NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3

Question 16

Step 1: Since \(D, E, F\) are the midpoints of the sides of \(\triangle ABC\), the triangle \(\triangle DEF\) is the medial triangle.

A known result: \(\text{Area}(\triangle DEF) = \dfrac{1}{4}\,\text{Area}(\triangle ABC)\).

So, \(\text{Area}(\triangle ABC) = 4\,\text{Area}(\triangle DEF)\).

Step 2: Find \(\text{Area}(\triangle DEF)\) using the coordinate area formula:

\(\text{Area} = \dfrac{1}{2}\left|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)\right|\).

Step 3: Take

• \(D(x_1,y_1)=\left(-\tfrac{1}{2},\tfrac{5}{2}\right)\)
• \(E(x_2,y_2)=(7,3)\)
• \(F(x_3,y_3)=\left(\tfrac{7}{2},\tfrac{7}{2}\right)\)

Step 4: Compute the differences:

• \(y_2-y_3 = 3 - \tfrac{7}{2} = -\tfrac{1}{2}\)
• \(y_3-y_1 = \tfrac{7}{2} - \tfrac{5}{2} = 1\)
• \(y_1-y_2 = \tfrac{5}{2} - 3 = -\tfrac{1}{2}\)

Step 5: Substitute in the formula:

\(\text{Area}(DEF) = \dfrac{1}{2}\left|\left(-\tfrac{1}{2}\right)\left(-\tfrac{1}{2}\right) + 7(1) + \left(\tfrac{7}{2}\right)\left(-\tfrac{1}{2}\right)\right|\)

Now simplify:

• \(\left(-\tfrac{1}{2}\right)\left(-\tfrac{1}{2}\right)=\tfrac{1}{4}\)
• \(7(1)=7\)
• \(\left(\tfrac{7}{2}\right)\left(-\tfrac{1}{2}\right)=-\tfrac{7}{4}\)

So inside absolute value:

\(\tfrac{1}{4} + 7 - \tfrac{7}{4} = 7 - \tfrac{6}{4} = 7 - \tfrac{3}{2} = \tfrac{11}{2}\)

Therefore,

\(\text{Area}(DEF) = \dfrac{1}{2}\times\left|\tfrac{11}{2}\right| = \dfrac{11}{4}\)

Step 6: Use the medial triangle area relation:

\(\text{Area}(ABC) = 4\times\dfrac{11}{4} = 11\)

Final Answer: \(\text{Area}(\triangle ABC)=11\) square units.