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A 10 m long ladder reaches a window 8 m high from a wall at a distance x. Find the distance 'x'.

Mathematics

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2.

  1. Find the length of AD using the Pythagorean theorem in △ABD\triangle ABD.

    In right-angled △ABD\triangle ABD:

    AB2=AD2+BD2AB^2 = AD^2 + BD^2 132=AD2+5213^2 = AD^2 + 5^2 169=AD2+25169 = AD^2 + 25 AD2=169−25AD^2 = 169 - 25 AD2=144AD^2 = 144 AD=144=12 cmAD = \sqrt{144} = 12 \text{ cm}
  2. Find the length of xx (CD) using the Pythagorean theorem in △ADC\triangle ADC.

    In right-angled △ADC\triangle ADC:

    AC2=AD2+CD2AC^2 = AD^2 + CD^2 152=122+x215^2 = 12^2 + x^2 225=144+x2225 = 144 + x^2 x2=225−144x^2 = 225 - 144 x2=81x^2 = 81 x=81=9 cmx = \sqrt{81} = 9 \text{ cm}

    The value of xx is 9 cm\boxed{\text{9 cm}}.

    3.

  3. Identify the components of the right-angled triangle.

    The ladder, the wall, and the ground form a right-angled triangle.
    Hypotenuse (ladder length) =10= 10 m
    One leg (height of window) =8= 8 m
    Other leg (distance from wall) =x= x m

  4. Apply the Pythagorean theorem.

    (ladder length)2=(height of window)2+(distance from wall)2(\text{ladder length})^2 = (\text{height of window})^2 + (\text{distance from wall})^2 102=82+x210^2 = 8^2 + x^2 100=64+x2100 = 64 + x^2 x2=100−64x^2 = 100 - 64 x2=36x^2 = 36 x=36=6 mx = \sqrt{36} = 6 \text{ m}

    The distance xx is 6 m\boxed{\text{6 m}}.

    4.
    A triplet (a,b,c)(a, b, c) is a Pythagorean triplet if a2+b2=c2a^2 + b^2 = c^2, where cc is the largest number.

    • i) 3, 7, 9
  5. Check if 32+72=923^2 + 7^2 = 9^2.

    $$
    3^2 + 7^2 = 9 + 49 = 58
    $$
    $$
    9^2 = 81
    $$
    Since $58 \neq 81$, (3, 7, 9) is **not** a Pythagorean triplet.
    
    • ii) 7, 24, 25
  6. Check if 72+242=2527^2 + 24^2 = 25^2.

    $$
    7^2 + 24^2 = 49 + 576 = 625
    $$
    $$
    25^2 = 625
    $$
    Since $625 = 625$, (7, 24, 25) **is** a Pythagorean triplet.  
    

    The Pythagorean triplet is (7, 24, 25)\boxed{\text{(7, 24, 25)}}.

    5.

  7. Identify the known sides of the right-angled triangle.

    Hypotenuse =25= 25 cm
    One leg =15= 15 cm
    Let the third side be yy cm.

  8. Apply the Pythagorean theorem.

    (hypotenuse)2=(leg1)2+(leg2)2(\text{hypotenuse})^2 = (\text{leg}_1)^2 + (\text{leg}_2)^2 252=152+y225^2 = 15^2 + y^2 625=225+y2625 = 225 + y^2 y2=625−225y^2 = 625 - 225 y2=400y^2 = 400 y=400=20 cmy = \sqrt{400} = 20 \text{ cm}

    The length of its third side is 20 cm\boxed{\text{20 cm}}.

    6.

  9. Visualize the movement as a right-angled triangle.

    The man's movement east (10 m) and north (24 m) form the two legs of a right-angled triangle. His displacement from the starting point is the hypotenuse.
    Leg 1 (East) =10= 10 m
    Leg 2 (North) =24= 24 m
    Let the displacement be dd m.

  10. Apply the Pythagorean theorem.

    d2=(East)2+(North)2d^2 = (\text{East})^2 + (\text{North})^2 d2=102+242d^2 = 10^2 + 24^2 d2=100+576d^2 = 100 + 576 d2=676d^2 = 676 d=676=26 md = \sqrt{676} = 26 \text{ m}

    The man's displacement from the starting point is 26 m\boxed{\text{26 m}}.

    7.

  11. Identify the sides of the triangle and the longest side.

    Sides are 15 cm, 36 cm, and 39 cm. The longest side is 39 cm.
    For a triangle to be right-angled, the square of the longest side must be equal to the sum of the squares of the other two sides (Pythagorean theorem).

  12. Check if the Pythagorean theorem holds.

    152+362=225+1296=152115^2 + 36^2 = 225 + 1296 = 1521 392=152139^2 = 1521

    Since 152+362=39215^2 + 36^2 = 39^2, the triangle is a right-angled triangle.
    The triangle is a right-angled triangle\boxed{\text{right-angled triangle}}.

    8.

  13. Identify the components of the right-angled triangle.

    The ladder, the wall, and the ground form a right-angled triangle.
    Hypotenuse (ladder length) =4.5= 4.5 m
    One leg (distance from wall) =2.7= 2.7 m
    Other leg (height on wall) =h= h m

  14. Apply the Pythagorean theorem.

    (ladder length)2=(distance from wall)2+(height on wall)2(\text{ladder length})^2 = (\text{distance from wall})^2 + (\text{height on wall})^2 4.52=2.72+h24.5^2 = 2.7^2 + h^2 20.25=7.29+h220.25 = 7.29 + h^2 h2=20.25−7.29h^2 = 20.25 - 7.29 h2=12.96h^2 = 12.96 h=12.96=3.6 mh = \sqrt{12.96} = 3.6 \text{ m}

    The height of the wall where the ladder reaches is 3.6 m\boxed{\text{3.6 m}}.

    9.

  15. Visualize the setup and form a right-angled triangle.

    Draw a horizontal line from the top of the shorter pole to the taller pole. This forms a right-angled triangle.
    The horizontal distance between the poles is one leg =12= 12 m.
    The difference in heights of the poles is the other leg =14 m−9 m=5= 14 \text{ m} - 9 \text{ m} = 5 m.
    The distance between their tops is the hypotenuse. Let this be dd.

  16. Apply the Pythagorean theorem.

    d2=(horizontal distance)2+(difference in heights)2d^2 = (\text{horizontal distance})^2 + (\text{difference in heights})^2 d2=122+52d^2 = 12^2 + 5^2 d2=144+25d^2 = 144 + 25 d2=169d^2 = 169 d=169=13 md = \sqrt{169} = 13 \text{ m}

    The distance between their tops is 13 m\boxed{\text{13 m}}.

    10.

  17. Identify the sides of the triangle and the longest side.

    Sides are 6.5 cm, 6 cm, and 2.5 cm. The longest side is 6.5 cm.
    For a triangle to be right-angled, the square of the longest side must be equal to the sum of the squares of the other two sides.

  18. Check if the Pythagorean theorem holds.

    62+2.52=36+6.25=42.256^2 + 2.5^2 = 36 + 6.25 = 42.25 6.52=42.256.5^2 = 42.25

    Since 62+2.52=6.526^2 + 2.5^2 = 6.5^2, the triangle is a right-angled triangle.
    If it is a right-angled triangle, the hypotenuse is always the longest side.
    The hypotenuse is 6.5 cm\boxed{\text{6.5 cm}}.

    11.

  19. Identify the components of the right-angled triangle formed by the diagonal.

    The adjacent sides of a rectangle form the legs of a right-angled triangle, and the diagonal is the hypotenuse.
    Leg 1 =3.6= 3.6 cm
    Leg 2 =1.5= 1.5 cm
    Let the diagonal be dd cm.

  20. Apply the Pythagorean theorem.

    d2=(side1)2+(side2)2d^2 = (\text{side}_1)^2 + (\text{side}_2)^2 d2=3.62+1.52d^2 = 3.6^2 + 1.5^2 d2=12.96+2.25d^2 = 12.96 + 2.25 d2=15.21d^2 = 15.21 d=15.21=3.9 cmd = \sqrt{15.21} = 3.9 \text{ cm}

    The length of its diagonal is 3.9 cm\boxed{\text{3.9 cm}}.

    12.

  21. Identify the properties of an isosceles right triangle.

    An isosceles right triangle has two equal sides (legs) and a right angle. Since it's right-angled at C, the legs are AC and BC.
    Given AC=4AC = 4 cm, then BC=AC=4BC = AC = 4 cm.
    AB is the hypotenuse.

  22. Apply the Pythagorean theorem.

    AB2=AC2+BC2AB^2 = AC^2 + BC^2 AB2=42+42AB^2 = 4^2 + 4^2 AB2=16+16AB^2 = 16 + 16 AB2=32AB^2 = 32 AB=32=16×2=42 cmAB = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2} \text{ cm}

    The length of AB is 42 cm\boxed{4\sqrt{2} \text{ cm}}.

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