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Question Number 4387 by Rasheed Soomro last updated on 17/Jan/16

Commented by Rasheed Soomro last updated on 17/Jan/16

$In the trapezium m ∠ A = m ∠ B = \frac{π}{2} rad .$ $m \overset{―}{AB} = m \overset{―}{AD} = x units and m \overset{―}{BC} = 2 x units .$ $The trapezium has been divided into two$ $sub - trapeziums of equal area .$ $Determine a and b, the heights of sub -$ $trapeziums .$
Commented by Yozzii last updated on 22/Jan/16
$Total figure area ,A_t =(1/2)(AD+BC)(AB) A_t =(1/2)(x+2x)(x)=((3x^2 )/2) Here, x=a+b=constant>0 Let A_1 =area of upper trapezium and A_2 =area of the lower trapezium. If q>0 is the length of the line dividing the two inner trapezia, A_1 =0.5(x+q)a and A_2 =0.5(2x+q)b. ∵ A_1 =A_2 ⇒b(2x+q)=(x+q)a q(b−a)=x(a−2b) q=((x(a−2b))/(b−a)) (a≠b) Since A_t =A_1 +A_2 ⇒((3x^2 )/2)=0.5({x+q}a+{2x+q}b) 3x^2 =ax(1+((a−2b)/(b−a)))+bx(2+((a−2b)/(b−a))) 3x=((a(b−a+a−2b))/(b−a))+((b(2b−2a+a−2b))/(b−a)) 3x=((−ab)/(b−a))+((−ab)/(b−a)) 3x=((2ab)/(a−b)) 3x(a−b)=2ab. ∵ a+b=x⇒a=x−b. ∴ 3x(x−2b)=2(x−b)b 3x^2 −6xb=2xb−2b^2 2b^2 −8xb+3x^2 =0 ∴b=((8x±(√(64x^2 −4×2×3x^2 )))/4) b=((8±2(√(10)))/4)x b=(((4±(√(10)))/2))x If b=((4+(√(10)))/2)x ⇒a=x(1−((4+(√(10)))/2))<0. But, a>0. ∴ b≠((4+(√(10)))/2)x. If b=((4−(√(10)))/2)x⇒a=x(1−((4−(√(10)))/2)) a=x(((2−4+(√(10)))/2))=x((((√(10))−2)/2)) ∴ (a,b)=(x((((√(10))−2)/2)),x(((4−(√(10)))/2)))$
$T o t a l f i g u r e a r e a, A_{t} = \frac{1}{2} (A D + B C) (A B)$ $A_{t} = \frac{1}{2} (x + 2 x) (x) = \frac{3 x^{2}}{2}$ $H e r e, x = a + b = c o n s t a n t > 0$ $L e t A_{1} = a r e a o f u p p e r t r a p e z i u m a n d$ $A_{2} = a r e a o f t h e l o w e r t r a p e z i u m .$ $I f q > 0 i s t h e l e n g t h o f t h e l i n e$ $d i v i d i n g t h e t w o i n n e r t r a p e z i a,$ $A_{1} = 0.5 (x + q) a$ $a n d A_{2} = 0.5 (2 x + q) b .$ $∵ A_{1} = A_{2} \Rightarrow b (2 x + q) = (x + q) a$ $q (b - a) = x (a - 2 b)$ $q = \frac{x (a - 2 b)}{b - a} (a \neq b)$ $S i n c e A_{t} = A_{1} + A_{2}$ $\Rightarrow \frac{3 x^{2}}{2} = 0.5 ({x + q} a + {2 x + q} b)$ $3 x^{2} = a x (1 + \frac{a - 2 b}{b - a}) + b x (2 + \frac{a - 2 b}{b - a})$ $3 x = \frac{a (b - a + a - 2 b)}{b - a} + \frac{b (2 b - 2 a + a - 2 b)}{b - a}$ $3 x = \frac{- a b}{b - a} + \frac{- a b}{b - a}$ $3 x = \frac{2 a b}{a - b}$ $3 x (a - b) = 2 a b .$ $∵ a + b = x \Rightarrow a = x - b .$ $∴ 3 x (x - 2 b) = 2 (x - b) b$ $3 x^{2} - 6 x b = 2 x b - 2 b^{2}$ $2 b^{2} - 8 x b + 3 x^{2} = 0$ $∴ b = \frac{8 x \pm \sqrt{64 x^{2} - 4 \times 2 \times 3 x^{2}}}{4}$ $b = \frac{8 \pm 2 \sqrt{10}}{4} x$ $b = (\frac{4 \pm \sqrt{10}}{2}) x$ $I f b = \frac{4 + \sqrt{10}}{2} x \Rightarrow a = x (1 - \frac{4 + \sqrt{10}}{2}) < 0 .$ $B u t, a > 0 . ∴ b \neq \frac{4 + \sqrt{10}}{2} x .$ $I f b = \frac{4 - \sqrt{10}}{2} x \Rightarrow a = x (1 - \frac{4 - \sqrt{10}}{2})$ $a = x (\frac{2 - 4 + \sqrt{10}}{2}) = x (\frac{\sqrt{10} - 2}{2})$ $∴ (a, b) = (x (\frac{\sqrt{10} - 2}{2}), x (\frac{4 - \sqrt{10}}{2}))$
Commented by Rasheed Soomro last updated on 22/Jan/16

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