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CBSE Class 10 Mathematics Solved Question Paper 2015

Long Answer Type

21. The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

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22.

Find the 60^th term of the AP 8, 10,12, ..., if it has a total of 60 terms and hence find the sum of its last 10 terms.

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23. A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?

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24.

Prove that the length of tangents drawn from an external point to a circle is equal.

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25. Prove that the tangent drawn at the midpoint of an arc of a circle is parallel to the chord joining the end points of the arc.

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26.

Construct a ΔABC in which AB = 6 cm, ∠A = 30^o and ∠B = 60^o. Construct another ΔAB'C' similar to ΔABC with base AB' = 8 cm

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27. At a point A, 20 metres above the level of water in a lake, the angle of elevation of a cloud is 30°. The angle of depression of the reflection of the cloud in the lake, at A is 60°. Find the distance of the cloud from A.

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28. A card is drawn at random from a well-shuffled deck of playing cards.

(i) a card of a spade or an ace.
(ii) a black king
(iii) neither a jack nor a king
(iv) either a king or queen

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29.
Find the values of k so that the area of the triangle with vertices (1,-1), (-4, 2k) and (-k, 5) is 24 sq. units

The vertices of the given ΔABC are A(1 ,-1), B( -4, 2k ) and C (-k, 5)

Area of ΔABC =

$equals 1 half space left square bracket straight x subscript 1 space left parenthesis straight y subscript 2 space minus straight y subscript 3 right parenthesis space plus space straight x subscript 2 space left parenthesis straight y subscript 3 minus straight y subscript 1 right parenthesis space plus space straight x subscript 3 left parenthesis straight y subscript 1 space minus straight y subscript 2 right parenthesis right square bracket equals 1 half left square bracket 1 space left parenthesis 2 straight k plus space 5 right parenthesis space minus 4 left parenthesis negative 5 space plus 1 right parenthesis space minus space straight k left parenthesis space minus 1 space minus 2 straight k right parenthesis right square bracket equals 1 half left parenthesis 2 straight k space plus space 5 space plus space 16 space plus straight k space plus 2 straight k squared right parenthesis equals 1 half left parenthesis 2 straight k squared space plus 3 straight k space plus 21 right parenthesis space saqure space units It space is space given space that Area space of space increment ABC space equals space 24 space square space units therefore 1 half space left parenthesis 2 straight k squared space plus space 3 straight k space plus 21 right parenthesis equals space 48 rightwards double arrow space 2 straight k squared space plus 3 straight k space plus 21 space equals space 48 space rightwards double arrow space 2 straight k squared space plus 9 straight k space minus 6 straight k space equals space 0 rightwards double arrow space straight k space left parenthesis 2 straight k space plus 9 right parenthesis space minus 3 space left parenthesis 2 straight k space plus 9 right parenthesis space equals space 0 rightwards double arrow left parenthesis straight k minus space 3 right parenthesis space left parenthesis 2 straight k space plus space 9 right parenthesis space equals space 0 rightwards double arrow straight k minus 3 space equals space 0 space or space 2 straight k space plus 9 space equals space 0 rightwards double arrow space straight k minus space 3 space equals space 0 space or space 2 straight k space plus 9 space equals space 0 rightwards double arrow straight k space equals 3 space or space straight k space equals space minus 9 over 2 Thus space value space of space straight k space is space 3 space or space fraction numerator negative 9 over denominator 2 end fraction$

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30. In Figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).