Question

In the given figure, ABCD is a parallelogram and L is the mid point of of DC. If ar(qua. ABCL) is 72 cm2, then find ar(triangleADC).

please give the answer quickly. It's urgent !

Answer 1

ar(ABCL) = 72 cm2
ar(ABC) + ar(ACL) = 72cm2
ar(ABC) + 1/2ar(ADC) = 72cm2 ( L is mid point DC so, AL is median of ∆ ADC)
ar(ABC) + 1/2ar(ABC) = 72cm2 (∆s have equal area of llgm)
3ar(ABC)/2 = 72cm2
ar(ABC) = 72*2/3
ar (ABC) = 48cm2
so, ar(ADC) = 48cm2
hence proved

Answer 2

• The Area of ∆ADC is equal to 48 cm² .

Given :- ABCD is a parallelogram and L is the mid point of of DC. Ar(qua. ABCL) is 72 cm² .

To Find :-

• Ar(∆ ADC) = ?

Concept used :-

• The median of a triangle divides a triangle into two triangles of equal area .
• The diagonal of a parallelogram divides it into two triangles of equal area .

Solution :-

given that,

→ Ar(qua. ABCL) = 72 cm²

So,

→ Ar(∆ABC) + Ar(∆CAL) = 72 cm² ----- Equation (1)

now,

→ Ar(∆ABC) = Ar(∆ADC) { The diagonal of a parallelogram divides it into two triangles of equal area } ------- Equation (2)

putting Equation (2) in Equation (1) we get,

→ Ar(∆ADC) + Ar(∆CAL) = 72 cm² ---- Equation (3)

Now,

→ Ar(∆ADC) = Ar(∆CAL) + Ar(∆LAD)

→ Ar(∆ADC) = Ar(∆CAL) + Ar(∆CAL) { since AL is median of ∆ADC and the median of a triangle divides a triangle into two triangles of equal area }

→ Ar(∆ADC) = 2•Ar(∆CAL)

→ Ar(∆CAL) = (1/2)•Ar(∆ADC) ------ Equation (4)

finally, putting Equation (4) in Equation (3) we get,

→ Ar(∆ADC) + (1/2)•Ar(∆ADC) = 72

→ Ar(∆ADC)[1 + (1/2)] = 72

→ Ar(∆ADC) × (3/2) = 72

→ Ar(∆ADC) = 72 × (2/3)

→ Ar(∆ADC) = 24 × 2

→ Ar(∆ADC) = 48 cm² (Ans.)

Hence, the Area of ∆ADC is equal to 48 cm² .

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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