ABC is an equilateral triangle of side 2a. Find each of its alti
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    Triangles

    Multiple Choice QuestionsShort Answer Type

    211. 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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    212. 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
    (A) 2 : 1   (B) 1 : 2  (C) 4 : 1   (D) 1 : 4.
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    213.

    Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
    (a) 2 : 3        (b) 4 : 9
    (c) 81 : 16    (d) 16 : 81.

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    Multiple Choice QuestionsLong Answer Type

    214. Sides of some triangles are given below. Determine which of them are right triangles. In case of a right angle, write the length of its hypotenuse.
    (i) 7 cm, 24 cm, 25 cm
    (ii) 3 cm, 8 cm, 6 cm
    (iii) 50 cm, 80 cm, 100 cm
    (iv) 13 cm, 12 cm, 5 cm.
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    215. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR show that PM2 = QM × MR.
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    216.

    In the given fig, ABD is a triangle right angled at A and AC ⊥ BD. Show that
    (i) AB2 = BC .BD
    (ii) AC2 = BC. DC
    (iii) AD2 = BD . CD

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    Multiple Choice QuestionsShort Answer Type

    217. ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2.
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    218. ABC is an isosceles triangle with AC = BC. If AB2 = 2AC2, prove that ∆ABC is a right triangle.
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    Multiple Choice QuestionsLong Answer Type

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    219. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

    ABC is an equilateral triangle in which AB = BC = CA = 2a.
    It is also given that AD ⊥ BC, BE ⊥ CA and CF ⊥ AB.
    Find : AD, BE and CF.
    ABC is an equilateral triangle in which AB = BC = CA = 2a.It is also
    Determination: In right triangles ADB and ADC.
    Hypotenuse AB = Hypotenuse AC [Given]

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    [Using RHS congruent condition]
    ∴                    BD = CD                 [CPCT]
                                = space 1 half BC
                            [∵   D is the mid-point of BC]
                                 equals 1 half left parenthesis 2 straight a right parenthesis space equals space straight a
    In right triangle ADB, we have (using Pythagoras Theorem)
                      AD squared plus BD squared space equals AB squared
    space rightwards double arrow space space space space space space space space space space space space space space space AD squared space equals space AB squared minus BD squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals left parenthesis 2 straight a right parenthesis squared minus left parenthesis straight a right parenthesis squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 4 straight a squared minus straight a squared space equals space 3 straight a squared rightwards double arrow space space space space space space space space space space space space space space space space AD space equals space square root of 3 straight a
    Similarly,      BE equals square root of 3 straight a
    and              CF equals square root of 3 straight a.
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    220. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
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