \(x : x : x \sqrt{2}\)

\(AD : DC : AC =\) \(x : x : x \sqrt{2}\) = \(x : x : 10\)

\(x \sqrt{2}\) \(= 10 ==> x = \frac{10}{\sqrt{2}}\)

Multiplying numerator and denominator by \(\sqrt{2}\) we get;

\(x =10\sqrt{2}/ 2 = 5\sqrt{2}\).

\(AD = BD = DC = 5\sqrt{2}\).

\(BC = 5\sqrt{2} + 5\sqrt{2}= 10\sqrt{2}\).

Area of \(\triangle ABC = 1/2 * base * height = 1/2 * BC * AD\)

Area of \(\triangle\) ABC = \(\frac{1}{2} * 10 \sqrt{2} * 5\sqrt{2}\).

Area of \(\triangle ABC = 5\sqrt{2} * 5\sqrt{2}\)

Area of \(\triangle ABC = 25 * 2 = 50\) . Answer (B) ...
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Re: In the figure, what is the area of Triangle ABC? [#permalink] New post   21 Jun 2017, 01:49
In Triangle ABC, Since AB = AC = 10
We know that angles opposite to equal sides are equal. Therefore, Angle ABC = 45 Degree.

Since angles is any triangle total 180 Degree, Angle BAC = 180 - (45+45) = 90 Degree
Now triangle ABC(right angled at BAC) will have area which is given by formula :
\(\frac{1}{2}\) * AB(Base) * AC(Height) = \(\frac{1}{2}\) * 10 * 10 = \(\frac{100}{2}\) = 50(Option B)
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Re: In the figure, what is the area of Triangle ABC? [#permalink] New post   21 Jun 2017, 16:31
Bunuel wrote:

In the figure, what is the area of Triangle ABC ?

(A) 25

(B) 50

(C) \(\frac{100}{\sqrt{2}}\)

(D) 100

(E) \(100\sqrt{2}\)

Show Spoiler
Attachment:
2017-06-21_1155.png





As per the diagram, Angle at A is 90
this is a rt. angled triangle with base and height as 10 each
hence area is 1/2*b*h
1/2*10*10
50
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Re: In the figure, what is the area of Triangle ABC? [#permalink] New post   22 Jun 2017, 03:23
Bunuel wrote:

In the figure, what is the area of Triangle ABC ?

(A) 25

(B) 50

(C) \(\frac{100}{\sqrt{2}}\)

(D) 100

(E) \(100\sqrt{2}\)

Show Spoiler
Attachment:
2017-06-21_1155.png



The question is very easy .. It can be solved by two methods.
The short way : To prove the triangle of a specific type like right angled , isosceles etc.
The long way : To calculate the specific sides and then find the area.

We will try to solve it using short process as required in GMAT for solving the problem faster.

/_DAC = 45 , /_ADC = 90
-> /_DAC = 180 - 45- 90 = 45

Also /_\ ABC is isosceles (AB = AC)

/_ABC = /_DAC = 45
-> /_BAC = 180 -45 - 45 = 90
So /_\BAC is right angle triangle, right angled at A.

Area (/_\ BAC) = 1/2*10*10 = 50

Answer B
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Re: In the figure, what is the area of Triangle ABC? [#permalink]
22 Jun 2017, 03:23
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