Question

3.
In the figure, ABC is an isosceles right-angled
triangle and PQRS is a rectangle.


find the shaded region

this sum is of class 10​

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

12cm

$\bold{Explanation:}$

ABC is an isosceles right triangle

AD = AE = 3 cm, DB = EC = 4 cm

⇒ AB = AC = (AD + DB) or (AE + EC) = 3 cm + 4 cm = 7 cm

Also since AB = AC

∴ ∠B = ∠C (angles opposite to equal sides) ....... (1)

But ∠A + ∠B + ∠C = 180° (Angle Sum property)

⇒ 90° + ∠B + ∠B = 180° (∵ ∠A = 90° and from (2))

⇒ 2∠B = 180° – 90° = 90°

Now on ∆BDG

Now since ABC is a right ∆

∴ BC2 = AB2 + BC2 = (7 cm)2 + (7 cm)2 = 49 cm2 + 49 cm2 = 98 cm2

Hence required shaded area = area of ∆ABC – Area of rectangle DEFG

= 24.5 cm2 – 12 cm2

= 12.5 cm2