>> Elementary Mathematics >> Geometry

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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
- Rectangular and Cartesian products
- Set theory
- Simple and Decimal fraction
- Simplification
- Statistics
- Straight lines
- Triangle
- Trigonometric ratio
- Trigonometry
- Volume and surface area

1). If every interior angle of regular octagon is \( \Large 135 ^{\circ} \), then find the external angle of it.
Every external angle of octagon = \( \Large 180 ^{\circ} - Interior\ angle \) = \( \Large 180 ^{\circ} - 135 ^{\circ} = 45 ^{\circ} \) | ||||

2). In a \( \Large \triangle ABC, \angle A : \angle B : \angle C = 2 : 4 : 3 \). The shortest side and the longest side of the triangles are respectively.
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3). In a \( \Large \triangle \)ABC, \( \Large \angle A = 90 ^{\circ} \), \( \Large \angle C = 55 ^{\circ} \) and \( \Large \overline{AD}\perp\overline{BC} \), What is the value of \( \Large \angle BAD \)?
In \( \Large \triangle BAC, \) \( \Large \angle B = 180 ^{\circ} - \left( 90 ^{\circ} + 55 ^{\circ} \right) = 35 ^{\circ} \) Now, in \( \Large \triangle ADB \), \( \Large \angle ADB = 90 ^{\circ} \) \( \Large \therefore \angle ADB + \angle DBA + \angle BAD = 180 ^{\circ} \) \( \Large \angle BAD = 180 ^{\circ} - 90 ^{\circ} - 35 ^{\circ} = 55 ^{\circ} \) | ||||

4). O is the circumcentre of the \( \Large \triangle \) ABC. If \( \Large \angle BAC \) = \( \Large 50 ^{\circ} \), then \( \Large \angle OBC \) is equal to
\( \Large \angle BOC = 2 \times 50 ^{\circ} = 100 ^{\circ} \) In \( \Large \triangle OBC, OB = OC \) => \( \Large \angle OBC = \angle OCB \) \( \Large \therefore Sum\ of\ three\ angles\ of\ a\ triangle = 180 ^{\circ} \) => \( \Large \angle OBC + \angle OCB + \angle BOC = 180 ^{\circ} \) => \( \Large 2 \angle OBC + 100 ^{\circ} = 180 ^{\circ} \) -> \( \Large \angle OBC = \frac{80 ^{\circ} }{2} = 40 ^{\circ} \) | ||||

5). ABC is a right angled triangle such that AB = a - b, BC = a and CA = a + b. D is a point on BC such that BD = AB. The ratio of BD : DC for any value of a and b is given by
In right angled \( \Large \triangle ABC \) \( \Large \left(a+b\right)^{2}= \left(a-b\right)^{2}+a^{2} \) => \( \Large a^{2}+b^{2}+2ab = a^{2}+b^{2}-2ab+a^{2} \) => \( \Large 4ab = a^{2} => 4b = a \) Now, \( \Large \frac{BD}{DC}=\frac{a-b}{b}=\frac{4b-b}{b} \) = \( \Large \frac{3b}{b} = \frac{3}{1} \) | ||||

6). ABC is a triangle, where BC = 2AB, \( \Large \angle B\) = \( \Large 30 ^{\circ} \) and \( \Large \angle A\) = \( \Large 90 ^{\circ} \). The magnitude of the side AC is
Given that, \( \Large \angle A = 90 ^{\circ} and \angle B = 30 ^{\circ} \) | ||||

7). The bisectors BI and CI of \( \Large \angle B\) and \( \Large \angle C\) of \( \Large \triangle ABC\) meet in I. What is \( \Large \angle BIC\) equal to?
Given that, BI and CI are angle bisectors of \( \Large \angle B \) and \( \Large \angle C\), respectively. | ||||

8). In the figure given below, \( \Large \angle PQR\) = \( \Large 90 ^{\circ} \) and QL is a median, PQ = 5 cm and QR = 12 cm. Then, QL is equal to
Given that, PQ = 5 cm, QR = 12 cm and QL is a median. | ||||

9). ABC and XYZ are two similar triangles with \( \Large \angle C\) = \( \Large \angle Z\), whose areas are respectively 32 and 60.5. If XY = 7.7 cm, then what is AB equal to?
For similar triangles, ratio of areas is equal to the ratio of the squares of any two corresponding sides. . | ||||

10). ABC is a triangle right angled at A and a \( \Large \perp AD \) is drawn on the hypotenuse BC. What is BC.AD equal to?
In case of a right angled triangle, if we draw a perpendicular from the vertex containing right angle to the hypotenuse, we get three triangles, two smaller and one original and these three triangles are similar triangles. So, \( \Large \triangle ABC - \triangle ABD - \triangle ADC \) \( \Large \therefore BC.AD = AB.AC \) |