Question

The area of the quadrilateral whose sides
measures 9 cm, 40 cm, 28 cm and 15 cm
and in which the angle between the first
two sides is a right angle, is
(a) 206 cm
by306 cm
(c) 356 cm
(d) 380 cm​

Answer 1

Step-by-step explanation:

Draw a diagonal in the the quadrilateral dividing it into 2 equal triangles.

therefore area of first triangle

1/2 × b × h

1/2 × 9 × 40

180cm²

now finding the hypotenuse of the triangle which is the diagonal of the quadrilateral.

By pythagoras property,

Hypotenuse² =Base² + Alt²

Hypotenuse² = 9² + 40²

Hypotenuse² = 81 + 1600

Hypotenuse² = 1681

Hypotenuse = 41 cm

Now finding area of the second triangle -

semiperimeter = a+b+c/2 = 28+15+40 = 55+28/2 = 83/2 =41.5 cm

By Heron's Formula,

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

root of 41.5 (41.5 - 28)(41.5 - 15)(41.5-40)

root of 41.5 × x 13.5 × 26 × 1.5

148 cm² (approx)

therefore total area= 180 + 148

= 328cm²

hope it helps :))