A child is looking for his father. He went 90 | vedclass

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(B) Let the starting point be $A$. The child moves $90 \, m$ east to point $B$. Then he turns right and moves $20 \, m$ to point $C$. Then he turns ri

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) Let the starting point be $A$. The child moves $90 \, m$ east to point $B$. Then he turns right and moves $20 \, m$ to point $C$. Then he turns right again and moves $30 \, m$ to point $D$ (uncle's place). From $D$,he turns right and moves $100 \, m$ north to point $E$ where he meets his father.Let's represent this on a coordinate plane:Starting point $A = (0, 0)$.Point $B = (90, 0)$.Point $C = (90, -20)$.Point $D = (90 - 30, -20) = (60, -20)$.Point $E = (60, -20 + 100) = (60, 80)$.The distance from the starting point $A(0, 0)$ to the final point $E(60, 80)$ is given by the distance formula:$AE = \sqrt{(60 - 0)^2 + (80 - 0)^2}$$AE = \sqrt{60^2 + 80^2}$$AE = \sqrt{3600 + 6400}$$AE = \sqrt{10000}$$AE = 100 \, m$.

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