Consider quadrilateral CREB made up of squares COAB and OREA, where AE=4. Assume squares COAB and OREA are congurent

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powrotnik5ld · Accepted Answer

Using the Pythagorean Theorem, it can be shown that the diagonal of a square measures 2 times the lenght of its sides. Then: OB=4⋅2=RA Which translates into a system of equations for m and n: 3m−2n=4⋅2 7(m−1)−4.8n=4⋅2 Solve the system by the elimination method. To do so, multiply the first equation by -2,4: −2.4(3m−2n)=−2.4(4⋅2) ⇒−7.2m+4.8n=−9.6⋅2 Then, −7.2m+4.8n=−9.6⋅2 7(m−1)−4.8n=4⋅2 Add both equations. Notice that the terms 4.8n and -4.8 cancel each other out: ⇒−7.2m+7(m−1)=−9.6⋅2+42 ⇒−7.2m+7m−7=−5.6⋅2 ⇒−0.2m=7−5.6⋅2 ⇒m=7−5.6⋅2−0.2 ⇒m=−35+28⋅2 ∴=28⋅2−35≈4.6 Replace value of m into the first original equation and solve for n: 3m−2n=4⋅−3m ⇒−2n=4⋅2−3m ⇒n=4⋅2−3m−2 ⇒n=32m−2⋅2 ⇒n=32(28⋅2−35)−2⋅2 ⇒n=42⋅21052−2⋅2 ∴n=40⋅2−1052≈4.1 Finally, to find the lenght of the swgment RB, notice that the segment EB has a lenght of 8, while RE has a lenght of 4. Since REB is a right triangle whose hypotenuse is RB, then: RB=EB2+RE2 =82+42 =64+16 =64+16 =80 16⋅5 165 =45=RB

indomanihle · Accepted Answer

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Consider quadrilateral CREB made up of squares COAB and OREA,

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