Quantitative Aptitude Ques 2243

Question: If $ A,(-,5,7), $ $ B,(-,4,-5), $ $ C,(-1,-,6) $ and $ D,(4,5) $ are the vertices of a quadrilateral, then find the area of the quadrilateral ABCD.

Options:

A) 72 sq units

B) 80 sq units

C) 90 sq units

D) 92 sq units

Show Answer

Answer:

Correct Answer: A

Solution:

  • By joining B and D, we get two triangles $ \Delta ABD $ and $ \Delta BCD. $ Given, $ x _1=-5, $ $ =x _2=-4, $ $ x _3=4, $ $ y _1=7 $ $ y _2=-5 $ and $ y _3=5 $ Area of $ \Delta ABD=| \frac{1}{2}[x _1(y _2-y _3)+x _2(y _3-y _1)+x _3(y _1-y _2)], | $ $ =\frac{1}{2}[-5(-,5-5)+(-4)(5-7)+4(7+5)] $ $ =\frac{1}{2}[50+8+48]=\frac{106}{2}=53squnits $ Now, in $ \Delta BCD $ $ x _1=-4, $ $ x _2=-1, $ $ x _3=4, $ $ y _1=-5, $ $ y _2=-6 $ and $ y _3=5 $ Area of $ \Delta BCD=| \frac{1}{2}[-4(-6-5)-1(5+5)+4(-5+6)]| . $ $ =\frac{1}{2}(44-10+4) $ $ =\frac{1}{2}\times 38=19 $ sq units

$ \therefore $ Area of quadrilateral $ \Delta BCD $ = Area of $ \Delta ABD $ + Area of $ \Delta BCD $ $ =53+19=72 $ sq units

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