Name: Figures on the Same Base
Uploaded: 2019-07-19
Duration: 6 min 34 s
Description: Two figures are said to be on one base and between the same parallels if they have a common side as the base and certain vertices lie on the line parallel to...

Question

$A B C$ is a right triangle right angled at $A . B C E D, A C F G$ and $A B M N$ are squares on the sides $BC, CA$ and $A B$ respectively. Line segments $AX ⊥ DE$ meets $B C$ at $Y,$ and $D E$ at $X$ . Join $A D, A E$ also $B F$ and $C M$ (See the figure).

Show that $ar (BCED) = ar (ABMN) + ar (ACFG)$

Can you write the result in words? This is a famous theorem of Pythagoras. You shall learn a simpler proof in this theorem in class $X$ .

Accepted Answer

Here, we have, $∠ A B M = ∠ D B C$ [both being $90 °$ because both are angles of squares]

Adding $∠ A B C$ to both sides, we get,

$∠ A B M + ∠ A B C = ∠ D B C + ∠ A B C$

$X$ .

Now, in $△ M B C$ and $△ A B D$ , we have,

$B C = B D$ [They are sides of the same square and hence equal]

$M B = A B$ [They are sides of the same square and hence equal]

and, $∠ M B C = ∠ A B D$ .

Thus, by SAS criterion of congruency, we get,

$Δ MBC ≅ Δ ABD .$

Thus, we have $ar (Δ MBC) = ar (Δ ABD) - - - - (i)$

We have, $AX ⊥ D E$ (given) and $B D ⊥ D E$ (Adjacent sides of square $B D E C$ ).
So, $B D I I A X$ (Two lines perpendicular to same line are parallel to each other)

We see, here that $△ A B D$ and rectangle $B D X Y$ are on the same base $B D$ and $B D & A X .$

Thus, we have $a r ∆ A B D = \frac{1}{2} \times □ B D X Y .$ [If there is a quadrilateral and a triangle on the same base and between same parallel lines, then the area of the triangle is half the area of the quadrilateral].

$\Rightarrow ar Δ M B C = \frac{1}{2} \times a r □ B D X Y$ {from (i)}

$\Rightarrow 2 \times ar Δ M B C = ar □ B D X Y$

$\Rightarrow ar □ B Y X D = 2 ar Δ M B C - - - - (i i)$

Since $A B M N$ is a square, we have $M B ∥ N A \Rightarrow M B ∥ N C .$

Thus, we see that $△ M B C$ and $□ M B A N$ are on the same base $M B$ and same parallel lines $M B$ and $N C$ ,

we have $a r △ M B C = \frac{1}{2} \times a r □ A B M N$

$\Rightarrow 2 a r Δ M B C = a r □ A B M N - - - - (i i i)$

From (ii) and (iii), we get,

$ar (B Y X D) = ar (A B M N) - - - - (i v)$

Similarly, we can prove that

$ar (CYXE) = ar (ACFG) - - - - (v)$

Adding (iv) and (v), we get,

$ar (B Y X D) + ar (CYXE) = ar (A B M N) + ar (ACFG)$

$ar (BCED) = ar (ABMN) + ar (ACFG)$

Now, $ar (BCED)$ can be written as $A B^{2}$ , $ar (ABMN)$ can be written as $A B^{2}$ and $ar (ACFG)$ can be written as $A C^{2} .$

Thus, $ar (BCED) = ar (ABMN) + ar (ACFG)$ can be written as $B C^{2} = A B^{2} + A C^{2}$ .

Therefore, the above result can be written in words as: The square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides.

This is called the Pythagoras theorem.

Important Points to Remember in Chapter -1 - Areas from Telangana Board Mathematics Class 9 Solutions

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