ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also AC = 5 cm, DE = 4 cm, and the area of ∆ AED = 6 cm². Find the perimeter and area of ABCD.
Solution:
It is given that
Area of ∆ AED = 6 cm²
AC = 5 cm
DE = 4 cm
We know that
Area of triangle = 1/2 × base × height
Area of triangle AED = 1/2 × DE × AE
Substituting the values
6 = 1/2 × 4 × AE
By further calculation
AE = (6 × 2)/4
AE = 3 cm
In right angled triangle AEC
AE = 3 cm
AC = 5 cm
Using the Pythagoras theorem
EC² = AC² - AE²
Substituting the values
EC² = 5² - 3²
EC² = 25 - 9
EC² = 16
So we get
EC = 4 cm
DE + EC = DC
DC = 4 + 4 = 8 cm
Here ABCD is a parallelogram
AB = DC = 8 cm
In right angled triangle AED
Using Pythagoras theorem
AD² = AE² + ED²
Substituting the values
AD² = 3² + 4²
AD² = 9 + 16
AD² = 25
So we get
AD = 5 cm
AD = BC = 5 cm
Here ABCD is a parallelogram
We know that
Perimeter of parallelogram ABCD = 2 (l + b)
= 2 (DC + AD)
Substituting the values
= 2 (8 + 5)
= 2 (13)
= 26 cm
Similarly
Area of parallelogram ABCD = base × height
= DC × AE
Substituting the values
= 8 × 3
= 24 cm²
Therefore, the perimeter of ABCD is 26 cm and the area is 24 cm².
✦ Try This: The perimeter of a rhombus is 30 cm and one diagonal is 3 cm long then the length of the other diagonal is
☛ Also Check: NCERT Solutions for Class 7 Maths Chapter 11
NCERT Exemplar Class 7 Maths Chapter 9 Problem 102
ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also AC = 5 cm, DE = 4 cm, and the area of ∆ AED = 6 cm². Find the perimeter and area of ABCD.
Summary:
ABCD is a parallelogram in which AE is perpendicular to CD (Fig. 9.54). Also AC = 5 cm, DE = 4 cm, and the area of ∆ AED = 6 cm². The perimeter of ABCD is 26 cm and the area is 24 cm².
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