Question

Select the appropriate alternative.
(1) In ∆ABC and ∆PQR, in a one to one correspondence $\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$ then

(A) ∆PQR ~ ∆ABC
(B) ∆PQR ~ ∆CAB
(C) ∆CBA ~ ∆PQR
(D) ∆BCA ~ ∆PQR

(2) If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

(A) $\frac{\mathrm{EF}}{\mathrm{PR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$ (B) $\frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{RP}}$

(C) $\frac{\mathrm{DE}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PQ}}$ (D) $\frac{\mathrm{EF}}{\mathrm{RP}}=\frac{\mathrm{DE}}{\mathrm{QR}}$



(3) In ∆ABC and ∆DEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true ?
(A)The triangles are not congruent and not similar
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.


(4) ∆ABC and ∆DEF are equilateral triangles, A (∆ABC) : A (∆DEF) = 1 : 2
If AB = 4 then what is length of DE?
(A) $2\sqrt{2}$
(B) 4
(C) 8
(D) $4\sqrt{2}$



(5) In the given figure, seg XY || seg BC, then which of the following statements is true?

(A) $\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AX}}{\mathrm{AY}}$ (B) $\frac{\mathrm{AX}}{\mathrm{XB}}=\frac{\mathrm{AY}}{\mathrm{AC}}$

(C) $\frac{\mathrm{AX}}{\mathrm{YC}}=\frac{\mathrm{AY}}{\mathrm{XB}}$ (D) $\frac{\mathrm{AB}}{\mathrm{YC}}=\frac{\mathrm{AC}}{\mathrm{XB}}$

Answer 1

(1)
Given: $\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{PR}}=\frac{\mathrm{CA}}{\mathrm{PQ}}$
By SSS test of similarity
∆PQR ~ ∆CAB
Hence, the correct option is (B).

(2)
In ∆DEF and ∆PQR
∠D ≅ ∠Q
∠R ≅ ∠E
By AA test of similarity
∆DEF~ ∆PQR
$\therefore \frac{\mathrm{DE}}{\mathrm{PQ}}=\frac{\mathrm{EF}}{\mathrm{QR}}=\frac{\mathrm{DF}}{\mathrm{PR}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}\right)$
$\therefore \frac{\mathrm{DE}}{\mathrm{PQ}}\ne \frac{\mathrm{EF}}{\mathrm{RP}}$
Hence, the correct option is (B).

(3)
In ∆ABC and ∆DEF
∠B = ∠E,
∠F = ∠C
By AA test of similarity
∆ABC ~ ∆DEF
Since, there is not any congruency criteria like AA.
Thus, ∆ABC and ∆DEF are not congruent.
Hence, the correct option is (B).

(4)
Given: ∆ABC and ∆DEF are equilateral triangles
Constrcution: Draw a perependicular from vertex A and D on AC and DF in both triangles.

In ∆ABX and ∆DEY
∠B = ∠C = 60^∘ (∆ABC and ∆DEF are equilateral triangles)
∠AXB = ∠DYB (By construction)
By AA test of similarity
∆ABX ~ ∆DEY
$\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AX}}{\mathrm{DY}}\left(\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}\right)$
$\therefore \frac{\mathrm{DE}}{\mathrm{PQ}}\ne \frac{\mathrm{EF}}{\mathrm{RP}}$

$$\begin{gathered} \frac{\mathrm{A}\left(△\mathrm{ABC}\right)}{\mathrm{A}\left(△\mathrm{DEF}\right)}=\frac{1}{2} \\ \Rightarrow \frac{{\displaystyle \frac{1}{2}\times \mathrm{AB}\times \mathrm{AX}}}{{\displaystyle \frac{1}{2}\times \mathrm{DE}\times \mathrm{DY}}}=\frac{1}{2} \\ \Rightarrow \frac{{\mathrm{AB}}^2}{{\mathrm{DE}}^2}=\frac{{\displaystyle 1}}{{\displaystyle 2}}\left(\because \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{AX}}{\mathrm{DY}}\right) \\ \Rightarrow {\mathrm{DE}}^2=32 \\ \Rightarrow \mathrm{DE}=4\sqrt{2} \end{gathered}$$
Hence, the correct option is (D).

(5)
Given: seg XY || seg BC
By basic proportionality theorem
$$\begin{gathered} \frac{\mathrm{AX}}{\mathrm{BX}}=\frac{\mathrm{AY}}{\mathrm{YC}} \\ \Rightarrow \frac{\mathrm{BX}}{\mathrm{AX}}+1=\frac{\mathrm{YC}}{\mathrm{AY}}+1 \\ \Rightarrow \frac{\mathrm{BX}+\mathrm{AX}}{\mathrm{AX}}=\frac{\mathrm{YC}+\mathrm{AY}}{\mathrm{AY}} \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{\mathrm{AB}}{\mathrm{AX}}=\frac{\mathrm{AC}}{\mathrm{AY}} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{AX}}{\mathrm{AY}} \end{gathered}$$
Hence, the correct option is (D).