Question

A quadrilateral whose sides are 3cm, 4cm, 4cm, 5cm and one of the diagonal is equal to 5cm as per the below figure. The area of the quadrilateral is:

a. 19.17 sq.cm

b. 15.17 sq.cm

c. 20.17 sq.cm

d. 22.17 sq.cm.​

Answer 1

Given : A quadrilateral whose sides are 3cm, 4cm, 4cm, 5cm and one of the diagonal is equal to 5 cm

To Find :  The area of the quadrilateral  

Solution:

Quadrilateral can be divided in 2 triangle

one with sides  3 , 4 , 5 cm

and another with sides 4 , 5 , 5 cm

Area of Quadrilateral = area of both triangle

Triangle with sides  3 , 4 , 5 cm  is right angle triangle as (3 ,4  , 5) is Pythagorean Triplet

Hence area = (1/2) * base * Height

= (1/2) * 3 * 4

= 6 cm²

Triangle with sides   4 , 5 , 5  cm

s = ( 4 + 5 + 5)/2 = 7

Area of triangle using Hern's formula

= √7(7 - 4)(7 - 5)(7 - 5)

= √7(3)(2)(2)

= 2√21

≈ 9.17 cm²

Area of the Quadrilateral ≈  6 + 9.17  ≈ 15.17 cm²

correct option is b

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Answer 2

Answer:

Option B.) 15.17

Step-by-step explanation:

Given:

A quadrilateral whose sides are 3cm, 4cm, 4cm, 5cm and one of the diagonal is equal to 5cm

To find:

The area of the quadrilateral

Solution:

We can divide this quadrilateral into two triangles:

So, Sides of 1st Triangle = 3 cm, 4 cm, 5cm

Sides of 2nd Triangle = 4 cm, 5 cm, 5 cm

Area of Quadrilateral = Sum of area of these two triangles

Area of 1st Triangle:

It is a right angled triangle so: area will be

$\frac{1}{2} \times base \times height$

$\frac{1}{2} \times 3 \times 4$

$1 \times 3 \times 2$

$6 \: cm {}^{2}$

Area of 2nd Triangle:

We can find the area of second triangle by Heron's formula:

$s = \frac{a + b + c}{2}$

$s = \frac{4 + 5 + 5}{2}$

$s = \frac{14}{2}$

$s = 7$

$\sqrt{s(s - a)(s - b)(s - c)}$

$\sqrt{7(7 - 5)(7 - 5)(7 - 4)}$

$\sqrt{7 \times 2 \times 2 \times 3}$

$2 \sqrt{7 \times 3}$

$2 \sqrt{21}$

$9.17$

Area of Quadrilateral will be:

$6 + 9.17$

$15.17$