Question

The length of one of the diagonals of a field in the form of quadrilateral is 54m. The perpendicular distances of the other two vertices from this diagonal are 14m and 9 m. Find the area of field.

Answer 1

Answer:

Area = 621 m²

Step-by-step explanation:

Given

BD = 54 m

AM = 14 m

CN = 9 m

Area of the field = Sum of area of the 4 triangles

i.e. ΔAMD + ΔAMB + ΔCNB + ΔCND

= ($\frac{1}{2}$ x DM x AM) + ($\frac{1}{2}$ x BM x AM) + ($\frac{1}{2}$ x BN x NC) + ($\frac{1}{2}$ x DN x NC)

= ($\frac{1}{2}$ x DM x 14) + ($\frac{1}{2}$ x BM x 14) + ($\frac{1}{2}$ x BN x 9) + ($\frac{1}{2}$ x DN x 9)

= (7 DM) + (7 BM) + ($\frac{9}{2}$ BN) + ($\frac{9}{2}$ DN)

Replacing

BM = BD - DM = 54 - DM

and

DN = DB - NB = 54 - NB

= (7 DM) + (7 (54 - DM)) + ($\frac{9}{2}$ BN) + ($\frac{9}{2}$ (54 - NB))

= (7 DM) + (378 - 7 DM) + ($\frac{9}{2}$ BN) + (243 - $\frac{9}{2}$ NB)

= 7 DM + 378 - 7 DM + $\frac{9}{2}$ BN + 243 - $\frac{9}{2}$ NB

= 378 + 243

= 621

Answer 2

Answer:

Step-by-step explanation:

Given

BD = 54 m

AM = 14 m

CN = 9 m

Area of the field = Sum of area of the 4 triangles

i.e. ΔAMD + ΔAMB + ΔCNB + ΔCND

= (\frac{1}{2}

2

1

x DM x AM) + (\frac{1}{2}

2

1

x BM x AM) + (\frac{1}{2}

2

1

x BN x NC) + (\frac{1}{2}

2

1

x DN x NC)

= (\frac{1}{2}

2

1

x DM x 14) + (\frac{1}{2}

2

1

x BM x 14) + (\frac{1}{2}

2

1

x BN x 9) + (\frac{1}{2}

2

1

x DN x 9)

= (7 DM) + (7 BM) + (\frac{9}{2}

2

9

BN) + (\frac{9}{2}

2

9

DN)

Replacing

BM = BD - DM = 54 - DM

and

DN = DB - NB = 54 - NB

= (7 DM) + (7 (54 - DM)) + (\frac{9}{2}

2

9

BN) + (\frac{9}{2}

2

9

(54 - NB))

= (7 DM) + (378 - 7 DM) + (\frac{9}{2}

2

9

BN) + (243 - \frac{9}{2}

2

9

NB)

= 7 DM + 378 - 7 DM + \frac{9}{2}

2

9

BN + 243 - \frac{9}{2}

2

9

NB

= 378 + 243

= 621.