Question

in a figure ABC is an isosceles right angled triangle and D E F G is a rectangle if a b is equal to 8 equal to 3 cm and B B equal to BC equal to 4 cm find the area of the shaded portion

Ouestion no 1 diagram

Answer 1

Answer:

17.9776

Step-by-step explanation:

Answer 2

The area of the shaded portion is 12 cm².

Step-by-step explanation:

It is given that,

ABC is an isosceles right-angled triangle at A.

DEFG is a rectangle

AD = AE = 3 cm

DB = EC = 4 cm

Step 1:

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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Also View:

ABC is an isosceles triangle whose equal sides AB and AC are 10 cm each and base BC = 8cm. AD perpendicular to BC and angle BPC = 90°. Find the area of the shaded portion?  

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In given figure , triangle ABC is an eqilateral triangle have each side equal to 10 cm and triangke PBC is a right angled triangle inside it the right angle at P in which PB = 8 cm. find area of shaded region ?

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