Question

find the perimeter and area of quadrilateral ABCD in which AB = 21cm ,<BAC= 90cm ,AC =20cm,CD= 42CM and AD=34CM​

Answer 1

$cb = \sqrt{ {20}^{2} + {21}^{2} } = \sqrt{441 + 400} \\ \\ = \sqrt{841} = 29$

perimeter=29+21+34+42=126

Answer 2

Perimeter of quadrilateral ABCD is 126 cm and area of quadrilateral ABCD is 384 sq. cm.

• In ΔABC ,

AC = 20 cm , AB = 21 cm and ∠CAB = 90°.

• Therefore , By Pythagoras Theorem

BC² = AC² + AB²

BC² = 20² + 21² = 400+441 = 841

BC = 29 cm

• Now Perimeter of quadrilateral ABCD = AB + BC + CD + AD

Perimeter = 21+29+42+34 = 126 cm.

• Now area of quadrilateral ABCD is equal to Area of ΔABC + Area of ΔCAD .

• Now area of ΔABC = $\frac{base . height}{2}$ = 20 × 21 / 2 = 210 sq cm.

• Area of ΔCAD = $\sqrt{s(s-a)(s-b)(s-c)}$   , Using Heron's formula

• Now s = (a+b+c)/2 = (42+34+20)/2 = 48 sq cm.

• Now, area = $\sqrt{48(48-42)(48-34)(48-20)}$ = $\sqrt{48(6)(14)(28)}$ = $\sqrt{112896}$ = 336 sq. cm

• Now area of quadrilateral ABCD = Area of ΔABC + Area of ΔCAD

• Area of quadrilateral ABCD = 48 + 336 = 384 sq cm.