Solutions: Triangles - Test 2

Q1. (a) 11.2 cm

$\frac{\text{Area}(ABC)}{\text{Area}(DEF)} = (\frac{BC}{EF})^2 \Rightarrow \frac{64}{121} = (\frac{BC}{15.4})^2$.
$\frac{8}{11} = \frac{BC}{15.4} \Rightarrow BC = \frac{8 \times 15.4}{11} = 8 \times 1.4 = 11.2$ cm.

Q2. (b) 7.2 cm

By BPT, $\frac{AD}{AB} = \frac{AE}{AC}$. Given $\frac{AD}{DB} = \frac{2}{3} \Rightarrow \frac{AD}{AB} = \frac{2}{5}$.
$\frac{2}{5} = \frac{AE}{18} \Rightarrow AE = \frac{36}{5} = 7.2$ cm.

Q3. (a) 10 cm

$\Delta ABC \sim \Delta QRP$. $\frac{\text{Area}(ABC)}{\text{Area}(QRP)} = (\frac{BC}{RP})^2 = \frac{9}{4}$.
$\frac{BC}{RP} = \frac{3}{2} \Rightarrow \frac{15}{RP} = \frac{3}{2} \Rightarrow RP = 10$ cm.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Reason defines the AAA similarity criterion which leads to similarity.

Q5. Let height of tower be $h$.
$\frac{\text{Height of stick}}{\text{Shadow of stick}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}$.
$\frac{20 \text{ cm}}{16 \text{ cm}} = \frac{h}{48 \text{ m}} \Rightarrow \frac{5}{4} = \frac{h}{48}$.
$h = \frac{5 \times 48}{4} = 60$ m. [2]

Q6. $\Delta ABE \cong \Delta ACD \Rightarrow AB = AC$ and $AE = AD$ (CPCT).
$\Rightarrow \frac{AD}{AB} = \frac{AE}{AC}$. Also $\angle A$ is common.
By SAS similarity, $\Delta ADE \sim \Delta ABC$. [2]

Q7. In $\Delta ABE$ and $\Delta CFB$:
$\angle A = \angle C$ (Opposite angles of parallelogram).
$\angle AEB = \angle CBF$ (Alternate interior angles, $AE \parallel BC$).
By AA similarity, $\Delta ABE \sim \Delta CFB$. [2]

Q8. In $\Delta ADC$ and $\Delta BAC$:
$\angle ADC = \angle BAC$ (Given). $\angle C = \angle C$ (Common).
$\Delta ADC \sim \Delta BAC$ (AA similarity).
$\frac{CA}{CB} = \frac{CD}{CA} \Rightarrow CA^2 = CB \cdot CD$. [3]

Q9. Given $\frac{AO}{BO} = \frac{CO}{DO} \Rightarrow \frac{AO}{CO} = \frac{BO}{DO}$. Also $\angle AOB = \angle COD$.
$\Delta AOB \sim \Delta COD$ (SAS). $\Rightarrow \angle OAB = \angle OCD$.
These are alternate interior angles, so $AB \parallel DC$. Hence $ABCD$ is a trapezium. [3]

Q10. Theorem: Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given: $\Delta ABC \sim \Delta PQR$. To Prove: $\frac{ar(ABC)}{ar(PQR)} = (\frac{AB}{PQ})^2 = (\frac{BC}{QR})^2 = (\frac{CA}{RP})^2$.
Construction: Draw $AM \perp BC$ and $PN \perp QR$. [1]
Proof: $ar(ABC) = \frac{1}{2} BC \times AM$, $ar(PQR) = \frac{1}{2} QR \times PN$. Ratio = $\frac{BC \times AM}{QR \times PN}$.
In $\Delta ABM$ and $\Delta PQN$: $\angle B = \angle Q$ (Similar triangles), $\angle M = \angle N = 90^\circ$.
$\Delta ABM \sim \Delta PQN \Rightarrow \frac{AM}{PN} = \frac{AB}{PQ}$.
Also $\frac{AB}{PQ} = \frac{BC}{QR}$. So $\frac{AM}{PN} = \frac{BC}{QR}$.
Ratio of areas = $\frac{BC}{QR} \times \frac{BC}{QR} = (\frac{BC}{QR})^2$. [4]

(i) AA Similarity (Angle of incidence = Angle of reflection, and 90 degree angles). [1]

(ii) Let height of tree be $H$. $\frac{H}{12} = \frac{1.5}{1.5} \Rightarrow H = 12$ m. [1]

(iii) Tree height $H=12$. Mirror distance from tree = 8 m.
$\frac{12}{8} = \frac{1.5}{x} \Rightarrow 1.5x = 12 \times 1.5 / 12$? No.
$\frac{12}{8} = \frac{1.5}{x} \Rightarrow \frac{3}{2} = \frac{1.5}{x} \Rightarrow 3x = 3 \Rightarrow x = 1$ m.
He must stand 1 m from the mirror. [2]