Home/ sourav209/Answers
  1. Asked: In:

    Find the area of the triangle whose vertices are : ( I ) (2, 3), (–1, 0), (2, – 4) ( ii) (–5, –1), (3, –5), (5, 2).

    Best Answer
    sourav209

    (i) (2,3),(-1,0),(2,-4) Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4) Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)] Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)] = 1/2[8+7+6] = 21/2 = 10.5 square units (ii) (-5,-1),(3,-5),(5,2) Vertices of triangle: A(Read more

    (i) (2,3),(-1,0),(2,-4)
    Vertices of triangle: A(2,3) ,B(-1,0) and C(2,-4)
    Using the formula for area of triangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
    Area of triangle ABC = 1/2[2{0-(-4)}+(-1){(-4)-3}+2(3-0)]
    = 1/2[8+7+6]
    = 21/2 = 10.5 square units
    (ii) (-5,-1),(3,-5),(5,2)
    Vertices of triangle: A(-5,-1), B(3,-5) and C(5,2)
    Using the formula for area of trangle ∆ = 1/2[x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)]
    Area of triangle ABC = 1/2 [(-5)(-5-2)+3{3(2-(-1)}+5{-1(-5)}]
    = 1/2[35+9+20]
    = 64/2 = 32 square units

    Here is the video explanation=>

    See less
    • 1
  2. Asked: In:

    In each of the following find the value of ‘k’, for which the points are collinear. (i) (7, –2), (5, 1), (3, k) (ii) (8, 1), (k, – 4), (2, –5)

    Best Answer
    sourav209

    (i) A(7,-2), B(5, 1), C(3,k) Area of triangle formed by three collinear points is zero. Therefore, the area of triangle ABC = 0 ⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0 ⇒ 7-7k+ 5k + 10 -9 = 0 ⇒ -2k = -8 ⇒ k = 4 (ii) P(8, 1), Q(k,-4), R(2,-5) Area of triangle formed by three collinear points is zero. TRead more

    (i) A(7,-2), B(5, 1), C(3,k)
    Area of triangle formed by three collinear points is zero.
    Therefore, the area of triangle ABC = 0
    ⇒ 1/2[7(1-k) +5{k- (-2)) +3(-2-1))] =0
    ⇒ 7-7k+ 5k + 10 -9 = 0
    ⇒ -2k = -8
    ⇒ k = 4
    (ii) P(8, 1), Q(k,-4), R(2,-5)
    Area of triangle formed by three collinear points is zero.
    Therefore, the area of triangle PQR = 0
    ⇒ 1/2[8{-4-(-5)+k(-5-1) + 2{1 -(-4)}] = 0
    ⇒ 8-6k+10 = 0
    ⇒ -6k = -18
    ⇒ k = 3

    See this Video explanation of this question😎

    See less
    • 1
  3. Asked: In:

    Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

    Best Answer
    sourav209

    Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0) Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2) Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1) = 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units Area of triangle ABC = 1/2[Read more

    Coordinates of mid-points of AB = P((x₁+x₂)/2), (y₁+y₂)/2) = P((0=2)/2), (-1+1)/2) = P(1,0)
    Coordinates of mid-points of BC = R((2+0)/2, (1+3)/2) = R(1,2)
    Coordinates of mid-points of AC = Q((0+0)/2, (-1+3)/2) = Q(0,1)
    = 1/2[1(2-1)+1(1-0)+0(0-1)] = 1/2[2] = 1 square units
    Area of triangle ABC
    = 1/2[0(1-3)+2{3-(-1)}+0(-1-1)] = 1/2[8] = 4 square units
    Hence, (Area of triangle PQR)/(Area of triangle ABC) = 1/4

    See this video for better explanation✌️

    See less
    • 1
  4. Asked: In:

    Find the area of the quadrilateral whose vertices, taken in order, are (– 4, – 2), (– 3, – 5), (3, – 2) and (2, 3).

    Best Answer
    sourav209

    Area of triangle ABC = 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}] = 1/2 [12+0+9] = 1/2 [21] = 10.5 square units Area of triangle ACD = 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}] = 1/2 [20+15+0] = 1/2[35] = 17.5 square units Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD ⇒ AreaRead more

    Area of triangle ABC
    = 1/2 [(-4){-5(-2)}+(-3){-2-(-2)}+3{-2-(-5)}]
    = 1/2 [12+0+9]
    = 1/2 [21] = 10.5 square units
    Area of triangle ACD
    = 1/2 [(-4)(-2-3)+3{3-(-2)}+2{-2-2(-2)}]
    = 1/2 [20+15+0] = 1/2[35] = 17.5 square units
    Area of quadrilateral ABCD = Area of triangle ABC +Area of triangle ACD
    ⇒ Area of quadrilateral ABCD = (10.5+17.5) square units = 28 square units

    See this Video Explanation😁

    See less
    • 1
  6. Asked: In:

    You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for engle ABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2).

    Best Answer
    sourav209

    Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0) Area of triangle ABD = 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}] = 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units Area of triangle ACD = 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)] = 1/2 [8+30-32] = 1/2[6] = 3 square units ⇒ Area oRead more

    Coordinates pf mid-points of BC = D((x₁+y₂)/2, (y₁+y₂)/2) = D((3+5)/2, (-2+2)/2) = D(4, 0)
    Area of triangle ABD
    = 1/2[4(-2-0)+3{0-(-6)}+4{-6-(-2)}]
    = 1/2[-8+18-16] = 1/2[-6] = -3 = 3 square units
    Area of triangle ACD
    = 1/2 [4(2-0)+5{0-(-6)}+4(-6-2)]
    = 1/2 [8+30-32] = 1/2[6] = 3 square units
    ⇒ Area of triangle ABD = Area of trinagle ACD = 3 square units
    Hence, a medium of a triangle divides it into two triangles of equal areas.

    See the video explanation here✌️

    See less
    • 1