Question

The points (-6, 1), (6, 10), (9, 6) and (-3, -3) are the vertices of a rectangle. If the area of the portion of this rectangle that lies above the x axis is a/b, find the value of (a + b), given a and b are coprime.​

Answer 1

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Answer 2

The value of (a+b) is 139.

Given:

Vertices of rectangle are (-6, 1), (6, 10), (9, 6) and (-3, -3).

Area of the rectangle which lies above x-axis= a/b

To Find:

Value of (a+b).

Solution:

Let ABCD be the rectangle with vertices A(-6, 1), B(6, 10), C(9, 6) and D(-3, -3).

Let the rectangle intersect the x-axis at points E and F.

Equation of line AD is ,

$y+3=\frac{4}{-3} (x+3)$

Now if it cuts x−axis then y = 0

or, 4x+12= −9

or, x = −21/4= −5.25

So coordinate of F = (−5.25,0) or (-5,0) approx.

Similarly equation of line CD is ,

$y+3=\frac{9}{12} (x+3)$

Now if it cuts the x−axis, then y =0

or, 12= 3x+9

or, x = 1

So the coordinate of E is (1,0)

$DF= \sqrt{(-5-(-3))^{2} +(0+3)^{2} } =\sqrt{(-2)^{2} +(3)^{2} } = \sqrt{13} units$

$DE= \sqrt{(-3+1)^{2} +(-3-0)^{2} } =\sqrt{4+9 } = \sqrt{13} units$

Now the area of triangle DEF using heron's formula,

$\frac{1}{2} X \sqrt{13} X\sqrt{13} =\frac{13}{2} unit^{2}$

Now, the Area of the rectangle ABCD= AB x AD

                                        $AB= \sqrt{(-6-6)^{2} +(1-10)^{2} } =\sqrt{(-12)^{2} +(-9)^{2} } = 15 units$

$AD=\sqrt{(-6+3)^{2}+(1+3)^{2} } =\sqrt{(-3)^{2} +(4)^{2} } = 5 units$

So, the area of ABCD= 15 x 5= 75 unit².

Area of rectangle above x-axis= area of ABCD- area of DEF

                                                   = 75- (13/2) unit²= (137/2) unit²

Therefore,

$\frac{a}{b} =\frac{137}{2}$

So, a+b= 137+2= 139

Hence, the value of a+b is 139.

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