Home
Class 9
MATHS
In Figure, A P || B Q\ || C R. Prove tha...

In Figure, `A P || B Q\ || C R`. Prove that `a r\ ( A Q C)=\ a r\ (P B R)`

Text Solution

Verified by Experts

Since `triangleABQ` and `trianglePBQ` lie on the same base `BQ` and are between the same parallels `AP` and `BQ`,
According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
`Area (triangleABQ) = Area (trianglePBQ)` ...(1)
Similarly, `triangleBCQ` and `triangleBRQ` lie on the same base `BQ` and are between the same parallels `BQ` and `CR`.
`Area (triangleBCQ) = Area (triangleBRQ`) ... (2)
On adding Equations (1) and (2), we obtain
`Area (triangleABQ) + Area (triangleBCQ) = Area (trianglePBQ) + Area (triangleBRQ)`
Hence, `Area (triangleAQC) = Area (trianglePBR)` is proved.
Promotional Banner

Similar Questions

Explore conceptually related problems

In figure, P S D A is a parallelogram in which P Q=Q R=R S\ a n d\ A P || B Q || C R || DS . Prove that a r\ (triangle \ P Q E)=\ a r\ (triangle \ C F D) .

In Figure P Q || C D and P R || C B. Prove that ( A Q)/(Q D)=(A R)/(R B)dot

In Figure, A B C D is a parallelogram. Prove that: a r( B C P)=a r( D P Q) CONSTRUCTION : Join A Cdot

In Fig. if P Q|| B C and P R|| C D . Prove that (i) (A R)/(A D)=(A Q)/(A B) (ii) (Q B)/(A Q)=(D R)/(A R) .

In Figure, X\ a n d\ Y are the mid-points of A C\ a n d\ A B respectively, Q P || B C\ a n d\ C Y Q\ a n d\ B X P are straight lines. Prove that a r\ (\ A B P)=\ a r\ (\ A C Q) .

In triangle A B C ,\ P\ a n d\ Q are respectively the mid-points of A B\ a n d\ B C and R is the mid-point of A P . Prove that: a r\ ( triangle R Q C)=3/8\ a r\ (triangle \ A B C) .

In Figure, A B C D\ a n d\ A E F D are two parallelograms? Prove that: a r\ (\ A P E):\ a r\ (\ P F A)=\ a r\ \ (Q F D)\ : a r\ (\ P F D)

In triangle A B C ,\ P\ a n d\ Q are respectively the mid-points of A B\ a n d\ B C and R is the mid-point of A P . Prove that: a r\ ( triangle P B Q)=\ a r\ (triangle \ A R C)

In Figure, /_P Q R=\ /_P R Q , then prove that /_P Q S=\ /_P R Tdot

In Figure, P S=Q R\ a n d\ /_S P Q=/_R Q Pdot Prove that P Q S\ ~=\ Q P R ,\ P R=Q S\ a n d\ /_Q P R=\ /_P Q S