Given an isosceles right angle triangle with side $s$ $s$ and a construction of inscribed rectangle MNOP such that PO // MN. Calculate the perimeter and area of rectangle MNOP in terms of $s$ $s$ ?

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Given an isosceles right angle triangle with side $s$ $s$ and a construction of inscribed rectangle MNOP such that PO // MN. Calculate the perimeter and area of rectangle MNOP in terms of $s$ $s$ ?

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Answer 1 · 2016-08-20T02:15:00

$p = \frac{3}{\sqrt{2}} s$ $p = 3/sqrt(2)s$

$A = \frac{s^{2}}{4}$ $A = s^2/4$

Explanation:

First, we will find $M P$ $MP$ .

Because $M N O P$ $MNOP$ is a rectangle, we know that $¯ ¯¯¯¯¯ ¯ M P$ $bar(MP)$ is parallel to $¯ ¯¯¯¯ ¯ O N$ $bar(ON)$ , and thus to $¯ ¯¯¯¯ ¯ B C$ $bar(BC)$ . This implies that $∠ A M P = ∠ A B C$ $angleAMP = angleABC$ and $∠ A P M = ∠ A C B$ $angleAPM = angle ACB$ , meaning $△ A M P$ $triangleAMP$ is similar to $△ A B C$ $triangleABC$ , and so is also isosceles.

As $A M = M B$ $AM = MB$ and $A M + M B = s$ $AM+MB = s$ , we know that $s = 2 A M$ $s = 2AM$ , or $A M = \frac{s}{2}$ $AM = s/2$ . Because $△ A M P$ $triangleAMP$ is isosceles, this also gives us $A P = \frac{s}{2}$ $AP = s/2$ . Using the Pythagorean theorem, then, we have $M P^{2} = A M^{2} + A P^{2} = 2 {(\frac{s}{2})}^{2} = \frac{s^{2}}{2}$ $MP^2 = AM^2 + AP^2 = 2(s/2)^2 = s^2/2$ , and so $M P = \frac{s}{\sqrt{2}}$ $MP = s/sqrt(2)$ .

Next, we will find $M N$ $MN$ .

Because $M N O P$ $MNOP$ is a rectangle, we know $∠ M N O = 90^{\circ}$ $angleMNO=90^@$ . Then, as $∠ B N M$ $angleBNM$ is its compliment, we also have $∠ B N M = 90^{\circ}$ $angleBNM = 90^@$ .

As the non-right angles of an isosceles right triangle are $45^{\circ}$ $45^@$ , we know $∠ A B C = 45^{\circ}$ $angleABC = 45^@$ , implying $∠ M B N = 45^{\circ}$ $angleMBN = 45^@$ . Thus $△ B N M$ $triangleBNM$ is also an isosceles right triangle, and so $B N = N M$ $BN = NM$ .

Applying the Pythagorean theorem again, we have $B M^{2} = B N^{2} + M N^{2} = 2 M N^{2}$ $BM^2 = BN^2 + MN^2 = 2MN^2$ . But, as $B M = \frac{s}{2}$ $BM = s/2$ , we can substitute that in and solve for $M N$ $MN$ to obtain $M N = \frac{s}{2 \sqrt{2}}$ $MN = s/(2sqrt(2))$

Now that we have the side lengths of the rectangle, we can easily find its perimeter $p$ $p$ and area $A$ $A$ .

$p = 2 (\frac{s}{\sqrt{2}}) + 2 (\frac{s}{2 \sqrt{2}}) = \frac{2 s}{\sqrt{2}} + \frac{s}{\sqrt{2}} = \frac{3}{\sqrt{2}} s$ $p = 2(s/sqrt(2)) + 2(s/(2sqrt(2))) = (2s)/sqrt(2)+s/sqrt(2) = 3/sqrt(2)s$

$A = (\frac{s}{\sqrt{2}}) (\frac{s}{2 \sqrt{2}}) = \frac{s^{2}}{4}$ $A = (s/sqrt(2))(s/(2sqrt(2))) = s^2/4$

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Given an isosceles right angle triangle with side $s$ $s$ and a construction of inscribed rectangle MNOP such that PO // MN. Calculate the perimeter and area of rectangle MNOP in terms of $s$ $s$ ?

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Science

Math

Humanities

... and beyond

Given an isosceles right angle triangle with side ss and a construction of inscribed rectangle MNOP such that PO // MN. Calculate the perimeter and area of rectangle MNOP in terms of ss?

1 Answer

Explanation:

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Impact of this question

Given an isosceles right angle triangle with side $s$ $s$ and a construction of inscribed rectangle MNOP such that PO // MN. Calculate the perimeter and area of rectangle MNOP in terms of $s$ $s$ ?