NCERT Solution For Class 10, Maths, Chapter 7 Coordinate Geometry, Exercise 7.3

NCERT Solution For Class 10, Maths, Chapter 7 Coordinate Geometry, Exercise 7.3

NCERT Solution For Class 10, Maths, Chapter 7 Coordinate Geometry, Exercise 7.3 includes study of area of triangles when its base and corresponding height (altitude) are given. We have to use the formula :

if the coordinates of the vertices of a triangle are given, we can find the area of given triangle by using above formula. Class 10, maths, Exercise 7.3 has five questions to discuss which may be easy then Heron’s formula which is used to find the area of a triangle.

Class 10, Maths, Chapter 7, Exercise 7.3 Solutions

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

(i) (2, 3), (–1, 0), (2, – 4)

(ii) (–5, –1), (3, –5), (5, 2)

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Q.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)

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Q.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.

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

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Q.5. 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 Δ ABC whose vertices are A (4, – 6), B (3, –2) and C (5, 2).

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Class 10 , Maths, Chapter 7 Coordinate Geometry (All exercise)