Question

in the figure, area of the rectangle ABCD is 36 cm². a) Find the area of △ABP b) If △ABP is an isosceles triangle, then find the areas of △ADP and △BCP​.. plz help mee

Answer 1

Answer:

a) 18 cm²

b) 9cm²

Step-by-step explanation:

( Refer to attachment for figure)

Construction : Draw a perpendicular line to AB from P and name that as Q

a) Area of the rectangle ABCD = Length × Breadth = AB × CB = 36 cm²

AB × CB = 36 cm² ------ ( 1 )

Area of the ∆ABP = 1/2 × Base × Height

= 1/2 × AB × PQ

Since height of the triangle = Breadth of the rectangle, PQ = CB

= 1/2 × AB × CB

From eq( 1 )

= 36/2

= 18 cm²

Therefore the area of the ∆ ABP is 18 cm².

b) Consider ∆ ADP and ∆ BCP

∠D = ∠C ( Since Angles of rectangle are equal to 90° )

Hypotenuse AP = Hypotenuse BP ( Sides of an isosceles triangle APB)

AD = CB ( Opposite sides of the rectangle are equal)

By RHS congruence rule ∆ ADP ≅ ∆ BCP

Area of two congruent triangles are equal

Area of ∆ ADP = Area of ∆ BCP ---- ( 2 )

We know that

Area of rectangle ABCD = Area of ∆ ABP + Area of ∆ ADP + Area of ∆ BCP

From ( 2 )

36 = 18 + Area of ∆ ADP + Area of ∆ ADP

36 - 18 = 2( Area of ∆ ADP )

2( Area of ∆ ADP ) = 18

Area of ∆ ADP = 18/2

Area of ∆ ADP = 9 cm²

Area of ∆ ADP = Area of ∆ BCP = 9 cm²

Therefore area of ∆ ADP and ∆ BCP are 9 cm² and 9 cm² respectively.