Question

In the adjoining figure, ABC is an isosceles right-
angled triangle and DEFG is a rectangle
AD = AE =3 cm and DB = EC = 4 cm, find the area of
the shaded region
​

Answer 1

Answer:

Step-by-step explanation:

In ∆ADE, applying Pythagoras theorem, we get

DE² = AD² + AE²

⇒ DE = √[3² + 3²]

⇒ DE = 3√2 cm

∴ DE = GF = 3√2 cm ……. [opposite sides of the rectangle are congruent] ….. (i)

We know from the figure that,

AB = AD + DB = 3 + 4 = 7 cm

AC = AE + EC = 3 + 4 = 7 cm

In ∆ABC, applying Pythagoras theorem, we get

BC² = AB² + AC²

⇒ BC = √ [7² + 7²]

⇒ BC = 7√2 cm …. (ii)

Step 2:

In ∆DBG and ∆EFC, we have

DB = EC ….. [given]

DG = EF …… [opposite side of the rectangle are congruent]

∠DGB = ∠EFC = 90° …… [angles of rectangle is a right angle]

∴ By SAS congruence, ∆DBG ≅ ∆EFC

∴ By CPCT, BG = FC ….. (iii)

From (i), (ii) & (iii), we get

BC = BG + GF + FC

⇒ BG + FC =  7√2 - 3√2

⇒ 2 BG = 4√2

⇒ BG = 2√2  cm

∴ BG = FC = 2√2 cm …. (iv)

Step 3:

Now, in ∆EFC, applying Pythagoras theorem, we get

EC² = EF² + FC²

⇒ 4² = EF² + (2√2)² ……. [from (iv)]

⇒ EF = √ [16 – 8]

⇒ EF = 2√2  

∴ EF = DG = 2√2 cm …… (v)

Thus,  

The area of the shaded portion i.e., area of the rectangle DEFG is,

= length * breadth

= DE * EF

= 3√2  * 2√2 ….. [substituting from (i) & (v)]

= 3 * 2 * 2

= 12 cm²

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