in rectangle ABCD ,AB=8,BC=6 and P is mid point of DC.
a) what is the area of triangle APB ?
b) what is the area of rectangle ABCD ?
c)what is the area of triangle ADP ?
d) what is the sum of areas of triangle ADP and triangle PBC ?
Answers
Answer:
Step-by-step explanation:
A quadrilateral ABCD with DC = 17 cm, AD = 9 cm and BC = 8 cm (Given)
In right ΔABD,
Using Pythagorean Theorem,
AB2 + AD2 = BD2
152 = AB2 + 92
AB2 = 225−81=144
AB = 12
Area of ΔABD = 1/2(12×9) cm2 = 54 cm2
In right ΔBCD:
Using Pythagorean Theorem,
CD2 = BD2 + BC2
172 = BD2 + 82
BD2 = 289 – 64 = 225
or BD = 15
Area of ΔBCD = 1/2(8×17) cm2 = 68 cm2
Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= 54 cm2 + 68 cm2
= 112 cm2
Hope it helps
Answer:
Answer:
Step-by-step explanation:
A quadrilateral ABCD with DC = 17 cm, AD = 9 cm and BC = 8 cm (Given)
In right ΔABD,
Using Pythagorean Theorem,
AB2 + AD2 = BD2
152 = AB2 + 92
AB2 = 225−81=144
AB = 12
Area of ΔABD = 1/2(12×9) cm2 = 54 cm2
In right ΔBCD:
Using Pythagorean Theorem,
CD2 = BD2 + BC2
172 = BD2 + 82
BD2 = 289 – 64 = 225
or BD = 15
Area of ΔBCD = 1/2(8×17) cm2 = 68 cm2
Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= 54 cm2 + 68 cm2
= 112 cm2
Hope it helps