Question

11. In Fig. 12.35, squares are drawn on the side AB and
the hypotenuse AC of the right triangle ABC. If BH
is perpendicular to FG, prove that area of the square
ABDE = area of the rectangle ARHF. (ICSE)
​

Answer 1

Given:

• From the given diagram square are drawn on sides of AB.
• In the given right angle ΔABC the hypotenuses is AC.

• Here, BH║FG.

To Find:  

• Area of the square ABDE = Area of the rectangle ARHF.

Solution:

Given, ΔABC is a right triangle.

Using Pythagoras theorem in ΔABC

We get,

$AC^{2}=AB^{2}+BC^{2}$

⇒$AB^{2}=AC^{2}-BC^{2}$

In the diagram, $AC=AR+RC$

⇒$AB^{2}=(AR+RC)^{2}-BC^{2}$

Using Pythagoras theorem in ΔBRC

We get,

⇒ $AB^{2}=(AR+RC)^{2}-(BR^{2}+RC^{2})$

Now, we will expand using $(A+B)^{2}=A^{2}+B^{2}+2AB$

⇒$AB^{2}=AR^{2}+2(AR.RC)+RC^{2}-(BR^{2}+RC^{2})$

Using Pythagoras theorem in ΔABR

⇒$AB^{2}=AR^{2}+2(AR.RC)+RC^{2}-(AB^{2}-AR^{2}+RC^{2})$

Here, we need to solve the terms

⇒$AB^{2}-AR^{2}+2(AR.RC)+RC^{2}+AB^{2}+AR^{2}-RC^{2}$

Now, we need to cancel the terms

$2AB^{2}=2AR^{2}+2AR.RC$  

Take "2" in common

$2AB^{2}=2(AR^{2}+AR.RC)$

Here, cancel the term "2" and take AR in common.

$AB^{2}=AR(AR+RC)$

∴ AC = AR + RC

⇒$AB^{2}=AR+AC$

⇒$AB^{2}=AR.AF$

⇒ Area of square ABDE = Area of rectangle ARHF

Hence proved the theorem.