Question

Palace of Versailles' rectangular garden has a length of $48 m$ and breadth of $32 m$ . It is divided into $4$ regions as shown below. It is known that $E B = D G = 24 m$ and $A H = F C = 32 m$ . Given the area of region $E B F K = 512 m^{2}$ , what is the area of region $H K G D$ ?

Accepted Answer

Given,

$(1)$ Dimensions of garden: $32 m \times 48 m$ .

$(2)$ Garden is divided into $4$ regions as in the figure below.

$(3)$ $E B = D G = 24 m$

$(4)$ $A H = F C = 32 m$ .

$(5)$ Area of region $E B F K = 512 m^{2}$

We need to find the area of region $H K G D$

Let us look at figure below:

In the figure, the right-angled triangles, $∆ A B F ≅ C D H$ by $S A S$ congruence rule.

$\Rightarrow A F = C H$ by CPCT.

Similarly, $∆ A F H ≅ ∆ C H F$ , by SSS congruence rule.

$a r (∆ A B F) + a r ∆ (A F H) = a r (∆ C D H) + a r ∆ (C H F)$

$\Rightarrow$ HF divides the rectangular garden into two equal areas: $a r (A B F H), a r (C D H F)$

$\Rightarrow a r (A B F H) = a r (C D H F) = \frac{1}{2} a r (□ A B C D) = \frac{1}{2} \times 48 \times 32 = 768 m^{2}$

$a r (A E K H) = a r (A B F H) - a r (E B F K) = 768 - 512 = 256 m^{2}$

Let us now divide the given area into convenient regular shapes that we can formulate for their area.

From the figure, $a r (E B F K) = a r (M B F K) + a r (∆ E M K)$

$M B F K$ is a trapezium, with $M K ∥ B F$ and altitude $B M$

$a r (M B F K) = \frac{1}{2} \times B M \times (B F + M K)$

$a r (∆ E M K) = \frac{1}{2} \times E M \times M K = \frac{1}{2} \times (E B - B M) \times M K$

Let $B M = x, M K = y$ . Given, $E B = 24 m$ , $F C = 32 m, BC = 48 m$

$B F = B C - F C = (48 - 32) m = 16 m$

$a r (E B F K) = \frac{1}{2} \times [x (16 + y) + (24 - x) y] = \frac{1}{2} (16 x + x y + 24 y - x y)$

$= \frac{1}{2} (16 x + 24 y) = 8 x + 12 y$

Given $a r (E B F K) = 512 m^{2}$

$\Rightarrow 8 x + 12 y = 512 . . . . . . . . . . . . . . . . . . . . (1)$

We have $a r (A E K H) = a r (A M K H) - a r (∆ E M K)$

$A M K H$ is a trapezium, with $A H ∥ M K$ and altitude $A M$

We have, $A H = 32 m$ , $M K = y$ , $A M = A B - B M$

$A B = 32 m$ , $\Rightarrow A M = 32 - x$

$a r (A M K H) = \frac{1}{2} \times A M \times (A H + M K)$

$= \frac{1}{2} \times (32 - x) (32 + y)$

$= \frac{1}{2} \times (1024 + 32 y - 32 x - x y)$

$a r (E M K) = \frac{1}{2} \times (24 - x) y$ , we arrived at this expression before.

$a r (A E K H) = \frac{1}{2} [(1024 + 32 y - 32 x - x y) - (24 - x) y]$

$= \frac{1}{2} (1024 + 32 y - 32 x - 24 y)$

$= \frac{1}{2} (1024 - 32 x + 8 y)$

$= 512 - 16 x + 4 y$

But $a r (A E K H) = 256 m^{2}$

$\Rightarrow 512 - 16 x + 4 y = 256$

$\Rightarrow 16 x - 4 y = 256$

$\Rightarrow 8 x - 2 y = 128 . . . . . . . . . . . . . . . . (2)$

We have the equations,

$8 x + 12 y = 512 . . . . . . . . . . . . . . . . . . . . (1)$

$8 x - 2 y = 128 . . . . . . . . . . . . . . . . (2)$

Equation $(1) - (2)$ $\to$ $14 y = 384$

$\Rightarrow y = \frac{384}{14} = \frac{192}{7} = 27 \frac{3}{7} m$

Substituting for $y$ in equation $(2)$ we get,

$8 x - 2 \times \frac{192}{7} = 128$

$\Rightarrow 8 x = 128 + \frac{384}{7} = \frac{896 + 384}{7} = \frac{1280}{7}$

$\Rightarrow x = \frac{160}{7} = 22 \frac{6}{7} m$

$a r (H K G D) = a r (H K N D) + a r (∆ K G N)$

$H K N D$ is a parallelogram, with $H D ∥ K N$ and altitude $D N$

$H D = A D - A H = (48 - 32) m = 16 m$

$K N = M N - M K = M N - y = 48 - \frac{192}{7} = \frac{144}{7} = 20 \frac{4}{7} m$

$D N = A M = A B - B M = A B - x = 32 - \frac{160}{7}$

$= \frac{64}{7} = 9 \frac{1}{7} m$

$a r (H K N D) = \frac{1}{2} \times D N \times (H D + K N)$

$= \frac{1}{2} \times \frac{64}{7} \times (16 + \frac{144}{7})$

$= \frac{32}{7} \times \frac{256}{7} = \frac{8192}{49} m^{2} = 167 \frac{9}{49} m^{2}$

$a r (∆ K G N) = \frac{1}{2} \times N G \times K N$

$K N = \frac{144}{7} m$ and we have $D G = 24 m$ and $D N = \frac{64}{7}$

$N G = D G - D N = 24 - \frac{64}{7} = \frac{104}{7} = 14 \frac{6}{7} m$

$a r (∆ K G N) = \frac{1}{2} \times \frac{104}{7} \times \frac{144}{7} = \frac{7488}{49} = 152 \frac{40}{49} m^{2}$

$a r (H K G D) = 167 \frac{9}{49} + 152 \frac{40}{49} = 320 m^{2}$

Palace of Versailles' rectangular garden has a length of $48 m$ and breadth of $32 m$ . It is divided into $4$ regions as shown below. It is known that $E B = D G = 24 m$ and $A H = F C = 32 m$ . Given the area of region $E B F K = 512 m^{2}$ , what is the area of region $H K G D$ ?

Important Questions on Pair of Linear Equations in Two Variables

Learn and Prepare for any exam you want !

Palace of Versailles' rectangular garden has a length of 48 m and breadth of 32 m. It is divided into 4 regions as shown below. It is known that EB=DG=24 m and AH=FC=32 m. Given the area of region EBFK=512 m2, what is the area of region HKGD?

Important Questions on Pair of Linear Equations in Two Variables

Learn and Prepare for any exam you want !

Palace of Versailles' rectangular garden has a length of $48 m$ and breadth of $32 m$ . It is divided into $4$ regions as shown below. It is known that $E B = D G = 24 m$ and $A H = F C = 32 m$ . Given the area of region $E B F K = 512 m^{2}$ , what is the area of region $H K G D$ ?