Abcd is a trapezium in which ab parallel to dc

Transcript Ex 6.2 ,9 ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that / = / Given: ABCD is a trapezium where AB II DC & diagonals AC & BD intersect at O To prove: / = / Construction: Let us draw a line EF II AB II DC passing through point O. Proof: Now , in EO II DC So, / = / Similarly , In DBA EO II AB / = / From (1) & (2) / = / / = / Hence proved Show More

In trapezium ABCD with $AB\parallel DC$, drawing a line $EF\parallel CD$ Now according to Basic Proportionality Theorem which states that "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio". Now in $\vartriangle ADC$, Since $EO\parallel CD$ ( from construction ) $ \Rightarrow \dfrac$ ( cross multiplying ) Hence proved. Note: Recall Basic Proportionality Theorem to solve such types of questions. Construction becomes important in solving such questions in a simple manner. We should make constructions wherever required.

Solution Given, ABCD is a trapezium and whose parallel sides in the figure are AB and DC. Since, A B ∣ ∣ C D and B C is transversal, then sum of two cointerior angles is 180 ∘, ∴ ∠ B + ∠ C = 180 ∘ ⇒ ∠ C = 180 ∘ − ∠ B = 180 ∘ − 45 ∘ [ ∵ ∠ B = 45 ∘ given ] ⇒ ∠ C = 135 ∘ Similarly, ∠ A + ∠ D = 180 ∘ [sum of cointerior angles is 180 ∘] ⇒ ∠ D = 180 ∘ − 45 ∘ [ ∵ ∠ A = 45 ∘ given ] ⇒ ∠ D = 135 ∘ Hence, angles C and D are 135 ∘ each.

Hint: As given in the question that AB is parallel to CD, we take two triangles namely triangle AOB and triangle COD. Now, we have to prove that the triangles are similar. From the result obtained, we can easily say that the ratio of AB and CD is equal to the ratio of AO and CO and the ratio of OB and OD are equal. Now, we can easily find the ratio of 2CD and CD and by squaring the ratio we get the ratio of the area of both triangles. Complete step-by-step solution: The figure of the question based on data provided is, As given in the question that they AB is parallel to CD. Now we take triangle AOB and triangle COD to prove that both the triangles are similar. So, we have to prove both the triangles $\Delta AOB$ and $\Delta COD$ are similar. In $\Delta AOB$ and $\Delta COD$, $ \Rightarrow \angle OAB = \angle OCD$ (Using alternate interior angles) $ \Rightarrow \angle OBA = \angle ODC$ (Using alternate interior angles) By AA similarity, $\Delta AOB \sim \Delta COD$ We know that, $\dfrac = 4$ Thus, the ratio of the area of triangle AOB and COD is $4:1$. Hence, the option(D) is the correct answer. Note: The key step for solving this problem is the proof of similarity of both the triangles. Once both the triangles are similar then by using the ratio of sides, we can easily evaluate the ratio of the area of the mentioned triangles. So, the knowledge of similar figures is required to solve this problem.