The linear transformation T: R2 2 points
R3 such that T(1,1) = (1,0,2) &
T(2,3) = (1, -1, 4). What is
T(8,11)?
O a) (5,-3, 16)
O b) (1, 3, 10)
Answers
Answer:
Subscribed ✌
\huge \underline\textbf{Solution}
Solution
Let ABCD be the given quadrilateral such that
∠\green{\bold{ABC \:=\:90dgree}}ABC=90dgree
\green{\bold{AB \:=\:9\:m}}AB=9m
\green{\bold{BC\:=\:40\:m}}BC=40m
\green{\bold{CD\:=\:28\:m}}CD=28m
\green{\bold{AD\:=\:m}}AD=m
In △ ABC, we have
\blue{\bold{ {AC}^{2}\:=\: {AB}^{2}\:+\: {BC}^{2}}}AC
2
=AB
2
+BC
2
\blue{\bold{ {AC}^{2}\:=\: {9}^{2}\:+\: {40}^{2}\:=\:1681}}AC
2
=9
2
+40
2
=1681
\blue{\bold{AC\:=\:41\:m}}AC=41m
Now,
\blue{\bold{Area\:of\:ABC\:=\: \frac{1}{2}(base\:×\:Height)}}AreaofABC=
2
1
(base×Height)
\blue{\bold{Area\:of\:ABC\:=\:\frac{1}{2}(AB\:×\:BC)}}AreaofABC=
2
1
(AB×BC)
\blue{\bold{Area\:of\:ABC\:=\:\frac{1}{2}(9\:×\:40)\: {m}^{2}\:=\:180\: {m}^{2}}}AreaofABC=
2
1
(9×40)m
2
=180m
2
In △ACD, we have
\green{\bold{AC\:=\:41\:m}}AC=41m
\green{\bold{CD\:=\:28\:m}}CD=28m
\green{\bold{DA\:=\:15\:m}}DA=15m
\blue{\bold{Let\:a\:=\:AC\:=\:41\:m}}Leta=AC=41m
\blue{\bold{b\:=\:CD\:=\:28\:m}}b=CD=28m
\blue{\bold{c\:=\:DA\:=\:15\:m}}c=DA=15m
Then,
\blue{\bold{s\:=\: \frac{1}{2}(41\:+\:28\:+\:15)\:=\:42}}s=
2
1
(41+28+15)=42
\blue{\bold{Area\:of\:ACD\:=\: \sqrt{s(s-a)(s-b)(s-c)}}}AreaofACD=
s(s−a)(s−b)(s−c)
\blue{\bold{Area\:of\:ACD\:=\: \sqrt{42 (42-41)(42-28)(42-15)}}}AreaofACD=
42(42−41)(42−28)(42−15)
\blue{\bold{Area\:of\:ACD\:=\: \sqrt{42×1×14×27}}}AreaofACD=
42×1×14×27
\blue{\bold{ \sqrt{14×3×14×27}}}
14×3×14×27
\blue{\bold{14\:×\:9\:=\: 126\:{m}^{2}}}14×9=126m
2
Area of quadrilateral ABCD = (Area of △ABC) + (Area of △ACD)
\sf\huge\underline\pink{(180\:+\:126)\:{m}^{2}\:=\:306\:{m}^{2}}
(180+126)m
2
=306m
2