Test Index

Properties of Triangle

Section: Mathematics

Question : 6 of 15

Marks: +1, -0

p=\sqrt{3},q=1

Solution:

\;\;\frac{\sqrt{3}}{\sin p}=2\Rightarrow \sin p=\;\;\frac{\sqrt{3}}{2}\;\text{so}\;P=60^{\circ},120^{\circ}

\;\;\frac{1}{\sin Q}=2\Rightarrow \sin Q=\;\;\frac{1}{2}\;\text{so}\;Q=30^{\circ},150^{\circ}

Clearly P = 60º not possible. So P = 120º, Q = 30º, R = 30º

PQ=PR=1,QE=RE=\;\;\frac{\sqrt{3}}{2}

O must be centroid so

OE=\;\;\frac{1}{3}\cdot PE=\;\;\frac{1}{3}\sqrt{1-\;\;\frac{3}{4}}=\;\;\frac{1}{6}

option 1 is correct

Area of

\Delta PQR=\;\;\frac{1}{2}\times\sqrt{3}\times\;\;\frac{1}{2}=\;\;\frac{\sqrt{3}}{4}

Semi-perimeter

=\frac{\sqrt{3}+1+1}{2}=\;\;\frac{\sqrt{3}+2}{2}

So in radius of

\Delta PQR=\frac{\;\;\frac{\sqrt{3}}{4}}{\;\;\frac{\sqrt{3}+2}{2}}=\frac{\sqrt{3}(2-\sqrt{3})}{2}

option 2 is correct

RS=\sqrt{\;\;\frac{PR^{2}+QR^{2}-2PS^{2}}{2}}

=\sqrt{\;\;\frac{1+3-2\times\;\;\frac{1}{4}}{2}}=\;\;\frac{\sqrt{7}}{2}

option 4 is correct

Area of

\Delta SOE=\;\;\frac{1}{3}.

Area of ΔSPE

=\;\;\frac{1}{3}\times\;\;\frac{1}{4}5

area of ΔPQR

=\;\;\frac{1}{6}\times\;\;\frac{1}{2}\times\;\;\frac{\sqrt{3}}{4}=\;\;\frac{\sqrt{3}}{48}

option 3 is incorrect

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