NCERT Exemplar Solutions Class 10 Mathematics Solutions for Coordinated Geometry - Exercise 7.4 in Chapter 7 - Coordinated Geometry
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A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of △ ADE.
According to the question,
The three vertices of a parallelogram ABCD are A (6, 1), B (8, 2) and C (9, 4)
Let the fourth vertex of parallelogram = (x, y),
We know that, diagonals of a parallelogram bisect each other
Since, mid – point of a line segment joining the points (x1, y1) and (x2, y2) is given by,
\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)\\ \text { Mid - point of } \mathrm{BD}=\text { Mid - point of } \mathrm{AC}\\ \left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{6+9}{2}, \frac{1+4}{2}\right)\\ \left(\frac{8+x}{2}, \frac{2+y}{2}\right)=\left(\frac{15}{2}, \frac{5}{2}\right)
So we have,
\begin{aligned} &\frac{8+x}{2}=\frac{15}{2}\\ &8+x=15\\ &\mathrm{x}=7\\ &\text { And }\\ &\frac{2+y}{2}=\frac{5}{2}\\ &2+y=5 \rightarrow y=3 \end{aligned}
So, fourth vertex of parallelogram is D(7, 3)
Now,
Midpoint of side
D C=\left(\frac{7+9}{2}, \frac{3+4}{2}\right) \\ E=\left(8, \frac{7}{2}\right)
\text{∵ Area of ΔABC with vertices } (x_1, y_1), (x_2, y_2)\text{ and }(x_3, y_3);\\ = \frac12[x_1(y_2 – y_3) + x2(y_3 – y_1) + x_3(y_1 – y_2)]
∴ Area of ΔADE with vertices A (6, 1), D (7, 3) and E (8, (7/2))
\Delta =\frac{1}{2}\left[6\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-1\right)+8(1-3)\right] \\ =\frac{1}{2}\left[6 \times\left(\frac{-1}{2}\right)+7\left(\frac{5}{2}\right)+8(-2)\right] \\ =\frac{1}{2}\left(\frac{35}{2}-19\right) \\ =\frac{1}{2}\left(\frac{-3}{2}\right)\\ =-\frac{3}{4}\\ \text{But area can’t be negative.}\\ \text{Hence, the required area of ΔADE is }\frac{3}{4}\text{ sq. units.}
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