>> Elementary Mathematics >> Loci and concurrency

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Contents:

- Elementary Mathematics
- Area and perimeter
- Circles
- Clocks
- Factorisation
- Geometry
- Height and Distance
- Indices and Surd
- LCM and HCF
- Loci and concurrency
- Logarithms
- Polynomials
- Quadratic Equations
- Quadrilateral and parallelogram
- Rational expression
- Real Analysis
- Rectangular and Cartesian products
- Set theory
- Simple and Decimal fraction
- Simplification
- Statistics
- Straight lines
- Triangle
- Trigonometric ratio
- Trigonometry
- Volume and surface area

1). The maximum number of tangents which can be drawn from an external point to a circle is
Correct Answer: two
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2). A circular running track is 10 m wide. The difference in the length of the outer boundary and the inner boundary I. depends on the length of the track II. depends on the radius of the inner boundary III. depends on the area of the track IV is \( \Large 20 \pi m \). Select the correct answer using the codes below: Codes :
If \( \Large r_{1} \) be the radius of the inner boundary, then radius of outer boundary \( \Large r_{2} = \left(r_{1}+10\right) \) Difference between length of outer boundary and inner boundary = \( \Large 2 \pi \left(r_{2}-r_{1}\right) = 2 \pi \left(r_{2}-r_{1}\right) \) = \( \Large 2 \pi \times 10 = 20 \pi m \) | ||||

3). The boundary of the shaded region in the given diagram consists of five semicircular areas. If AB=7cm, BC=3.5cm, CD=7cm and DE=7 cm, then area of the shaded region is
Required area = (Area of semi circle AC - Area of semi-circle AB) + (Area of semi-circle BE - Area of semi circle CD) + (Area of semi-circle DE) | ||||

4). If C is a circle passing through three non-collinear points D,E, F such that DE = EF = DF = 3 cms, then radius of the circle C is
\( \Large FM = \sqrt{ \left(3\right)^{2}- \left(\frac{3}{2}\right)^{2} } = \sqrt{9-\frac{9}{4}} = \frac{\sqrt{27}}{2} \) Now FO : OM = 2 : 1 \( \Large \therefore\ \frac{\sqrt{27}}{2}=2x+1x \) \( \Large \frac{3\sqrt{3}}{2}=3x \) \( \Large \therefore\ x = \frac{\sqrt{3}}{2} \) \( \Large \therefore\ FO = 2x = 2 \times \frac{\sqrt{3}}{2} \) => \( \Large FO = \sqrt{3} = r \) Therefore, Hene radius of circle is \( \Large \sqrt{3}\ cm \) | ||||

5). In a circle of radius 7 cm, an arc subtends an angle of \( \Large 108 ^{\circ} \) at the centre. The area of the sector is
Area of setor = \( \Large \frac{\theta}{360} \times \pi r^{2} \) | ||||

6). Area of the shaded portion in the given figure, where the arcs are quadrants of a circle, is
Required area = Area of square -4 x area of one sector = \( \Large 14 \times 14-4 \times \frac{1}{4} \pi \times 7^{2} \) = \( \Large 196 - 4 \times \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 \) = \( \Large 196\ -\ 154 \) = \( \Large 42\ m^{2} \) | ||||

7). In the given figure, PQ is tangent at A; BC is the diameter. If \( \Large \angle ABC = 42 ^{\circ}.\ then\ \angle PAB \) is equal to
AP is tangent to the circle at A. Therefore OA and AP are perpendicular to each other. | ||||

8). If length of the tangent from origin to the circle \( \Large x^{2}+y^{2}-26x+K=0\ is\ 5 \), then K is equal to
Given circle is
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9). If area of the given circle is \( \Large 100 \pi \) square cm, then side of the square inscribed in the circle is
If r is the radius of the circle, then area of the circle = \( \Large \pi r^{2} = 100 \pi cm^{2} \) => \( \Large r^{2} = 100 \) ...(i) => \( \Large r = 10 cm \) From the given figure, \( \Large x^{2}+x^{2}=r^{2} \) => \( \Large 2x^{2}=r^{2} \) ...(ii) side of the square = 2x =\( \Large 2\sqrt{\frac{r^{2}}{2}}=\sqrt{2}r=10\sqrt{2} cm \) | ||||

10). The middle points of all chords (each having the same length) of a circle lie on a
Let 0 \( \Large \left(\alpha ,\ \beta\right) \) and r be the centre and radius of the given circle respectively. |