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# In the gure above, triangle PQO is an isosceles triangle with sides of

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Math Expert
Joined: 02 Sep 2009
Posts: 52343
In the gure above, triangle PQO is an isosceles triangle with sides of  [#permalink]

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10 Jan 2019, 02:01
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Difficulty:

25% (medium)

Question Stats:

100% (00:35) correct 0% (00:00) wrong based on 23 sessions

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In the gure above, triangle PQO is an isosceles triangle with sides of lengths 5, 5, and 6. What is the area of rectangle MNOP?

A. 12
B. 18
C. 24
D. 30
E. 36

Attachment:

2019-01-10_1400.png [ 17.68 KiB | Viewed 250 times ]

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In the gure above, triangle PQO is an isosceles triangle with sides of  [#permalink]

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10 Jan 2019, 02:11
My answer is C ie 24.

The Altitude of Isosceles triangle is \sqrt{(L^2)-((b/2)^2)}

hence altitude= \sqrt{25-9}= \sqrt{16}= 4

Hence area of rectangle = 6*4= 24
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Re: In the gure above, triangle PQO is an isosceles triangle with sides of  [#permalink]

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10 Jan 2019, 05:57

Solution

Given:
PQO is an isosceles triangle
• PQ = 5
• QO = 5
• OP = 6

To find:
• Area of rectangle MNOP

Approach and Working:
Let h be the height of the triangle = width of the rectangle
• $$h^2 + (\frac{6}{2})^2 = 5^2$$
• $$h^2 = 25 – 9 = 16$$
• Thus, h = 4

Therefore, area of rectangle = OP * h = 6 * 4 = 24

Hence, the correct answer is Option C

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Re: In the gure above, triangle PQO is an isosceles triangle with sides of  [#permalink]

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10 Jan 2019, 06:24
1
Bunuel wrote:

In the gure above, triangle PQO is an isosceles triangle with sides of lengths 5, 5, and 6. What is the area of rectangle MNOP?

A. 12
B. 18
C. 24
D. 30
E. 36

Attachment:
2019-01-10_1400.png

triangle 3:4:5
altitude is 4
which is width
so 4*6 = 24 IMO C
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Re: In the gure above, triangle PQO is an isosceles triangle with sides of &nbs [#permalink] 10 Jan 2019, 06:24
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# In the gure above, triangle PQO is an isosceles triangle with sides of

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