Loci and concurrency Questions and answers

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Elementary Mathematics

21). If area of circle is 13.86 hectares, then cost of fencing it at the rate of 60 paise per metre is

A). Rs.784.00

B). Rs.788.00

C). Rs.792.00

D). Rs.796.00

View Answer

Correct Answer: Rs.792.00

Area of the circle = 1386 hectares

= \( \Large 13.86 \times 100 \times 100m^{2} \)

\( \Large \left(\because 1\ hectare-10000m^{2}\right) \)

= \( \Large 1386 \times 100m^{2} \)

Let r be the radius of the circle. Then

\( \Large \frac{22}{7}r^{2}=1386 \times 100 \)

=> \( \Large r^{2}=\frac{1386 \times 100 \times 7}{22} \)

= \( \Large 63 \times 100 \times 7 \)

=> \( \Large r = 3 \times 7 \times 10=210 m \)

Circumference of the circle = \( \Large 2 \pi r \)

= \( \Large 2 \times \frac{22}{7} \times 210 \)

= \( \Large 44 \times 30 metres \)

Therefore, Cost of fencing at the rate of 60 paise per metre

= \( \Large \frac{44 \times 30 \times 60}{100} \) = Rs.792

22). If chords AP and AQ of a circle of radius 6 cm are at distance 3 cm and \( \Large 3\sqrt{2} \) cm respectively from the centre O of the circle, then area of the smaller sector POQ is

A). \( \Large 24 \pi cm^{2} \)

B). \( \Large 21 \pi cm^{2} \)

C). \( \Large 15 \pi cm^{2} \)

D). \( \Large 12 \pi cm^{2} \)

View Answer

Correct Answer: \( \Large 15 \pi cm^{2} \)

\( \Large \sin \angle OAM = \frac{3}{6}=\frac{1}{2}= \sin 30 ^{\circ} \)

=> \( \Large \angle OAM = 30 ^{\circ} \)

Again, \( \Large \sin \angle OAN = \frac{3\sqrt{2}}{6}=\frac{1}{\sqrt{2}}= \sin 45 ^{\circ} \)

=> \( \Large \angle OAN = 45 ^{\circ} \)

Therefore, \( \Large \angle PAQ = 30 ^{\circ} + 45 ^{\circ} = 75 ^{\circ} \)

Therefore, \( \Large \angle POQ = 2 \times \angle PAQ \)

= \( \Large 2 \times 75 ^{\circ} = 150 ^{\circ} \)

\( \Large \alpha = \frac{150 \pi }{180}=\frac{5}{6} \pi \)

Therefore, Area of the smaller sector,

\( \Large POQ = \frac{1}{2} \alpha r^{2} \)

= \( \Large \frac{1}{2} \times \frac{5 \pi }{6}6^{2} = 15 \pi cm^{2} \)

23). If C be a circle with centre O and radius x cm, then locus of midpoints of equal chords of C each of length y cm will be

A). a circle with centre O and radius \( \Large \frac{1}{2}\sqrt{4x^{2}+y^{2}} cm \)

B). a circle with centre O and radius \( \Large \frac{1}{2}\sqrt{4x^{2}-y^{2}} cm \)

C). diameter of the circle C

D). a circle with centre O and radius \( \Large \frac{1}{2}\sqrt{4x^{2}-x^{2}} cm \)

View Answer

Correct Answer: a circle with centre O and radius \( \Large \frac{1}{2}\sqrt{4x^{2}-y^{2}} cm \)

Let AB be one of the equal chords, and M be middle point, then

\( \Large AM =\frac{1}{2}AB = \frac{1}{2}y\ cm \)

and OM is perpendicular to AB

In right angled \( \Large \triangle OMA, \)

\( \Large OM = \sqrt{OA^{2}-AM^{2}} \)

= \( \Large \sqrt{x^{2}-\frac{y^{2}}{4}} = \frac{1}{2}\sqrt{4x^{2}-y^{2}} \)

which is constant as x, y are constants and the centre O is fixed.

Hence, required locus is a circle of radius

\( \Large \frac{1}{2}\sqrt{4x^{2} - y^{2}} \)

24). The straight lines AB and CD intersect at E. If EF and EG are bisectors of \( \Large \angle DEA\ and\ \angle AEC \) respectively and if \( \Large \angle AEF = x\ and\ \angle AEG = y \), then

A). \( \Large x + y > 90 ^{\circ} \)

B). \( \Large x + y < 90 ^{\circ} \)

C). \( \Large x + y = 90 ^{\circ} \)

D). \( \Large x + y = 180 ^{\circ} \)

View Answer

25). If AB is parallel to CD in the given figure, then \( \Large \angle x,\ \angle y\ and\ \angle z \) respectively are

A). \( \Large 60 ^{\circ} , 50 ^{\circ} ,\ 70 ^{\circ} \)

B). \( \Large 50 ^{\circ} , 60 ^{\circ} ,\ 70 ^{\circ} \)

C). \( \Large 60 ^{\circ} , 60 ^{\circ} ,\ 60 ^{\circ} \)

D). \( \Large 70 ^{\circ} , 50 ^{\circ} ,\ 60 ^{\circ} \)

View Answer

Correct Answer: \( \Large 60 ^{\circ} , 50 ^{\circ} ,\ 70 ^{\circ} \)

\( \Large \angle x = 60 ^{\circ} \left(corresponding\ angle\right) \)

\( \Large \angle EGF = 180 ^{\circ} - 110 ^{\circ} = 70 ^{\circ} \)

\( \Large \angle z = \angle EGF=70 ^{\circ} \left(corresponding\ angle\right) \)

\( \Large \angle y = 180 ^{\circ} - 60 ^{\circ} - \angle z = 180 ^{\circ} - 60 ^{\circ} - 70 ^{\circ} = 50 ^{\circ} \)

Therefore, \( \Large \angle x = 60 ^{\circ} ,\ \angle y = 50 ^{\circ} ,\ \angle z = 70 ^{\circ} \)

26). A line parallel to BC in \( \Large \triangle ABC \) meets AB. AC in D and E respectively. If \( \Large AD= \left(4x-3\right)cm, \) \( \Large DB= \left(3x-1\right)cm,\) \( \Large AE= \left(8x-7\right)cm\ and\ EC= \left(5x-3\right)cm \), then admissible value of x is

A). 0

B). 1

C). 2

D). 3

View Answer

Correct Answer: 1

Since, \( \Large DE\perp BC. \therefore \)

\( \Large \frac{AD}{DB} = \frac{AE}{EC} \)

=> \( \Large \frac{4x-3}{3x-1} = \frac{8x-7}{5x-3} \)

=> \( \Large 20x^{2}-15x-12x+9=24x^{2}-21x-8x+7 \)

=> \( \Large 20x^{2}-27x+9 = 24x^{2}- 29x+7 \)

=> \( \Large 4x^{2}-2x-2=0 \)

\( \Large 2x^{2}-x-1=0 \)

\( \Large \therefore\ x = 1,

-\frac{1}{2} \)

But \( \Large x \neq -\frac{1}{2},\ since\ length\ can't\ be\ -ve, \therefore x = 1 \)

27). If radius of a circle is increased by 75%, then its circumference will increase by

A). 50%

B). 75%

C). 25%

D). 100%

View Answer

Correct Answer: 75%

Let the radius be x units.

Therefore, Increased radius = \( \Large x + x \times \frac{75}{100} = \frac{7x}{4} \)

Original circumference = \( \Large 2 \pi x \)

Increased circumference = \( \Large 2 \pi \times \frac{7x}{4}=\frac{14 \pi x}{4} \)

= \( \Large \frac{7 \pi x}{2} \)

Therefore, Increase in circumference = \( \Large \frac{7 \pi x}{2}-2 \pi x \)

= \( \Large \frac{3 \pi x}{2} \)

Therefore, Percentage increase = \( \Large \frac{3 \pi \frac{x}{2}}{2 \pi x} \times 100 \)

\( \Large = 75\% \)

28). The earth makes a complete revolution about its axis in 24 hours. The angle through which it will turn in 4 hours and 40 minutes, is equal to

A). \( \Large 50 ^{\circ} \)

B). \( \Large 60 ^{\circ} \)

C). \( \Large 70 ^{\circ} \)

D). \( \Large 120 ^{\circ} \)

View Answer

29). If measurement of each of the interior angles of a polygon is \( \Large 160 ^{\circ} \), then number of sides of the polygon would be equal to

A). 9

B). 12

C). 18

D). 27

View Answer

Correct Answer: 18

If number of sides of the polygon is n. then by hypothesis,

\( \Large \frac{2n-4}{n} \times 90=160 \)

=> \( \Large 2n-4 = \frac{160}{90} \times n \)

=> \( \Large 18n - 36 = 16n \)

=> \( \Large 2n = 36 \)

=> \( \Large n = 18 \)

30). \( \Large \angle ABC \) is equal to \( \Large 45 ^{\circ} \) as shown in the figure. If \( \Large \frac{AC}{AB} = \sqrt{2}\) , then \( \Large \angle BAC \) is equal to

A). \( \Large 95 ^{\circ} \)

B). \( \Large 100 ^{\circ} \)

C). \( \Large 105 ^{\circ} \)

D). \( \Large 110 ^{\circ} \)

View Answer

Correct Answer: \( \Large 105 ^{\circ} \)

Let AB =x, therefore, \( \Large AC = \sqrt{2}x \)

\( \Large \frac{\sin 45 ^{\circ} }{\sqrt{2}x} = \frac{\sin \left[ 180 ^{\circ} - \left(45 ^{\circ} + \theta \right) \right]}{x} \)

=> \( \Large \frac{1}{\sqrt{2} \times \sqrt{2}} = \sin \left(135 ^{\circ} - \theta \right) \)

=> \( \Large \sin \left(135 ^{\circ} - \theta \right) = \frac{1}{2} = \sin 30 ^{\circ} \)

=> \( \Large 135 ^{\circ} - \theta = 30 ^{\circ} \)

=> \( \Large \theta = 135 ^{\circ} -30 ^{\circ} =105 ^{\circ} \)

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