Basic Concepts of Points and their Coordinates Class 11 OP Malhotra Exe-15C ISC Maths Solutions Ch-15 Solutions. In this article you would learn about Area of Triangle and Quadrilateral. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.
Basic Concepts of Points and their Coordinates Class 11 OP Malhotra Exe-15C ISC Maths Solutions Ch-15
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-15 | Basic Concepts of Points and their Coordinates |
| Writer | O.P. Malhotra |
| Exe-15(C) | Area of Triangle and Quadrilateral |
Area of Triangle and Quadrilateral
Basic Concepts of Points and their Coordinates Class 11 OP Malhotra Exe-15C Solution.
Que-1: Find the area of the triangle whose vertices are
(i) (4, 2),(4, 5) and (-2, 2);
(ii) (0, 0),(-2, 3) and (10, 7);
(iii) (a, 0),(0, b) and (x, y).
Sol: (i) Required area = (1/2) | (20 – 8) + (8 + 10) + (- 4 – 8)|
= (1/2) | 12 + 18 – 12| = 9 sq. units
(ii) Required area = 1/2 | (0 – 0) + (- 14 – 30) + (0 + 0)|
= 22 sq. units
(iii) Required area = 1/2 | (ab – 0) + (0 + bx) + (0 – ay)|
= 1/2 | ab – bx – ay|
= 1/2 (bx + ay – ab) sq. units
Que-2: Find the area of the quadrilateral whose vertices are
(i) (1, 1),(7,- 3),(12, 2) and (7, 21);
(ii) (1, 1),(3, 4),(5, – 2) and (4, – 7).
Sol: (i) Required area = 1/2 | (- 3 – 7) + (14 + 36) + (252 – 14) + (7 – 21)|
= 1/2 | – 10 + 50 + 238 – 14 | = 132 sq. units
(ii) Required area = 1/2 | (4 – 3) + (- 6 – 20) + (- 35 + 8) + (4 + 7) |
= 1/2 | 1 – 26 – 27 + 11 |
= 41/2 sq. units
Que-3: Show that the following points are collinear :
(i) (1, 4),(3, – 2) and (- 3, 16);
(ii) (- 5, 1), (5, 4) and (10, 7).
Sol: (i) Given vertices of △ABC are A(1, 4); B(3, – 2) and C(- 3, 16)
∴ area of △ABC = 1/2 | (- 2 – 12) + (48 – 6) + (- 12 – 16)|
= 1/2 | – 14 + 42 – 28 | = 0 sq. units
Hence the points A, B and C are lies on same straight line.
∴ given points A, B and C are collinear.
(ii) Let the given points are A(- 5, 1); B(5, 4) and C(10, 7).
Here area of △ABC = 1/2 |(- 20 – 5) + (35 – 40) + (10 + 35) |
= 1/2 | – 25 – 5 + 45 | = 152 ≠ o
Thus given points are not collinear.
Que-4: If (7, a),(- 5, 2) and (3, 6) are collinear, find a.
Sol: Since given points are A(7, a), B(- 5, 2) and C(3, 6) are collinear.
So area of △ABC having vertices A, B and C be equal to 0.
area of △ABC =0
⇒ 1/2 | (14 + 5a) + (- 30 – 6) + (3a – 42) | = 0
⇒ | 14 + 5a – 36 + 3a – 42 | = 0
⇒ | 8a – 64 | = 0
⇒ 8a = 64
⇒ a = 8
Que-5: If the area of the quadrilateral whose angular points A, B, C, D taken in order are (1, 2), (- 5, 6), (7, – 4) and (-2, k) be zero, find the value of k.
Sol: area of quadrilateral ABCD = 1/2 | (6 + 10) + (20 – 42) + (7k – 8) + (- 4 – k) |
= 1/2 | 16 – 22 + 7k – 8 – 4 – k | = 1/2 | – 18 + 6k |
given area of quadrilateral ABCD = 0 ⇒ 1/2 | – 18 + 6k | = 0
⇒ 6k = 18
⇒ k = + 3
Que-6: A, B, C are the points (-1, 5), (3, 1) and (5, 7) respectively. D, E, F, are the mid-points of BC, CA, AB respectively. Prove that △ABC = 4 ΔDEF.
Sol: Since D be the mid-point of BC.
∴ Coordinates of D are [{(3+5)/2}, {(1+7)/2}] i.e. D(4, 4)
Also E be the mid-point of AC.
∴ Coordinates of E are [{(−1+5)/2}, {(7+5)/2}] i.e. (2, 6)
Further F be the mid-point of AB.
Thus, coordinates of F are [{(3−1)/2}, {(1+5)/2}] i.e. (1, 3)
∴ area of △ABC = 1/2 | (- 1 – 15) + (21 – 5) + (25 + 7) |
= 1/2 | 16 + 0 – 8 |
= 4 sq. units
∴ area of △ABC = 4 area of △DEF
Que-7: The straight lines y = m1x + c1, y = m2x + c2, and x = 0 intersect in the three points P, Q, and R. Find the area of the triangle PQR. What is the value of the area if c1 = c1?
Sol: Given eqns. of straight lines are ;
y = m1x + c1 …(1)
y = m2x + c2 …(2)
x = 0 …(3)
Let eqn. (1) and eqn. (2) intersect at point P
–: End of Basic Concepts of Points and their Coordinates Class 11 OP Malhotra Exe-15C ISC Maths Ch-15 Solutions. :–
Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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