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  1. Asked: In:

    D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB²+ DE².

    somvir jha

    In ΔACE, by pythagoes thoeorem AC² + CE² AE² ...(1) In ΔBCD, by Pythagoras theorem BC² + CD² = DB² ...(2) From the equation (1) and (2), we have AC² + CE² + BC² + CD² = AE² + DB ² ...(3) In ΔBCD, by pythagoras theorem DE² = CD² + CE² ...(4) In ΔABC, by pythagoras theorem AB² = AC² + CB² ...(5) FromRead more

    In ΔACE, by pythagoes thoeorem
    AC² + CE² AE² …(1)
    In ΔBCD, by Pythagoras theorem
    BC² + CD² = DB² …(2)
    From the equation (1) and (2), we have
    AC² + CE² + BC² + CD² = AE² + DB ² …(3)
    In ΔBCD, by pythagoras theorem
    DE² = CD² + CE² …(4)
    In ΔABC, by pythagoras theorem
    AB² = AC² + CB² …(5)
    From the equation (3), (4) and (5), we have
    DE² + AB² = AE² + DB²

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  2. Asked: In:

    In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

    Best Answer
    somvir jha

    Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC. ∴ BE = EC = BC/2 = a/2 In ΔABE, by pythagoras theorem AB² = AE² + BE² ⇒ a² = AE² + (a/2)² = AE² + (a²/4) ⇒ AE² = a² - (a²/4) = 3a²/4 4AE² = 3a² ⇒ 4 × (Altitude) = 3 × (Side)

    Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC.
    ∴ BE = EC = BC/2 = a/2
    In ΔABE, by pythagoras theorem
    AB² = AE² + BE²
    ⇒ a² = AE² + (a/2)² = AE² + (a²/4)
    ⇒ AE² = a² – (a²/4) = 3a²/4
    4AE² = 3a²
    ⇒ 4 × (Altitude) = 3 × (Side)

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  3. Asked: In:

    Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

    Best Answer
    somvir jha

    Let AB and CD are the two pole with height 6 m and 11 m respectively. Therefore, CP = 11 - 6 = 5 m and AP = 12 m In triangle APC, by pythagoras theorem AP² + PC² = AC² ⇒ 12² = 5² = AC² ⇒ AC² = 144 + 25 = 169 ⇒ AC = 13m Hence, the distance between the tops of two poles is 13 m.

    Let AB and CD are the two pole with height 6 m and 11 m respectively.
    Therefore, CP = 11 – 6 = 5 m and AP = 12 m
    In triangle APC, by pythagoras theorem
    AP² + PC² = AC²
    ⇒ 12² = 5² = AC²
    ⇒ AC² = 144 + 25 = 169
    ⇒ AC = 13m
    Hence, the distance between the tops of two poles is 13 m.

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  4. Asked: In:

    D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB²+ DE².

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    somvir jha
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  6. Asked: In:

    Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.

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  7. Asked: In:

    If the areas of two similar triangles are equal, prove that they are congruent.

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  8. Asked: In:

    D, E and F are respectively the mid-points of sides AB, BC and CA of Triangles ABC. Find the ratio of the areas of Triangles DEF and Triangles ABC.

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  9. Asked: In:

    Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

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  11. Asked: In:

    Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

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  12. Asked: In:

    ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is

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