RS Aggarwal Solutions Chapter 16 Coordinate Geometry Exercise 16C Class 10 Maths
Chapter Name  RS Aggarwal Chapter 16 Coordinate Geometry 
Book Name  RS Aggarwal Mathematics for Class 10 
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Exercise 16C Solutions
1. Find the area of △ABC whose vertices are:
(i) A(1, 2), B(2, 3) and C(3, 4)
(ii) A(5, 7), B( 4, 5) and C(4, 5)
(iii) A(3, 8), B(4, 2) and C(5, 1)
(iv) A(10, 6), B(2, 5) and C(1, 3)
Solution
(i) A(1, 2) , B(2, 3) and C(3, 4) are the vertices of △ABC. Then,
(x_{1 }= y, y_{1} = 2), (x_{2} = 2, y_{2 }= 3) and (x_{3} = 3, y_{3} = 4)
(ii) A(5, 7), B(4, 5) and C(4, 5) are the vertices of △ABC. Then,
(x_{1} = 5, y_{1 }= 7), (x_{2} = 4, y_{2} =  5) and (x_{3} = 4, y_{3 }= 5)
(iii) A(3, 8), B(4, 2) and C(5, 1) are vertices of △ABC. Then,
(x_{1} = 3, y_{1} = 8), (x_{2} = 4, y_{2} = 2) and (x_{3 }= 5, y_{3 }= 1)
(iv) A(10, 6), B(2, 5) and C(1, 3) are the vertex of △ABC. Then,
(x_{1} = 10, y_{1} = 6), (x_{2} = 2, y_{2} = 5) and (x_{3} = 1, y_{3 }= 3)
2. Find the area of a quadrilateral ABCD whose vertices area A(3, 1), B(9, 5) C(14, 0) and D(9, 19).
Solution
By joining A and C, we get two triangles ABC and ACD.
Let
A(x_{1}, y_{1}) = A(3, 1), B(x_{2}, y_{2}) = B(9, 5), C(x_{3}, y_{3}) = C(14, 0) and D(x_{4}, y_{4}) = D(9, 19)
Then,
Area of △ABC = 1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})]
= 1/2[3(5 – 0) + 9(0 + 1) + 14(1 + 5)]
= 1/2[15 + 9 + 56]
= 25 sq. units.
Area of △ACD = 1/2[x_{1}(y_{3} – y_{4}) + x_{3}(y_{4} – y_{1}) + x_{4}(y_{1} – y_{3})]
= 1/2[3(0 – 19) + 14(19 + 1) + 9( 1 – 0)]
= ½[57 + 280 – 9]
= 107 sq. units
So, the area of the quadrilateral is 25 + 107
= 132 sq. units.
3. Find the area of quadrilateral PQRS whose vertices are P(5, 3), Q(4, 6), R(2, 3) and S(1, 2).
Solution
By joining P and R, we get two triangles PQR and PRS.
Let P(x_{1}, y_{1}) = P(5, 3), Q(x_{2}, y_{2}) = Q(4, 6), R(x_{3}, y_{3}) = R(2, 3). Then,
S(x_{4}, y_{4}) = S(1, 2)
Area of △PQR = 1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[5(6 + 3) – 4(3 + 3) + 2(3 + 6)]
= ½[15 – 0 + 6]
= 21/2 sq. units
Area of △PRS = 1/2[x_{1}[(y_{3} – y_{4}) + x_{3}(y_{4 }– y_{1}) + x_{4}(y_{1 }– y_{3})]
= ½[5 (3 – 2) + 2(2 + 3) + 1( 3 + 3)]
= ½[25 + 10 + 0]
= 35/2 sq. units.
So, the area of the quadrilateral PQRS is 21/2 + 35/2
= 28 sq. units units.
4. Find the area of quadrilateral ABCD whose vertices are A(3, 1), B(2, 4) C(4, 1) and D(3, 4)
Solution
By joining A and C, we get two triangles ABC and ACD.
Let A(x_{1}, y_{1}) = A(3, 1), B(x_{2}, y_{2}) = B(2, 4), C(x_{3}, y_{3}) = C(4, 1) and Then D(x_{4}, y_{4}) = D(3, 4)
Area of △ABC = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[3(4 + 1) – 2(1 + 1) + 4(1 + 4)]
= ½[9 = 0 + 12]
= 21/2 sq. units.
Area of △ACD = ½(x_{1}(y_{3} – y_{4}) + x_{3}(y_{4} – y_{1}) + x_{4}(y_{1} – y_{3})]
= ½[3(1 – 4) + 4(4 + 1) + 3(1 + 1)]
= ½[15 + 20 + 0]
= 35/2 sq. units
So, the area of the quadrilateral ABCD is 21/2 + 35/2
= 28 sq. units
5. Find the area of quadrilateral ABCD whose vertices are A(5, 7), B(4, 5) C(1, 6) and D(4, 5)
Solution
By joining A and C, we get two triangles ABC and ACD.
Let A(x_{1}, y_{1}) = A(5, 7), B(x_{2}, y_{2}) = B(4, 5), C(x_{3}, y_{3}) = C(1, 6) and D(x_{4}, y_{4}) = D(4, 5)
Then,
Area of △ABC = ½[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[5(5 + 6) – 4(6 – 7) – 1(7 + 5)]
= 1/2[ 5 + 52 – 12]
= 35/2 sq. units.
Area of △ACD = ½[x_{1}(y_{3 }– y_{4}) + x_{3}(y_{4} – y_{1}) + x_{4}(y_{1} – y_{3})]
= ½[5(6 – 5) – 1(5 – 7) + 4(7 + 6)]
= ½[55 + 2 + 52]
= 109/2 sq. units
So, the area of the quadrilateral ABCD is 35/2 + 109/2
= 72 sq. units.
6. Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are A(2, 1) B(4, 3) and C(2, 5)
Solution
The verticals of the triangle are A(2, 1), B(4, 3) and C(2, 5)
Coordinates of midpoint of AB = P((x_{1}, y_{1}) = (2 + 4)/2, (1 + 3)/2 = (3, 2)
Coordinates of midpoint of BC = Q(x_{2}, y_{2}) = (4 + 2)/2, (3 + 5)/2 = (3, 4)
Coordinates of midpoint of AC = R(x_{3}, y_{3}) = (2 + 2)/2, (1 + 5)/2 = (2, 3)
Now,
Area of △PQR = ½[x_{2}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[3(4 – 3) + 3(3 – 2) + 2(2 – 4)]
= ½[3 + 3 – 4]
= 1 sq. unit
Hence, the area of the quadrilateral triangle is 1 sq. unit.
7. A(7, 3), B(5, 3) and C(3, 1) are the vertices of a △ABC and AD is its median. Prove that the median AD divides △ABC into two triangles of equal areas.
Solution
The vertices of the triangle are A(7, 3), B(5, 3), C(3, 1)
Coordinates of D = (5 + 3)/2, (3 – 1)/2 = (4, 1)
For the area of the triangle ADC, let
A(x_{1}, y_{1}) = A(7, 3), D(x_{2}, y_{2}) = D(4, 1) and C(x_{3}, y_{3}) = C(3, 1). Then
Area of △ADC = 1/2[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[7(1 + 1) + 4(1 + 3) + 3(3 – 1)]
= ½[14 + 8 – 12]
= 5 sq. unit
Now, for the area of triangle ABD, let
A(x_{1}, y_{1}) = A(7, 3), B(x_{2}, y_{2}) = B(5, 3) and D(x_{3}, y_{3}) = D(4, 1). Then
Area of △ADC = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[7(3 – 1) + 5(1 + 3) + 4( 3 – 3)]
= ½[14 + 20 – 24]
= 5 sq. unit
Thus, Area (△ADC) = Area(△ABD) = 5 sq. units.
Hence, AD divides △ABC into two triangles of equal areas.
8. Find the area △ABC with A(1, 4) and midpoints of sides through A being (2, 1) and (0, 1).
Solution
Let (x_{2}, y_{2}) and (x_{3}, y_{3}) be the coordinates of B and C respectively. Since, the coordinates of A are (1, 4), therefore
(1 + x_{2})/2 ⇒ x_{2} = 3
(4 + y_{2})/2 =  1
⇒ y_{2} = 2
(1 + x_{2})/2 = 0
⇒ x_{3 }= 1
(4 + y_{3})/2 =  1
⇒ y_{3} = 2
Let A(x_{1}, y_{1}) = A(1, 4), B(x_{2}, y_{2}) = B(3, 2) and C(x_{3}, y_{3}) = C(1, 2) Now
Area(△ABC) = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})]
= ½[1(2 – 2) + 3(2 + 4) – 1(4 – 2)]
= ½[0 + 18 + 6]
= 12 sq. units
Hence, the area of the triangle △ABC is 12 sq. units.
9. A(6, 1), B(8, 2) and C(9, 4) are the vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of △ADE
Solution
Let(x, y) be the coordinates of D and (x’, y’) be the coordinates of E. Since, the diagonals of a parallelogram bisect each other at the same point, therefore
(x + 8)/2 = (6 + 9)/2 ⇒ x = 7
(y + 2)/2 = (1 + 4)/2 ⇒ y = 3
Thus, the coordinates of D are (7, 3) E is midpoint of DC, therefore
x’ = (7 + 9)/2
⇒ x’ = 8
y’ = (3 + 4)/2
⇒ y’ = 7/2
Thus, the coordinates of E are (8, 7/2)
Let A(x_{1}, y_{1}) = A(6, 1), E(x_{2}, y_{2}) = E(8, 7/2) and D(x_{3}, y_{3}) = D(7, 3) Now
Area (△ABC) = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
= ½[6(7/2 – 3) + 8(3 – 1) + 7(1 – 7/2)]
= ½[3/2]
= ¾ sq. unit
Hence, the area of the triangle △ADE is 3/4 sq. units.
10. If the vertices of △ABC be A(1, 3), B(4, p) and C(9, 7) and its area is 15 square units, find the values of p.
Solution
Let A(x_{1}, y_{1}) = A(1, 3), B(x_{2}, y_{2}) = B(4, p) and C(x_{3}, y_{3}) = C(9, 7)
Now
Area of (△ABC) = ½[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})]
⇒ 15 = 1/2 [1(p – 7) + 4(7 + 3) – 9(3 – p)]
⇒ 15 = ½[10p + 60]
⇒ 10p + 60 = 30
Therefore,
⇒ 10p + 60 = 30 or 30
⇒ 10p =  90 or – 30
⇒ p = 9 or 3
Hence, p = 9 or p = 3
11. Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, 3) and C(7, k) is 6 square units.
Solution
Let A(x_{1}, y_{1}) = A(k + 1, 1), B(x_{2}, y_{2}) = B(4, 3) and C(x_{3}, y_{3}) = C(7, k) Now
Area of (△ABC) = ½[x_{1}(y_{2} – y_{1}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})]
⇒ 6 = ½[(k + 1)(3 + k) + 4(k – 1) + 7(1 + 3)]
⇒ 6 = ½[k^{2} – 2k – 3 – 4k – 4 + 28]
⇒ k^{2} – 6k + 9 = 0
⇒ (k – 3)^{2} = 0
⇒ k = 3
Hence, k = 3.
12. For what value of k(k > 0) is the area of the triangle with vertices (2, 5), (k, 4) and (2k + 1, 10) equal to 53 square units ?
Solution
Let A(x_{1} = 2, y_{1} = 5), B(x_{2 }= k, y_{2} = 4) and C(x_{3} = 2k + 1, y_{3} = 10) be the vertices of the triangle, So
Area (△ABC) = 1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
⇒ 53 = ½[(2)(4 – 10) + k(10 – 5) + (2k + 1)(5 + 4)]
⇒ 53 = ½[28 + 5k + 9(2k + 1)]
⇒ 28 + 5k + 18k + 9 = 106
⇒ 37 + 23k = 106
⇒ 23k = 106 – 37 = 69
⇒ k = 69/23 = 3
Hence, k = 3
13. Show that the following points are collinear:
(i) A(2, 2), B(3, 8) and C(1, 4)
(ii) A(5, 1), B(5, 5) and C(10, 7)
(iii) A(5, 1), B(1, 1) and C(11, 4)
(iv) A(8, 1), B(3, 4) and C(2, 5)
Solution
(i) Let A(x_{1} = 2, y_{1} = 2), B(x_{2} = 3, y_{2} = 8) and C(x_{3 }= 1, y_{3} = 4) be the given points.
Now, x_{1}(y_{2 }– y_{1}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})
= 2(8 – 4) + (3)(4 + 2) + (1)(2 – 8)
= 8 – 18 + 10
= 0
Hence, the given points are collinear.
(ii) Let A(x_{1} = 5, y_{1} = 1), B(x_{2 }= 5, y_{2 }= 5) and C(x_{3} = 10, y_{3 }= 7) be the given points.
Now, x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})
= (5)(5 – 7) + 5(7 – 1) + 10(1 – 5)
= 5(2) + 5(6) + 10(4)
= 10 + 30 – 40
= 0
Hence, the given points are collinear.
(iii) Let A(x_{1} = 5, y_{1} = 1), B(x_{2} = 1, y_{2} = 1) and C(x_{3 }= 11, y_{3 }= 4) be the given points.
Now x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1 }– y_{2})
= 5(1 – 4) + 1(4 – 1) + 11(1 + 1)
= 25 + 3 + 22
= 0
Hence, the given points are collinear.
(iv) Let A(x_{1} = 8, y_{1 }= 1), B(x_{2 }= 3, y_{2} = 4) and C(x_{3} = 2, y_{3} = 5) be the given points.
Now, x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})
= 8(4 + 5) + 3(5 – 1) + 2(1 + 4)
= 8 – 18 + 10
= 0
Hence, the given points are collinear.
14. Find the value of x for which points A(x, 2), B(3, 4) and C(7, 5) are collinear.
Solution
Let A(x_{1}, y_{1}) = A(x, 2), B(x_{2}, y_{2}) = B(3, 4) and C(x_{3}, y_{3}) = C(7, 5). So the condition for three collinear points is
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1 }– y_{2}) = 0
⇒ x(4 + 5) – 3(5 – 2) + 7(2 + 4) = 0
⇒ x + 21 + 42 = 0
⇒ x =  63
Hence, x = 63.
15. For what value of x are the points A(3, 12), B(7, 6) and C(x, 9) are collinear.
Solution
A(3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x_{1 }= 3, y_{1} = 12), (x_{2 }= 7, y_{2} = 6) and (x_{3} = x, y_{3} = 9)
It is given that points A, B and C are collinear. Therefore,
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2}) = 0
⇒ (3)(6 – 9) + 7(9 – 12) + x(12 – 6) = 0
⇒ (3)(3) + 7(3) + x(6) = 0
⇒ 9 – 21 + 6x = 0
⇒ 6x – 12 = 0
⇒ 6x = 12
⇒ x = 12/6 = 2
Therefore, when x = 2, the given points are collinear
16. For what value of y, are the points P(1, 4), Q(3, y) and R(3, 16) are collinear ?
Solution
P(1, 4), Q(3, y) and R(3, 16) are the given points. Then:
(x_{1} = 1, y_{1} = 4), (x_{2} = 3, y_{2} = y) and (x_{3} = 3, y_{3} = 16)
It is given that the points P, Q and R are collinear.
Therefore,
x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ 1(y – 16) + 3(16 – 4) + (3)(4 – y) = 0
⇒ 1(y – 16) + 3(16 – 4) + (3)(4 – y) = 0
⇒ 1(y – 16) + 3(12) – 3(4 – y) = 0
⇒ y – 16 + 36 – 12 + 3y = 0
⇒ 8 + 4y = 0
⇒ 4y =  8
⇒ y = (8/4) = 2
When y = 2, the given points are collinear.
17. Find the value of y for which the points A(3, 9),B(2, y) and C(4, 5) are collinear.
Solution
Let A(x_{1} = 3, y_{1} = 9), B(x_{2 }= 2, y_{2} = y) and C(x3 = k, y_{3} = 5) be the given points
The given points are collinear if
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ (3)(y + 5) + 2(5 – 9) + 4(9 – y)= 0
⇒ 3y – 15 – 28 + 36 – 4y = 0
⇒ 7y = 36 – 43
⇒ y = 1
18. For what values of k are the points A(8, 1), B(3, 2k) and C(k, 5) collinear.
Solution
Let A(x_{1} = 8, y_{1} = 1), B(x_{2} = 3, y_{2 }= 2k) and C(x_{3 }= k, y_{3} = 5) be the given points
The given points are collinear if
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ 8(2k + 5) + 3( 5 – 1) + k(1 + 2k) = 0
⇒  16k + 40 – 18 + k + 2k^{2} = 0
⇒ 2k^{2 }– 15k + 22 = 0
⇒ 2k^{2} – 11k – 4k + 22 = 0
⇒k(2k – 11) – 2(2k – 11) = 0
⇒ (k – 2)(2k – 11) = 0
⇒ k = 2 or k = 11/22
Hence, k = 2 or k = 11/22
19. Find a relation between x and y, if the points A(2, 1), B(x, y) and C(7, 5) are collinear.
Solution
Let A(x_{1} = 2, y_{1} = 1), B(x_{2} = x, y_{2} = y) and C(x_{3} = 7, y_{3} = 5) be the given points
The given points are collinear if
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ 2(y – 5) + x(5 – 1) + 7(1 – y) = 0
⇒ 2y – 10 + 4x + 7 – 7y = 0
⇒ 4x – 5y – 3 = 0
Hence, the required relation is 4x – 5y – 3 = 0
20. Find a relation between x and y, if the points A(x, y), B(5, 7) and C(4, 5) are collinear.
Solution
Let A(x_{1} = x, y_{1} = y), B(x_{2} =  5, y_{2} = 7) and C(x_{3} = 4, y_{3} = 5) be the given points
The given points are collinear if
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ x(7 – 5) + (5)(5 – y) + (4)(y – 7) = 0
⇒ 7x – 5x – 25 + 5y – 4y + 28 = 0
⇒ 2y – 10 + 4x + 7 – 7y = 0
⇒ 2x + y + 3 = 0
Hence, the required relation is 2x + y + 3 = 0
21. Prove that the points A(a, 0), B(0, b) and C(1, 1) are collinear, if (1/a + 1/b) = 1.
Solution
Consider the points A(a, 0), B(0, b) and (x_{3} = 1, y_{3 }= 1)
Here, (x1 = a, y1 = 0), (x2 = 0, y2 = b) and (x3 = 1, y3 = 1).
It is given that the points are collinear. So,
x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒a(b – 1) + 0(1 – 0) + 1(0 – b) = 0
⇒ ab – a – b = 0
Dividing the equation by ab:
⇒ 1 – 1/b – 1/a = 0
⇒ 1 – (1/a + 1/b) = 0
⇒ (1/a + 1/b) = 1
Therefore, the given points are collinear if (1/a + 1/b) = 1.
22. If the points P(3, 9), Q(a, b) and R(4, 5) are collinear and a + b = 1, find the value of a and b.
Solution
Let A(x_{1} = 3, y_{1} = 9), B(x_{2} = a, y_{3} = b) and C(x_{3} = 4, y_{1} = 5) be the given points.
The given points are collinear if
x_{1}(y_{2} – y_{3}) + x_{3}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2}) = 0
⇒ (3)(b + 5) + a(5 – 9) + 4(9 – b) = 0
⇒ 3b – 15 – 14a + 36 – 4b = 0
⇒ 2a + b = 3
Now, solving a + b = 1 and 2a + b = 3, we get a = 2 and b = 1
Hence, a = 2 and b = 1
23. Find the area of △ABC with vertices A(0, 1), B(2, 1) and C(0, 3). Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is 4: 1.
Solution
Let A(x_{1} = 0, y_{1} = 1), B(x_{2}, y_{2} = 1) and C(x_{3} = 0, y_{3} = 3) be the given points. Then
Area (△ABC) = ½[x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1 }– y_{2})]
= ½[0(1 – 3) + 2(3 + 1) + 0( 1 – 1)]
= 1/2 × 8
= 4 sq. units
So, the area of the triangle △ABC is 4sq. units.
Let D(a_{1}, b_{1}), E(a_{2}, b_{2}) and F(a_{3}, b_{3}) be the midpoints of AB, BC and AC respectively
Then
a_{1} = (0 + 2)/2 = 1, b_{1} = (1 + 1)/2 = 0
a_{2} = (2 + 0)/2 = 1, b_{2} = (1 + 3)/2 = 2
a_{3} = (0 + 0)/2 = 0, b_{3 }= (1 + 3)/2 = 1
Thus, the coordinates of D, E and F are D(a_{1} = 1, b_{1} = 0), E(a_{2} = 1, b_{2} = 2) and
F(a_{3 }= 0, b_{3} = 1). Now
Area (△DEF) = ½[a1(b_{2} – b_{2}) + a_{2}(b_{3} – b_{1}) + a_{3}(b_{1} – b_{2})]
= ½[1(2 – 1) + 1(1 – 0) + 0(0 – 2)]
= ½[1 + 1 + 0]
= 1 sq. unit
So, the area of triangle △DEF is 1 sq. unit.
Hence, △ABC : △DEF = 4 :1