Pertanyaan

Diketahui O( 0,0) , A( 2,0) , B( 2, y) , C( 0, y) dan D( 0, 2 1 y) . Nilai y → 2 lim keliling OABD keliling BCD adalah ....

Diketahui O(0,0), A(2,0), B(2,y), C(0,y) dan D(0, $\frac{1}{2}$ y). Nilai $y \to 2 lim \frac{keliling BCD}{keliling OABD}$ adalah ....

$begin mathsize 14px style fraction numerator 5 plus 2 square root of 5 over denominator 5 end fraction end style$
$begin mathsize 14px style fraction numerator 5 plus square root of 5 over denominator 10 end fraction end style$
$begin mathsize 14px style fraction numerator 5 minus 2 square root of 5 over denominator 5 end fraction end style$
$begin mathsize 14px style fraction numerator 5 minus square root of 5 over denominator 10 end fraction end style$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah B.

Pembahasan

Panjang BD = Keliling ∆ BCD = Keliling Oleh karena itu, nilai dari adalah sebagai berikut. Jadi, jawaban yang tepat adalah B.

Panjang BD = $begin mathsize 14px style square root of 2 squared plus open parentheses y over 2 close parentheses squared end root equals square root of 4 plus y squared over 4 end root equals square root of fraction numerator 16 plus y squared over denominator 4 end fraction end root equals 1 half square root of 16 plus y squared end root end style$

Keliling ∆BCD = $begin mathsize 14px style 2 plus y over 2 plus 1 half square root of 16 plus y squared end root equals fraction numerator 4 plus y plus square root of 16 plus y squared end root over denominator 2 end fraction end style$

Keliling $begin mathsize 14px style OABD equals 2 plus y plus 1 half square root of 16 plus y squared end root plus y over 2 equals fraction numerator 4 plus 3 y plus square root of 16 plus y squared end root over denominator 2 end fraction end style$

Oleh karena itu, nilai dari $begin mathsize 14px style limit as straight y rightwards arrow 2 of invisible function application fraction numerator keliling blank BCD over denominator keliling blank OABD end fraction end style$ adalah sebagai berikut.

$begin mathsize 14px style limit as y rightwards arrow 2 of invisible function application space fraction numerator fraction numerator 4 plus y plus square root of 16 plus y squared end root over denominator 2 end fraction over denominator fraction numerator 4 plus 3 y plus square root of 16 plus y squared end root over denominator 2 end fraction end fraction equals limit as straight y rightwards arrow 2 of invisible function application blank fraction numerator 4 plus y plus square root of 16 plus y squared end root over denominator 4 plus 3 y plus square root of 16 plus y squared end root end fraction equals fraction numerator 4 plus 2 plus square root of 16 plus y squared end root over denominator 4 plus 3 times 2 plus square root of 16 plus y squared end root end fraction equals fraction numerator 6 plus square root of 20 over denominator 4 plus 6 plus square root of 20 end fraction equals fraction numerator 6 plus 2 square root of 5 over denominator 10 plus 2 square root of 5 end fraction equals fraction numerator up diagonal strike 2 open parentheses 3 plus square root of 5 close parentheses over denominator up diagonal strike 2 open parentheses 5 plus square root of 5 close parentheses end fraction equals fraction numerator blank over denominator open parentheses 5 plus square root of 5 close parentheses end fraction cross times fraction numerator open parentheses 5 minus square root of 5 close parentheses over denominator open parentheses 5 minus square root of 5 close parentheses end fraction equals fraction numerator 15 minus 3 square root of 5 plus 5 square root of 5 minus 5 over denominator 25 minus 5 end fraction equals fraction numerator 10 plus 2 square root of 5 blank over denominator 20 end fraction equals fraction numerator 5 plus square root of 5 over denominator 10 end fraction end style$

Jadi, jawaban yang tepat adalah B.

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