Solutions: Triangles - Test 1

Q1. (b) 18 cm

Ratio of sides = $AB/DE = 4/6 = 2/3$. Perimeter of $\Delta DEF = 6+9+12 = 27$.
Perimeter of $\Delta ABC = (2/3) \times 27 = 18$ cm.

Q2. (a) 4

By BPT: $\frac{x}{x-2} = \frac{x+2}{x-1} \Rightarrow x^2 - x = x^2 - 4 \Rightarrow x = 4$.

Q3. (c) 2:3

Ratio of areas = (Ratio of medians)$^2$. Ratio of medians = $\sqrt{4/9} = 2:3$.

Q4. (c) A is true but R is false.

Congruent figures are always similar, but similar figures are not necessarily congruent.

Q5. By BPT: $\frac{AD}{DB} = \frac{AE}{EC}$.
$\frac{1.5}{3} = \frac{1}{EC} \Rightarrow \frac{1}{2} = \frac{1}{EC} \Rightarrow EC = 2$ cm. [2]

Q6. $\Delta AOB \sim \Delta COD$ (AA similarity).
$\frac{\text{Area}(\Delta AOB)}{\text{Area}(\Delta COD)} = (\frac{AB}{CD})^2 = (\frac{2CD}{CD})^2 = (\frac{2}{1})^2 = 4:1$. [2]

Q7. Ratio of perimeters = Ratio of corresponding sides.
$\frac{32}{48} = \frac{AC}{6} \Rightarrow \frac{2}{3} = \frac{AC}{6} \Rightarrow AC = 4$ cm. [2]

Q8. In $\Delta PQR$, $\angle 1 = \angle 2 \Rightarrow PQ = PR$ (Sides opposite to equal angles).
Given $\frac{QR}{QS} = \frac{QT}{PR}$. Substitute $PR = PQ$: $\frac{QR}{QS} = \frac{QT}{PQ}$.
In $\Delta PQS$ and $\Delta TQR$: $\frac{QS}{QR} = \frac{PQ}{QT}$ (Reciprocal) and $\angle Q$ is common.
By SAS similarity, $\Delta PQS \sim \Delta TQR$. [3]

Q9. Let height of tower be $h$.
$\Delta ABC \sim \Delta PQR$ (Sun's elevation is same).
$\frac{\text{Height of pole}}{\text{Shadow of pole}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}$.
$\frac{6}{4} = \frac{h}{28} \Rightarrow h = \frac{6 \times 28}{4} = 6 \times 7 = 42$ m. [3]

Q10. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. [1]

Given: $\Delta ABC$, $DE \parallel BC$. To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.

Construction: Join BE and CD. Draw $DM \perp AC$ and $EN \perp AB$. [1]

Proof: Area($\Delta ADE$) = $\frac{1}{2} \times AD \times EN$. Area($\Delta BDE$) = $\frac{1}{2} \times DB \times EN$.
Ratio = $\frac{AD}{DB}$. Similarly, Ratio of Area($\Delta ADE$) to Area($\Delta DEC$) = $\frac{AE}{EC}$.
Since $\Delta BDE$ and $\Delta DEC$ are on same base DE and between parallel lines DE and BC, their areas are equal.
Hence, $\frac{AD}{DB} = \frac{AE}{EC}$. [3]

(i) Scale 1:50. $\frac{\text{Model}}{\text{Actual}} = \frac{1}{50}$.
$\frac{20}{H} = \frac{1}{50} \Rightarrow H = 1000$ cm = 10 m. [1]

(ii) Scale = $\frac{\text{Drawing}}{\text{Actual}} = \frac{5 \text{ cm}}{1000 \text{ cm}} = \frac{1}{200}$ or 1:200. [1]

(iii) Ratio of areas = (Scale factor)$^2 = (3/5)^2 = 9/25$.
$\frac{81}{\text{Area}} = \frac{9}{25} \Rightarrow \text{Area} = \frac{81 \times 25}{9} = 9 \times 25 = 225$ sq cm. [2]