# 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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