ABCD is rectangle and lines DX- DY and XY are drawn as shown- Area of -AXD is 5- Area of - BXY is 4 and area of - CYD is 3- If the area of -DXY can be expressed as -x where x - N then x is equal to -

The correct option is C 84Area of Δ AXD = 1/2 AD . AX = 5 ; #160; AX = 10/a XB = b 10/a Area of Δ CYD = 1/2 b #160; YC =3; #160; YC = 6/ ...

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Byju's Answer

Standard VI

Mathematics

Area of Square

ABCD is recta...

Question

ABCD is rectangle and lines DX, DY and XY are drawn as shown. Area of $Δ$ AXD is 5, Area of $Δ$ BXY is 4 and area of $Δ$ CYD is 3. If the area of $Δ$ DXY can be expressed as $\sqrt{x}$ where x $\in$ N then x is equal to -

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Solution

The correct option is C 84
Area of $Δ A X D = \frac{1}{2} A D . A X = 5; A X = \frac{10}{a}$
$X B = b - \frac{10}{a}$
Area of $Δ C Y D = \frac{1}{2} b Y C = 3; Y C = \frac{6}{b}; B Y = a - \frac{6}{b}$
Area of $Δ B X Y = \frac{1}{2} (b - \frac{10}{a}) (a - \frac{6}{b}) = 4$
$a^{2} b^{2} - 24 a b + 60 = 0$
Solving quadratic equation
$a b = 12 + \sqrt{84}$
Now, Area of $Δ D X Y =$ Area of Rectangle $A B C D - (5 + 4 + 3) = \sqrt{84} = a b - (12) = \sqrt{84}$ .
Thus, $x = 84$ .

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ABCD is rectangle and lines DX, DY and XY are drawn as shown. Area of ΔAXD is 5, Area of Δ BXY is 4 and area of Δ CYD is 3. If the area of ΔDXY can be expressed as √x where x ∈ N then x is equal to -

The correct option is C 84Area of ΔAXD=12AD.AX=5;AX=10aXB=b−10aArea of ΔCYD=12bYC=3;YC=6b;BY=a−6bArea of ΔBXY=12(b−10a)(a−6b)=4a2b2−24ab+60=0Solving quadratic equationab=12+√84Now, Area of ΔDXY= Area of Rectangle ABCD−(5+4+3)=√84=ab−(12)=√84 .Thus, x=84.

ABCD is rectangle and lines DX, DY and XY are drawn as shown. Area of $Δ$ AXD is 5, Area of $Δ$ BXY is 4 and area of $Δ$ CYD is 3. If the area of $Δ$ DXY can be expressed as $\sqrt{x}$ where x $\in$ N then x is equal to -