Math, asked by ckrp26, 4 months ago

{2n | n€Z ,n<0} list the four elements of the set​

Answers

Answered by masafemeajee90
1

Answer:

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\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

mfhb4673

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