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  1. Asked: In:

    Triangles : ∆ABC ~ ∆DEF. If AB = 4 cm, BC = 3.5 cm, CA = 2.5 cm and DF = 7.5 cm, find the perimeter of ∆DEF.

    Deepak Bora Newbie
    This answer was edited.

    Because triangle ABC ∼ triangle DEF ∴ perimeter of triangle ABC : perimeter of triangle DEF = AC : DF Perimeter of triangle ABC = 4+3.5+2.5 = 10cm 10 : perimeter of triangle DEF = 2.5 : 7.5 Cross multiplying Perimeter of triangle DEF = 10× 7.5/2.5 ∴ perimeter of triangle DEF = 30cm

    Because triangle ABC ∼ triangle DEF

    ∴ perimeter of triangle ABC : perimeter of triangle DEF = AC : DF

    Perimeter of triangle ABC = 4+3.5+2.5 = 10cm

    10 : perimeter of triangle DEF = 2.5 : 7.5

    Cross multiplying

    Perimeter of triangle DEF = 10× 7.5/2.5

    ∴ perimeter of triangle DEF = 30cm

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  2. Asked: In:

    Triangles : The lengths of the diagonals of a rhombus are 24 cm and 32 cm. Calculate the length of the altitude of the rhombus.

    Deepak Bora Newbie
    This answer was edited.

    The diagonals of a rhombus bisect each-other at right angles,with the side of the rhombus being the hypotenuse . Half the lengths of the diagonals AO = 12 cm , BO= 16 cm. and the hypotenuse(AB) in right triangle ABO in  rhombus ABCD. Using Pyhagoras Theorem for the right-angled triangle ABO, h² = 12Read more

    The diagonals of a rhombus bisect each-other at right angles,with the side of the rhombus being the hypotenuse . Half the lengths of the diagonals AO = 12 cm , BO= 16 cm. and the hypotenuse(AB) in right triangle ABO in  rhombus ABCD.

    Using Pyhagoras Theorem for the right-angled triangle ABO,

    h² = 12² + 16² = 400.

    ∴ The length of the side of the rhombus is 20 cm.

    area of a rhombus = ½×d1×d2

    are of a rhombus= base × height

    ∴ comparing both formulas

    ½×24×32 = 20×h    (base is side of the rhombus)

    ∴ h= 19.2 cm.

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  3. Asked: In:

    Given 15 cot A = 8, find sin A and sec A.

    Deepak Bora Newbie

    Given.   15cotA = 8                  CotA = 8/15                  CotA = B/P         using Pythagoras theorem we can find the third side, Therefore H = 17        sinA = P/H = 15/17        SecA = H/B =17/8

    Given.   15cotA = 8

                     CotA = 8/15

                     CotA = B/P

            using Pythagoras theorem we can find the third side,

    Therefore H = 17

           sinA = P/H = 15/17

           SecA = H/B =17/8

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  4. Asked: In:

    A man invests Rs.10400 in 6% shares at Rs.104 and Rs.11440 in 10.4% shares at Rs.143.How much income would he get in all?

    Deepak Bora Newbie

    total investment = Rs. 10400 dividend % = 6% market value = Rs. 104 face value = Rs. 100 number of shares = investment ⁄  market value = 10400/104 = 100 dividend = dividend %(face value)× number of shares = 6/100×(100)×100 = Rs. 600 similarly in second case total investment = Rs. 11440 dividend % =Read more

    total investment = Rs. 10400

    dividend % = 6%

    market value = Rs. 104

    face value = Rs. 100

    number of shares = investment ⁄  market value = 10400/104 = 100

    dividend = dividend %(face value)× number of shares

    = 6/100×(100)×100

    = Rs. 600

    similarly in second case

    total investment = Rs. 11440

    dividend % = 10.4%

    M. V. = Rs. 143

    F. V. = Rs. 100

    n= 11440/143= 80

    dividend = 10.4/100 × 100 ×80 = Rs. 832

    total dividend = Rs. (600+832)

    = Rs. 1432.

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