In the given figure- AB - DC and AD - BP- If DC - 3AB- then ar- APB - ar- ADB - ar- ACB is equal to

Given that nbsp; AB||DC and AD||BP. Let PQ ⊥ AB. Now, AB||DC. ⇒ AB||DP Thus, ABPD is a parallelogram We know that triangles on the same base and between t ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Figures on the Same Base and Between the Same Parallels

In the given ...

Question

In the given figure, $A B | | D C a n d A D | | B P . I f D C = 3 A B, t h e n a r (Δ A P B) + a r (Δ A D B) + a r (Δ A C B)$ is equal to

\frac{1}{2} a r (A B P D)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{3}{4} a r (A B P D)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a r (Δ B P C)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a r (Δ D B C)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
$a r (Δ D B C)$
Given that $A B | | D C a n d A D | | B P . L e t P Q ⊥ A B . N o w, A B | | D C . \Rightarrow A B | | D P$

Thus, ABPD is a parallelogram
We know that triangles on the same base and between the same parallels are equal in area.

$∴ A r (Δ A P B) = A r (Δ A D B) = A r (Δ A C B) \dots . . (i)$

Also, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

$∴ A r (Δ A P B) = \frac{1}{2} A r (A B P D) \dots . . (i i) N o w, A r (Δ A P B) + A r (Δ A D B) + A r (Δ A C B) = 3 \times A r (Δ A P B) [From (i)] = 3 \times \frac{1}{2} A r (A B P D) [From (ii)] = \frac{3}{2} A r (A B P D) \dots . . (i i i) A l s o, \frac{3}{4} [A r (Δ A P C)] = \frac{3}{4} \times (\frac{1}{2} \times B a s e \times H e i g h t) = \frac{3}{8} \times P C \times P Q = \frac{3}{8} \times (D C - D P) \times P Q = \frac{3}{8} \times (3 A B - D P) \times P Q (∵ D C = 3 A B) = \frac{3}{8} \times (3 A B - A B) \times P Q (A B = D P a s A B P D i s a p a r a l l e l o g r a m) = \frac{3}{8} \times 2 A B \times P Q = \frac{3}{4} \times A r (A B P D) N o w, A r (Δ B P C) = \frac{1}{2} \times B a s e \times H e i g h t = \frac{1}{2} \times P C \times P Q = \frac{1}{2} \times 2 A B \times P Q (∵ P C = D C - D P = 3 A B - A B = 2 A B) = A B \times P Q = A r (A B P D) A r (Δ D B C) = \frac{1}{2} \times B a s e \times H e i g h t = \frac{1}{2} \times D C \times P Q = \frac{1}{2} \times 3 A B \times P Q (G i v e n) = \frac{3}{2} \times A B \times P Q = \frac{3}{2} A r (A B P D) = A r (Δ A P B) + A r (Δ A D B) + A r (Δ A C B) [From (iii)]$

Hence, the correct answer is option (d).

Suggest Corrections

In the given figure, AB||DC and AD||BP. If DC=3AB, then ar(ΔAPB)+ar(ΔADB)+ar(ΔACB) is equal to

In the given figure, $A B | | D C a n d A D | | B P . I f D C = 3 A B, t h e n a r (Δ A P B) + a r (Δ A D B) + a r (Δ A C B)$ is equal to