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
Answers
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.
