Question

Areas of Triangles and Quadrilaterals
535
C
7
42 cm
23. Find the perimeter and area of the
quadrilateral ABCD in which AB = 21 cm,
ZBAC = 90°, AC = 20 cm, CD = 42 cm and
AD = 34 cm.
D
cm
34 cm
B
A 21 cm​

Answer 1

Explanation:

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}

2

base.height

= 20 × 21 / 2 = 210 sq cm.

Area of ΔCAD = \sqrt{s(s-a)(s-b)(s-c)}

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)}

48(48−42)(48−34)(48−20)

= \sqrt{48(6)(14)(28)}

48(6)(14)(28)

= \sqrt{112896}

112896

= 336 sq. cm

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

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