Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Exercise 9.11. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.We connect diagonal AC such thatarea (ABCD) = (ΔABC) + (ΔADC)Now,(ΔABC)=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=12|−4[7−(−5)]+0[−5−5]+5[5−7]| unit2=12|−4(7+5)+0+5(5−7)| unit2=12|(−4)×12+5×(−2)|unit 2=12|−48−10|unit 2=12|−58| unit2=582unit 2=29 unit2Similarly, (ΔACD)=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)|=12|−4[−5−(2)]+5[−2−5]+(−4)[5−(−5)]| unit2=12|−4(−5+2)+5(−2−5)−4(5+5)|unit 2=12|(−4)×(−3)+5×(−7)−4×(10)|unit 2=12|12−35−40|unit 2=12|−63| unit2=632 unit2Hence, area (ABCD) = 29+632 unit2=58+632 unit2=1212 unit2

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Ncert Solutions Maths class 11th Maths Class 11th Straight Lines

Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

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alok kumar singh | Contributor-Level 10

4 months ago

Exercise 9.1

1. Let the given points be A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2).

Then quadrilated ABCD can be plotted on the graph by joining the points A, B, C and D.

We connect diagonal AC such that

area (ABCD) = (ΔABC) + (ΔADC)

Now,

$(ΔABC) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [7 - (- 5)] + 0 [- 5 - 5] + 5 [5 - 7] | unit^{2}$

$= \frac{1}{2} | - 4 (7 + 5) + 0 + 5 (5 - 7) | unit^{2}$

$= \frac{1}{2} | (- 4) \times 12 + 5 \times (- 2) |unit^{2}$

$= \frac{1}{2} | - 48 - 10 |unit^{2}$

$= \frac{1}{2} | - 58 | unit^{2}$

$= \frac{58}{2} unit^{2} = 29 unit^{2}$

Similarly, $(ΔACD) = \frac{1}{2} | x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) |$

$= \frac{1}{2} | - 4 [- 5 - (2)] + 5 [- 2 - 5] + (- 4) [5 - (- 5)] | unit^{2}$

$= \frac{1}{2} | - 4 (- 5 + 2) + 5 (- 2 - 5) - 4 (5 + 5) |unit^{2}$

$= \frac{1}{2} | (- 4) \times (- 3) + 5 \times (- 7) - 4 \times (10) |unit^{2}$

$= \frac{1}{2} | 12 - 35 - 40 |unit^{2}$

$= \frac{1}{2} | - 63 | unit^{2}$

$= \frac{63}{2} unit^{2}$

Hence, area (ABCD) = $29 + \frac{63}{2} unit^{2}$

$= \frac{58 + 63}{2} unit^{2} = \frac{121}{2} unit^{2}$

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Q1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

1 Answer

Exercise 9.1

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