Question

(1) The diagonal AC of a quadrilateral ABCD is 6 cm and the perpendiculars, BM from the vertex B is

3 cm and DN from D is 5 cm, dropped on the same diagonal. Find the area of the quadrilateral

ABCD.​

Answer 1

Answer:

Area of Quadrilateral ABCD = 24 $cm^{2}$

Step-by-step explanation:

Given:

ABCD is a Quadrilateral.

Length of Diagonal AC is 6cm.

Perpendicular BM from Vertex B is 3cm.

Perpendicular DN from Vertex D is 5cm.

To Find:

Area of Quadrilateral ABCD.

Concept:

Here, we are given a Quadrilateral which has a diagonal and two perpendiculars of different lengths are dropped on the same diagonal. The diagonal cuts the quadrilateral into two different triangles. Finding the area of both triangles and adding them will give us the Area of Quadrilateral needed.

Formula Used:

Area of a Triangle = 1/2 * b * h...….(I)

where b = base of the triangle and h = height of the triangle

Solution:

Let's firstly find out the area of triangle ABC.

» Base of ∆ABC= 6cm.

» Height of ∆ABC= 3cm.

Equating the values in Formula (I) . We get,

Area of the ∆ABC = (1/2 * 6 * 3) $cm^{2}$

                              = 9 $cm^{2}$

Now, Let's find the Area of triangle ADC.

» Base of ∆ADC= 6cm.

» Height of ∆ADC= 5cm.

Equating the values in Formula (I) . We get,

Area of the ∆ADC = (1/2 * 6 * 5) $cm^{2}$

                              = 15 $cm^{2}$

Now, adding the Areas of ∆ABC and ∆ADC gives Area of Quadrilateral ABCD. So,

Area of the Quadrilateral ABCD = Area of the ∆ABC + Area of the ∆ADC

                                                     = 9 $cm^{2}$ + 15 $cm^{2}$ = 24 $cm^{2}$