ABCD is a parallelogram where the base DC is 6 cm and height is 4 cm. A rhombus AEFD is drawn on the sides AB and DC as shown in the figure. AF = 4.5 cm and DE = 7.5 cm. What is the area of the parallelogram EBCF?
Answers
Step-by-step explanation:
From the question it is given that,
ABCD is a parallelogram, AM⊥DC and AN⊥CB
AM=6cm
AN=10cm
The area of parallelogram ABCD is 45cm
2
Then, area of parallelogram ABCD=DC×AM=BC×AN
45=DC×6=BC×10
(i) DC=45/6
Divide both numerator and denominator by 3 we get,
=15/2
=7.5cm
Therefore, AB=DC=7.5cm
(ii) BC×10=45
BC=45/10
BC=4.5cm
(iii) Now, consider △ADM and △ABN
∠D=∠B … [because opposite angles of a parallelogram]
∠M=∠N … [both angles are equal to 90
∘
]
Therefore, △ADM∼△ABN
Therefore, area of △ADM/area of △ABN=AD
2
/AB
2
=BC
2
/AB
2
=4.52/7.52
=20.25/56.25
=2025/5625
=81/225
=9/25
Therefore, area of △ADM : area of △ANB is 9:25.
Step-by-step explanation:
From the question it is given that,
ABCD is a parallelogram, AM⊥DC and AN⊥CB
AM=6cm
AN=10cm
The area of parallelogram ABCD is 45cm
2
Then, area of parallelogram ABCD=DC×AM=BC×AN
45=DC×6=BC×10
(i) DC=45/6
Divide both numerator and denominator by 3 we get,
=15/2
=7.5cm
Therefore, AB=DC=7.5cm