Question types

P-2 Similarity question types

211 questions across 6 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.

211
Questions
6
Question groups
5
Question types
Sample Questions

P-2 Similarity questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

Select the appropriate alternative.
In figure 1.71, seg XY || seg BC, then which of the following statements is true?

$A.\frac{A B}{A C}=\frac{A X}{A Y}$
$B.\frac{A X}{X B}=\frac{A Y}{A C}$
$C.\frac{A X}{Y C}=\frac{A Y}{X B}$
$D.\frac{A B}{Y C}=\frac{A C}{X B}$
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Select the appropriate alternative.
ΔABC and ΔDEF are equilateral triangles, A(ΔABC) : A(ΔDEF) = 1 : 2. If AB = 4 then what is length of DE?

 
  • A
    $2 \sqrt2$
  • B
    4
  • C
    8
  • $4 \sqrt2$

Answer: D.

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Select the appropriate alternative.
In Δ and ΔDEF ∠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.
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Select the appropriate alternative.
If in ΔDEF and ΔPQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

$A.\frac{E F}{P R}=\frac{D F}{P Q}$
$B.\frac{D E}{P Q}=\frac{E F}{R P}$
$C.\frac{D E}{Q R}=\frac{D F}{P Q}$
$D.\frac{E F}{R P}=\frac{D E}{Q R}$
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Select the appropriate alternative.
In Δ ABC and ΔPQR, in a one to one correspondence $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$ then

A. Δ PQR ~ Δ ABC
B. Δ PQR ~ Δ CAB
C. Δ CBA ~ Δ PQR
D. Δ BCA ~ Δ PQR
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In figure 1.66, seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by completing the following activity.
A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x
$DF = DP +\square=\square+\square=3 x$
In Δ FDE and Δ FPQ,
∠FDE ≅ ∠ .......... corresponding angles
∠FED ≅ ∠ ......... corresponding angles
∴ Δ FDE ~ Δ FPQ .......... AA test
$\therefore \frac{ A (\Delta FDE )}{ A (\Delta FPQ )}=\frac{\square}{\square}=\frac{(3 x )^2}{(2 x )^2}=\frac{9}{4}$
$A (\Delta FDE )=\frac{9}{4} A(\Delta FPQ )=\frac{9}{4} \times \square=\square$
$A (\square DPQE )= A (\Delta FDE )- A (\Delta FPQ )$
$\quad=\square-\square$
$\quad=\square$
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In $\triangle A B C$, ray $B D$ bisects $\angle A B C . A-D-C$, side $D E \|$ side $B C, A-E-B$, then prove that $\frac{A B}{B C}=\frac{A E}{E B}$
Image
In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]

[From (i) and (ii)]
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Prove The Theorem : The ratio of the intercepts made on a transversal by three parallel lines is equal to the ratio of the corrosponding intercepts made on any other transversal by the same parallel lines.
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Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle?
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Areas of two similar triangles are 225 sq.cm & 81 sq.cm. If a side of the smaller triangle is 12 cm, then find corresponding side of the bigger triangle.
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Q 163 Marks Question3 Marks
Prove The Theorem : If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion.
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Q 183 Marks Question3 Marks
In Δ ABC, B - D – C and BD = 7, BC = 20 then find following ratios.

$(1)\frac{ A (\Delta ABD )}{ A (\Delta ADC )}$
$(2)\frac{ A (\Delta ABD )}{ A (\Delta ABC )}$
$(3)\frac{ A (\Delta ADC )}{ A (\Delta ABC )}$
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Prove The Theorem : When two triangles are similar, the ratio of areas of those triangles is equal to the ratio of the squares of their corresponding sides.
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In fig 1.78, bisectors of ∠B and ∠C of ΔABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6 then find $\frac{AX}{XY}.$
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In the figure 1.76, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD.
AB = 60, BC = 70, CD = 80, PS = 280 then find PQ, QR and RS.
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If ΔABC ~ ΔPQR and AB: PQ = 2:3, then fill in the blanks.
$\frac{ A (\triangle ABC )}{ A (\Delta PQR )}=\frac{ AB ^2}{\square}=\frac{2^2}{3^2}=\frac{\square}{\square}$
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