Question

In the given Figure, $\Delta ABC\sim \Delta PQR$ and $\text{quad }ABCD\sim \text{quad }PQRS$ . Determine the values of $x,y,z$ in each case.

Answer 1

Hint: For answering this question we will use this point similar figures have corresponding sides as proportional. And find the values of $x,y,z$ using this by observing which two figures are similar and what are the corresponding sides.

Complete step by step answer:
Now considering from the question we have 2 triangles similar to each other represented as $\Delta ABC\sim \Delta PQR$ and two quadrilaterals similar to each other represented as $\text{quad ABCD}\sim \text{quad PQRS}$ .
Given that from the diagram for $\Delta ABC$ , $AB=12,BC=7,AC=10$ and for $\Delta PQR$, $PQ=9,QR=x,PR=y$ .
Given that from the diagram for $\text{quad ABCD}$ , $AB=16,BC=50,DC=\dfrac{50}{3},AD=20$ and for $\text{quad PQRS}$ , $PQ=x,QR=y,RS=z,PS=7$ .
As the given figures are similar their corresponding sides will be proportional.
So we can write mathematically as in the case of $\Delta ABC\sim \Delta PQR$ can be represented as: $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AC}{PR}$ and can be simply written as $\dfrac{12}{9}=\dfrac{7}{x}=\dfrac{10}{y}$.
By simplifying these proportions we can give $\dfrac{12x}{9}=7$ and $\dfrac{12y}{9}=10$.
By further simplifying we will have it as $12x=7\times 9$ and $12y=10\times 9$ .
By performing further calculations we will have $x=\dfrac{63}{12}$ and $y=\dfrac{90}{12}$ .
Hence we will have $x=\dfrac{21}{4}$ and $y=\dfrac{15}{2}$ .
So we can write mathematically as in the case of $\text{quad ABCD}\sim \text{quad PQRS}$ can be represented as: $\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{CD}{RS}=\dfrac{DA}{SP}$ and can be simply written as $\dfrac{20}{7}=\dfrac{16}{x}=\dfrac{50}{y}=\dfrac{50}{3z}$.
By simplifying this proportions we can give $\dfrac{20x}{7}=16$ and $\dfrac{20y}{7}=50$and $\dfrac{20\times 3z}{7}=50$ .
By further simplifying we will have it as $20x=7\times 16$ and $20y=50\times 7$ and $20\times 3z=50\times 7$.
By performing further calculations we will have $x=\dfrac{16\times 7}{20}$ and $y=\dfrac{50\times 7}{20}$ and $z=\dfrac{50\times 7}{20\times 3}$.

Hence we will have $x=\dfrac{21}{4}$ and $y=\dfrac{15}{2}$ and $z=\dfrac{35}{6}$.

Note: While answering this type of question we should carefully observe the corresponding sides of the similar figures. If in the case of the similar triangles we had made a mistake and taken the wrong sides as corresponding sides then we will have $\dfrac{12}{9}=\dfrac{7}{y}=\dfrac{10}{x}$ which will give us $x=\dfrac{15}{2}$ and $y=\dfrac{15}{2}$ which is a mistake.