Solve the followings Questions.

1. Find the area of the triangle whose vertices are:

(i) (2, 3), (–1, 0), (2, –4)

(ii) (–5, –1), (3, –5), (5, 2)

Answer:

(i) (2, 3), (–1, 0), (2, –4)

Area of Triangle is given by

Area of Triangle =

Area of  the given triangle = ½[2 {0 − (−4)} – 1 (−4 − 3) + 2 (3 − 0)]

=½ (8 + 7 + 6) = 21/2 sq. units

(ii) (–5, –1), (3, –5), (5, 2)

Area of the given triangle = ½ [−5 (−5 − 2) + 3 {2 − (−1)} + 5 {−1 − (−5)}]

= ½(35 + 9 + 20)

= 32 sq. units

2. 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)

Answer:

(i) (7, –2), (5, 1), (3, k)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Therefore, for points (7, -2) (5, 1), and (3,k), area = 0

⇒ ½[7 (1 − k) + 5 {k − (−2)} + 3 (−2 − 1)]

= ½(7 − 7k + 5k + 10 − 9) = 0

⇒ ½(7 − 7k + 5k + 1) = 0

⇒ ½(8 − 2k) = 0

⇒ 8 − 2k = 0

⇒ 2k = 8

⇒ k = 4

(ii) (8, 1), (k, –4), (2, –5)

Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.

Therefore, for points (8, 1) (k, -4), and (2,-5), area = 0

⇒ ½[8 {−4 − (−5)} + k (−5 − 1) + 2 {1 − (−4)}]

= ½(8 − 6k + 10) = 0

⇒ ½(18 − 6k) = 0

⇒18 − 6k = 0

⇒ 18 = 6k

⇒ k = 3

3. 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.

Answer:

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

D = (0+2/2 , -1+1/2) = (1,0)

E = (0+0/2 , -3-1/2) = (0,1)

Area of a triangle = 1/2 {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)}

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)}

                        = 1/2 (1+1) = 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)]

                         = 1/2 {8} = 4 square units

Therefore, the required ratio is 1:4.

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

Answer:

Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)}

Area of ΔABC =1/2[−4 (−5 − 2) – 3 {-2 − (−2)} + 3 {−2 − (−5)}]

=1/2 [12 + 0 + 9]

= 21/2 sq. units

Again using formula to find area of triangle:

Area of △ACD = [−4 (−2 − 3) + 3 {3 − (−2)} + 2 {−2 − (-2)}]

= 1/2 [20 + 15 + 0]

= 35/2 sq. units

Area of ☐ABCD = Area of ΔABC + Area of ΔACD

                           = 21/2 + 35/2 = 28 sq. units

5. We know that median of a triangle divides it into two triangles of equal areas. Verify this result for △ABC whose vertices are A (4, –6), B (3, –2) and C (5, 2).

Answer:

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2).

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = (3+5/2, -2+2/2) = (4,0)

Area of a triangle = 1/2 {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)}

Area of △ABD = 1/2[4 (−2 − 0) + 3 {0 − (−6)} + 4 {−6 − (−2)}]

=1/2(−8 + 18 −16)

=1/2 (−6) = −3 sq units

Area cannot be in negative.

Therefore, we just consider its numerical value.

Therefore, area of △ABD = 3 sq units

Again using formula to find area of triangle:

Area of △ABD = 1/2 [4 (0 − 2) + 4{2 − (−6)} + 5 {−6 −0 )}]

 = 1/2 (- 8 + 32 −30) = ½ (-6) = -3 sq units.

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas

Hence Proved.

CHAPTER NAME	OLD NCERT	NEW NCERT
Real Numbers	EXERCISE 1.1
	EXERCISE 1.2	1.1	CLICK HERE
	EXERCISE 1.3	1.2	CLICK HERE
	EXERCISE 1.4
Polynomials	EXERCISE 2.1	2.1	CLICK HERE
	EXERCISE 2.2	2.2	CLICK HERE
	EXERCISE 2.3
	EXERCISE 2.4
Pair of Linear Equations in Two Variables	EXERCISE 3.1
	EXERCISE 3.2	3.1	CLICK HERE
	EXERCISE 3.3	3.2	CLICK HERE
	EXERCISE 3.4	3.3	CLICK HERE
	EXERCISE 3.5
	EXERCISE 3.6
	EXERCISE 3.7
Quadratic Equations	EXERCISE 4.1	4.1	CLICK HERE
	EXERCISE 4.2	4.2	CLICK HERE
	EXERCISE 4.3
	EXERCISE 4.4	4.3	CLICK HERE
Arithmetic Progressions	EXERCISE 5.1	5.1	CLICK HERE
	EXERCISE 5.2	5.2	CLICK HERE
	EXERCISE 5.3	5.3	CLICK HERE
	EXERCISE 5.4	5.4 (Optional)	CLICK HERE
Triangles	EXERCISE 6.1	6.1	CLICK HERE
	EXERCISE 6.2	6.2	CLICK HERE
	EXERCISE 6.3	6.3	CLICK HERE
	EXERCISE 6.4
	EXERCISE 6.5
	EXERCISE 6.6
Coordinate Geometry	EXERCISE 7.1	7.1	CLICK HERE
	EXERCISE 7.2	7.2	CLICK HERE
	EXERCISE 7.3
	EXERCISE 7.4
Introduction to Trigonometry	EXERCISE 8.1	8.1	CLICK HERE
	EXERCISE 8.2	8.2	CLICK HERE
	EXERCISE 8.3
	EXERCISE 8.4	8.3	CLICK HERE
Some Applications of Trigonometry	EXERCISE 9.1	9.1	CLICK HERE
Circles	EXERCISE 10.1	10.1	CLICK HERE
	EXERCISE 10.2	10.2	CLICK HERE
Construction
Areas Related to Circles	EXERCISE 12.1
	EXERCISE 12.2	11.1	CLICK HERE
	EXERCISE 12.3
Surface Areas and Volumes	EXERCISE 13.1	12.1	CLICK HERE
	EXERCISE 13.2	12.2	CLICK HERE
	EXERCISE 13.3
	EXERCISE 13.4
	EXERCISE 13.5
Statistics	EXERCISE 14.1	13.1	CLICK HERE
	EXERCISE 14.2	13.2	CLICK HERE
	EXERCISE 14.3	13.3	CLICK HERE
	EXERCISE 14.4
Probability	EXERCISE 15.1	14.1	CLICK HERE
	EXERCISE 15.2