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4 years ago

In the circle, CP=x and PD=y. If xy=28 and AP=2, then what is PB?

Mathematics

1 answer:

mario62 [17]4 years ago

5 0

We know that
When two chords intersect each other inside a circle, the products of their segments are equal

so
CP*PD=AP*PB
x*y=2*PB-----------> 28=2*PB----------> PB=14

the answer is
PB=14

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Step-by-step explanation:

DIAGRAM :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large{14\ cm}}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large{14\ cm}}\put(-0.5,-0.4){\bf}\put(-0.5,3.2){\bf}\put(5.3,-0.4){\bf}\put(5.3,3.2){\bf}\end{picture}$

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SOLUTION :

Here's the required formula to find the length of rectangle :

${\longrightarrow{\pmb{\sf{A_{(Rectangle)} = l \times b}}}}$

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b = breadth

Substituting all the given values in the formula to find the length of rectangle :

$\begin{gathered}\qquad{\longrightarrow{\sf{A_{(Rectangle)} = l \times b}}}\\\\\qquad{\longrightarrow{\sf{322 = l \times 14}}}\\\\\qquad{\longrightarrow{\sf{322 = 14l}}}\\\\\qquad{\longrightarrow{\sf{l = \dfrac{322}{14}}}}\\\\\qquad{\longrightarrow{\sf{l = \cancel{\dfrac{322}{14}}}}}\\\\\qquad{\longrightarrow{\sf{l = 23 \: cm}}}\\\\\qquad{\star{\underline{\boxed{\sf{ \pink{l = 23 \: cm}}}}}}\end{gathered}$

Hence, the length of rectangle is 23 cm.

$\begin{gathered}\end{gathered}$

LEARN MORE :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

$\rule{300}{2.5}$

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