In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.
Mensuration
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In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

KnowledgeBoat · Accepted Answer

From figure, In right angle triangle ADE, Using pythagoras theorem, ⇒ DE² = AD² + AE² ⇒ DE² = 3² + 3² ⇒ DE² = 9 + 9 ⇒ DE² = 18 ⇒ DE = cm Since, DEFG is a rectangle. ∴ GF = DE = cm. In △DBG and △ECF, DB = EC = 4 cm DG = EF (Opposite sides of rectangle are equal) ∠DGB = ∠EFC = 90° (∵ DEFG is a rectangle) Hence, by RHS axiom △DBG ≅ △ECF. So, BG = FC (By C.P.C.T.) Let BG = FC = x. In right angle triangle ABC, ⇒ BC² = AB² + AC² ⇒ BC² = 7² + 7² ⇒ BC² = 49 + 49 ⇒ BC² = 98 ⇒ BC = cm From figure, BG + GF + FC = BC ⇒ BG + GF + FC = ⇒ x + + x = ⇒ 2x = ⇒ 2x = ⇒ x = cm. In right angle triangle DBG, ⇒ DB² = BG² + DG² ⇒ 4² = ² + DG² ⇒ 16 = 8 + DG² ⇒ DG² = 16 - 8 = 8 ⇒ DG = cm. Area of right angle triangle DBG = x BG x DG = x x = x 8 = 4 cm². Since, △DBG ≅ △ECF. ∴ Areas of both triangle are equal. Area of right angle triangle ADE = x AD x AE = x 3 x 3 = = 4.5 cm². Area of shaded region = Area of (△ADE + △DBG + △ECF) = 4.5 + 4 + 4 = 12.5 cm². Hence, area of shaded region = 12.5 cm².

Mathematics

In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

Mensuration

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In the figure (ii) given, ABC is an isosceles right-angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.

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