Question

In quadrilateral ABCD, AB = 40 cm, BC = 35 cm, CD = 48 cm
DA = 29 cm and BD = 29 cm. Find the area of quadrilateral
ABCD.​

Answer 1

Area of quadrilateral

Given :-

A quadrilateral ABCD with

• AB = 40 cm
• BC = 35 cm
• CD = 48 cm
• DA = 29 cm
• BD (Diagonal) = 29 cm

Now, area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BDC

Now,

We will calculate the area of triangles by Heron's formula.

In triangle ABD :-

s = (29 + 40 + 29)/2

= 49 cm

Area of triangle ABD = √s(s-a) (s-b) (s-c)

= √49(49-29) (49-40) (49-29)

= √49(20) (9) (20)

= √176400 = 420 cm²

In triangle BDC :-

s = (29 + 35 + 48)/2

= 56 cm

Area of triangle BDC = √s(s-a) (s-b) (s-c)

= √56(56-29) (56-35) (56-48)

= √56(27) (21) (8)

= √254016 = 504 cm²

So,

Area of quad. ABCD = 420 + 504 cm² = 924 cm²

Answer 2

Answer:

★ Area of quadrilateral= 924 cm² ★

Step-by-step explanation:

Given:

• AB = 40 cm , BC = 35 cm , CD = 48 cm , DA = 29 cm and BD = 29 cm

To Find:

• Area of quadrilateral ABCD

Solution: In ∆ABD of quadrilateral ABCD , by Heron's formula

→ Semiperimeter = a + b + c/ 2

→ S = 29 + 40 + 29 / 2 = 98/2 = 49 cm

★ Area of ∆ABD = √s(s–a)(s–b)(s–c) ★

→ √49(49–40)(49–29)(49–29)

→ √49(9)(20)(20)

→ √176400 = 420 cm²

Now, In ∆BCD of quadrilateral ABCD, by Heron's formula

→ Semiperimeter = a + b + c / 2

→ S = 35 + 29 + 48 / 2 = 112/2 = 56 cm

★ Area of ∆BCD = √s(s–a)(s–b)(s–c) ★

→ √56(56–35)(56–29)(56–48)

→ √56(21)(27)(8)

→ √254016 = 504 cm²

So, The area of quadrilateral ABCD will be = Area of ∆ ABD + Area of ∆BCD

→ Area of ABCD = (420+504)cm² = 924cm²

Hence, Area of ABCD will be 924 cm²