Question

ABCD is a quadrilateral in which diagonal BD =14cm if AL perpendicular BD and CM perpendicular such that AL =8 cm and CM=6cm find the area of quadrilateral ABCD

Answer 1

Answer:

Step-by-step explanAreaofΔABD=12×base×height

=(12×5×7)cm2=352cm2

AreaofΔABD=(12×5×7)cm2=352cm2

Since the diagonal BD divides ABCD into two triangles of equal area.

Therefore, Areaofparallelogram=AreaofΔABD+AreaofΔCBD

=(352+352)cm2=702cm2=35cm2

Therefore, Area of parallelogram = 35 cm2

Answer 2

Answer: $98 cm^{2}$

Step-by-step explanation:  

Given BD=14cm ,  CM=6cm , AL=8cm  and AL and CM are perpendicular

Area of triangle is  = $\dfrac{1}{2} \times base\times height$

In Δ DBC  BD is base and CM is height

so, area of  Δ DBC  = $\dfrac{1}{2} \times BD\times CM$ = $\dfrac{1}{2} \times 14cm\times 6cm$ = $42cm^{2}$

Now In Δ ABD BD is base and  AL is height

so, area of  Δ ABD = $\dfrac{1}{2}\times BD \times AL$ $= \dfrac{1}{2} \times 14cm \times 8cm = 56cm^{2}$

Area of quadrilateral is = Area of  Δ DBC + Area of  Δ ABD

                                      $=42cm^2+56cm^2 \\\\= 98cm^{2}$

So area of quadrilateral is $98 cm^{2}$