John is constructing a rectangular garden with side lengths of 3x+12 and 14x-2. What is the area?
Answers
Answer:
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Answer:
Now, In \triangle△ OAP
By using Pythagoras Theorem, we have
\begin{gathered}\rm \: {OA}^{2} = {OP}^{2} + {AP}^{2} \\ \end{gathered}
OA
2
=OP
2
+AP
2
\begin{gathered}\rm \: {r}^{2} = {(17 - x)}^{2} + {5}^{2} \\ \end{gathered}
r
2
=(17−x)
2
+5
2
On substituting the value of \rm \: r^2r
2
from equation (1), we get
\begin{gathered}\rm \: {x}^{2} + {12}^{2} = {17}^{2} + {x}^{2} - 34x + {5}^{2} \\ \end{gathered}
x
2
+12
2
=17
2
+x
2
−34x+5
2
\begin{gathered}\rm \: 144= 289 - 34x +25 \\ \end{gathered}
144=289−34x+25
\begin{gathered}\rm \: 144= 314 - 34x \\ \end{gathered}
144=314−34x
\begin{gathered}\rm \: 34x= 314 - 144 \\ \end{gathered}
34x=314−144
\begin{gathered}\rm \: 34x= 170 \\ \end{gathered}
34x=170
\begin{gathered}\rm\implies \:x = 5 \: \\ \end{gathered}
⟹x=5
On substituting x = 5 in equation (1), we get
\begin{gathered}\rm \: {r}^{2} = {12}^{2} + {5}^{2} \\ \end{gathered}
r
2
=12
2
+5
2
\begin{gathered}\rm \: {r}^{2} = 144 + 25 \\ \end{gathered}
r
2
=144+25
\begin{gathered}\rm \: {r}^{2} = 169 \\ \end{gathered}
r
2
=169
\begin{gathered}\rm \: {r}^{2} = {13}^{2} \\ \end{gathered}
r
2
=13
2
\begin{gathered}\bf\implies \:r \: = \: 13 \: cm \\ \end{gathered}
⟹r=13cm
Now, In \triangle△ OAP
By using Pythagoras Theorem, we have
\begin{gathered}\rm \: {OA}^{2} = {OP}^{2} + {AP}^{2} \\ \end{gathered}
OA
2
=OP
2
+AP
2
\begin{gathered}\rm \: {r}^{2} = {(17 - x)}^{2} + {5}^{2} \\ \end{gathered}
r
2
=(17−x)
2
+5
2
On substituting the value of \rm \: r^2r
2
from equation (1), we get
\begin{gathered}\rm \: {x}^{2} + {12}^{2} = {17}^{2} + {x}^{2} - 34x + {5}^{2} \\ \end{gathered}
x
2
+12
2
=17
2
+x
2
−34x+5
2
\begin{gathered}\rm \: 144= 289 - 34x +25 \\ \end{gathered}
144=289−34x+25
\begin{gathered}\rm \: 144= 314 - 34x \\ \end{gathered}
144=314−34x
\begin{gathered}\rm \: 34x= 314 - 144 \\ \end{gathered}
34x=314−144
\begin{gathered}\rm \: 34x= 170 \\ \end{gathered}
34x=170
\begin{gathered}\rm\implies \:x = 5 \: \\ \end{gathered}
⟹x=5
On substituting x = 5 in equation (1), we get
\begin{gathered}\rm \: {r}^{2} = {12}^{2} + {5}^{2} \\ \end{gathered}
r
2
=12
2
+5
2
\begin{gathered}\rm \: {r}^{2} = 144 + 25 \\ \end{gathered}
r
2
=144+25
\begin{gathered}\rm \: {r}^{2} = 169 \\ \end{gathered}
r
2
=169
\begin{gathered}\rm \: {r}^{2} = {13}^{2} \\ \end{gathered}
r
2
=13
2
\begin{gathered}\bf\implies \:r \: = \: 13 \: cm \\ \end{gathered}
⟹r=13cm
Now, In \triangle△ OAP
By using Pythagoras Theorem, we have
\begin{gathered}\rm \: {OA}^{2} = {OP}^{2} + {AP}^{2} \\ \end{gathered}
OA
2
=OP
2
+AP
2
\begin{gathered}\rm \: {r}^{2} = {(17 - x)}^{2} + {5}^{2} \\ \end{gathered}
r
2
=(17−x)
2
+5
2
On substituting the value of \rm \: r^2r
2
from equation (1), we get
\begin{gathered}\rm \: {x}^{2} + {12}^{2} = {17}^{2} + {x}^{2} - 34x + {5}^{2} \\ \end{gathered}
x
2
+12
2
=17
2
+x
2
−34x+5
2
\begin{gathered}\rm \: 144= 289 - 34x +25 \\ \end{gathered}
144=289−34x+25
\begin{gathered}\rm \: 144= 314 - 34x \\ \end{gathered}
144=314−34x
\begin{gathered}\rm \: 34x= 314 - 144 \\ \end{gathered}
34x=314−144
\begin{gathered}\rm \: 34x= 170 \\ \end{gathered}
34x=170
\begin{gathered}\rm\implies \:x = 5 \: \\ \end{gathered}
⟹x=5
On substituting x = 5 in equation (1), we get
\begin{gathered}\rm \: {r}^{2} = {12}^{2} + {5}^{2} \\ \end{gathered}
r
2
=12
2
+5
2
\begin{gathered}\rm \: {r}^{2} = 144 + 25 \\ \end{gathered}
r
2
=144+25
\begin{gathered}\rm \: {r}^{2} = 169 \\ \end{gathered}
r
2
=169
\begin{gathered}\rm \: {r}^{2} = {13}^{2} \\ \end{gathered}
r
2
=13
2
\begin{gathered}\bf\implies \:r \: = \: 13 \: cm \\ \end{gathered}
⟹r=13cm