**10.2.1 Transformations II, PT3 Focus Practice**

Question 1:

Question 1:

In the diagram, ∆

*P’Q’R’*is the image of ∆*PQR*under an enlargement. (a) State the scale factor of the enlargement.

*PQR*= 7 cm

^{2}, calculate the area of ∆

*P’Q’R’.*(b) If the area of ∆

Solution:Solution:

**(a)**

$\text{Scalefactor,}k=\frac{P\text{'}Q\text{'}}{PQ}=\frac{6}{3}=2$

**(b)**

Area of image =

*k*^{2 }× Area of objectArea of ∆

*P’Q’R’*= 2^{2 }× 7 =

**28 cm**^{2 }^{}**Question 2:**

In diagram below,

*OP’Q’*is the image of*OPQ*under an enlargement with centre*O*.Given

*PQ*= 4cm, calculate the length, in cm, of*P’Q’*.

Solution:Solution:

**Question 3:**

In diagram below,

*PQ’R’S’*is the image of*PQRS*under an enlargement.Calculate the length, in cm, of

*SS’*.

Solution:Solution:

**Question 4:**

On the Cartesian plane,

*Q’R’S’*is the image of ∆*QRS*under an enlargement of centre*T*.State the coordinates of

*T.*

Solution:Solution:

Coordinates of

*T*= (4, 4).**Question 5:**

In diagram below, quadrilateral

*AFGH*is the image of*ABCD*under an enlargement.*AFGH*is 15cm

^{2}. Find the area, in cm

^{2}, represented by the shaded region. (b) The area represented by the quadrilateral

Solution:Solution:

**(a)**

$\begin{array}{l}\text{Scale factor}\\ =\frac{3}{6+3}=\frac{3}{9}=\frac{1}{3}\end{array}$

**(b)**

Area of image =

*k*^{2 }× Area of object $\begin{array}{l}15={\left(\frac{1}{3}\right)}^{2}\times \text{Areaofobject}\\ \text{Areaofobject}=15\times 9=135\\ \\ \text{Areaofshadedregion}=135-15\\ \text{}=120c{m}^{2}\end{array}$