>> Elementary Mathematics >> Quadrilateral and parallelogram

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1). The side of a rhombus whose diagonals are 16 cm and 12 cm respectively, is
Given : \( \Large d_{1}=16\ cm,\ d_{2}=12\ cm \) | ||||

2). If diagonals of a parallelogram are perpendicular, then it is a
Correct Answer: rhombus
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3). If a triangle and a rectangle have equal areas and equal altitude, then base of the triangle is equal to
Let h be the common height of triangle and rectangle. If a and b be respectively the bases of triangle and rectangle, then by hypothesis \( \Large \frac{1}{2}a \times h = b \times h\) => \( \Large \frac{1}{2}a = b \) => \( \Large a = 2b \) | ||||

4). The number of sides of a regular polygon each of whose angles measures \( \Large 156 ^{\circ} \) is
Sum of the interior angles of an n-sided regular polygon = \( \Large \left(2n-4\right) \times 90 ^{\circ} \) Therefore, \( \Large \left(2n-4\right) \times 90 ^{\circ} = 156 \times n \) => \( \Large 180n - 360 = 156n \) => \( \Large 180n - 156n = 360 \) => \( \Large n = \frac{360}{24} = 15 \) | ||||

5). O is the point of intersection of the diagonals AC and BD of a rhombus ABCD. P, Q, R are points on OC, OB and OA respectively such that OP = 1 unit, OQ = 2 units and OR = 4 units. The angle PQR is
From \( \Large \triangle OPQ, PQ^{2}=1^{2}+2^{2} = 5 \) ...(i) | ||||

6). The area of a rhombus is \( \Large 120 cm^{2} \). If one of its diagonals is of length 10 cm, then length of one of its sides is
Area of a rhombus = \( \Large \frac{1}{2} \times \ products\ of\ its\ diagonals \)
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7). ABCD is a rhombus and O is the point of intersection of the diagonals AC and BD. The locus of a point which is equidistant from AB and AD is
Correct Answer: AC
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8). If one diagonal of a parallelogram is 70 cm and perpendicular distance of this diagonal from either of the outlying vertices is 27cm, then area of parallelogram (in \( \Large cm^{2} \) is
Given : AC = 70 cm and BM = 27 cm Therefore, Area of parallelogram ABCD = \( \Large Area\ of\ \triangle ABC + Area\ of\ \triangle ACD \) \( \Large \left(\frac{1}{2} \times 70 \times 27+\frac{1}{2} \times 70 \times 27\right)cm^{2} \) \( \Large = 1890\ cm^{2} \) | ||||

9). The area of a trapezium is \( \Large 275 cm^{2} \). If its parallel sides are in the ratio 2 : 3 and perpendicular distance between them is 5 cm, then smaller of the parallel sides is
Given : Area of trapezium = \( \Large 275\ cm^{2} \) Let parallel sides be 2x and 3x. Area of trapezium = \( \Large \frac{1}{2} \times sum\ of\ parallel\ side \times h \) Therefore, \( \Large \frac{1}{2} \left(2x+3x\right) \times 5 = 275 \) \( \Large x = \frac{2 \times 275}{25}\ =\ 22\ cm \) Therefore, Smaller of the parallel sides = 2x = 44 cm | ||||

10). If each angle of a polygon is \( \Large 165 ^{\circ} \), then number sides of the polygon is
Exterior angle = \( \Large 180 ^{\circ} - 165 ^{\circ} = 15 ^{\circ} \) Therefore, Number of sides = \( \Large \frac{360 ^{\circ} }{Exterior\ angle} \) = \( \Large \frac{360}{15} = 24 \) |