Solved: "Consider rectangel CREB that is formed of squares..."

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Answered question

2021-01-04

Consider rectangel CREB that is formed of squares COAB and OREA, where $A E = 4 f t$ . Assume squares COAB and OREA are congruent.
$B O = 3 m - 2 n$
$A R = 7 (m - 1) - 4.8 n$
Determine the values of m and n and the length of RB.

Answer & Explanation

Dora

Skilled2021-01-05Added 98 answers

The diagonal of each square divides it into two triangles with angles of $45^{\circ}, 45^{\circ} and 90^{\circ}$ . The diagonals (BO and AR) are the triangles' hypotenuses. The hypotenuses is $\sqrt{2}$ times the leg in such triangles. A leg is also a square side, which is 4ft, so
$B O = A R = 4 \sqrt{2}$
$3 m - 2 n = 7 (m - 1) - 4.8 n = 4 \sqrt{2}$
$3 m - 2 n = 4 \sqrt{2}$
$- 2 n = - 3 n + 4 \sqrt{2}$
$n = 1.5 m - 2 \sqrt{2}$
$7 (m - 1) - 4.8 n = 4 \sqrt{2}$
$7 (m - 1) - 4.8 (1.5 m - 2 \sqrt{2}) = 4 \sqrt{2}$
$7 m - 7 - 7.2 m + 9.6 \sqrt{2} = 4 \sqrt{2}$
$- 0.2 m = 4 \sqrt{2} + 7 - 9.6 \sqrt{2}$
$- 0.2 m = - 5.6 \sqrt{2} + 7$
$m = 28 \sqrt{2} - 35$
$n = 1.5 m - 2 \sqrt{2}$
$= 1.5 (28 \sqrt{2} - 35) - 2 \sqrt{2}$
$= 42 \sqrt{2} - 52.5 - 2 \sqrt{2}$
$= 40 \sqrt{2} - 52.5$
$R B^{2} = B A^{2} + A E^{2}$
$= 8^{2} + 4^{2}$
$= 64 + 16$
$= 80$
$R B = \sqrt{80}$
$= 4 \sqrt{5}$

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