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Home » In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of ∆APQ is 1/16 of the area of ∆ABC.
In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of ∆APQ is 1/16 of the area of ∆ABC.
CBSE | Class 10 | Exercise 4C | Maths | RS Aggarwal | Triangles
In the given figure, ABC is a triangle and PQ is a straight line meeting AB in P and AC in Q. If AP = 1cm, PB = 3cm, AQ = 1.5cm, QC = 4.5cm, prove that area of ∆APQ is 1/16 of the area of ∆ABC.

 

 

 

Answer:

Given,

\frac{A P}{A B}=\frac{1}{1+3}=\frac{1}{4}

\frac{A Q}{A C}=\frac{1.5}{1.5+4.5}=\frac{1.5}{6}=\frac{1}{4}

\frac{A P}{A B}=\frac{A Q}{A C}

∠???? = ∠????

By SAS similarity, we can conclude that ∆APQ – ∆ABC.

\frac{a r(\triangle A P Q)}{a \pi(\triangle A B C)}=\frac{A P^{2}}{A B^{2}}=\frac{1^{2}}{4^{2}}=\frac{1}{16}

\frac{\operatorname{ar}(\triangle A P Q)}{a r(\triangle A B C)}=\frac{1}{16}

\operatorname{ar}(\triangle A P Q)=\frac{1}{16} \times a r(\triangle A B C)

Hence, proved.

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