Question 11 Mark
All concentric circles are _________ to each other.
Answer
View full question & answer→All concentric circles are similar to each other.
Questions
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18 questions · self-marked practice — reveal the answer and mark yourself.
Question 21 Mark
Let $\triangle ABC \sim \triangle DEF$ and their areas be respectively $81 cm^2$ and $144 cm^2$. If $EF =24 cm$, then length of side BC is....... cm .
Answer
View full question & answer→Let $\triangle\text{ABC}\sim\triangle\text{DEF}$ and their areas be respectively $81cm^2$ and $144cm^2$. If EF = 24cm, then length of side BC is 18cm.Solution:
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\frac{\text{ar.}\triangle\text{ABC}}{\text{ar.}\triangle\text{DEF}}=\Big(\frac{\text{BC}}{\text{EF}}\Big)^2$
$\frac{81}{144}=\Big(\frac{\text{BC}}{24}\Big)^2$
$\frac{\text{BC}}{24}=\sqrt{\frac{81}{144}}$
$\frac{\text{BC}}{24}=\frac{9}{12}$
$\text{BC}=\frac{9}{12}\times24$
$\text{BC} = 18\text{cm}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\frac{\text{ar.}\triangle\text{ABC}}{\text{ar.}\triangle\text{DEF}}=\Big(\frac{\text{BC}}{\text{EF}}\Big)^2$
$\frac{81}{144}=\Big(\frac{\text{BC}}{24}\Big)^2$
$\frac{\text{BC}}{24}=\sqrt{\frac{81}{144}}$
$\frac{\text{BC}}{24}=\frac{9}{12}$
$\text{BC}=\frac{9}{12}\times24$
$\text{BC} = 18\text{cm}$
Question 31 Mark
In fig. MN || BC and AM : MB = 1 : 2, then $\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=$ _________. 
Answer
View full question & answer→In fig. MN || BC and AM : MB = 1 : 2, then $\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=\frac{1}{9}$ $$ Solution: Since we have given that AM : MB = 1 : 2 So, AB = 1 + 2 = 3 Since MN || BC So, Using "Area similarity theorem" ,we get that$\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AM}^2}{\text{AB}^2}=\frac{1}{3^2}=\frac{1}{9}$
Hence, the required ratio is 1 : 9.
Hence, the required ratio is 1 : 9.
Question 41 Mark
In fig. MN || BC and AM : MB = 1 : 2, then $\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=$ _________. 
Answer
View full question & answer→In fig. MN || BC and AM : MB = 1 : 2, then $\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=\frac{1}{9}$ $$ Solution: Since we have given that AM : MB = 1 : 2 So, AB = 1 + 2 = 3 Since MN || BC So, Using "Area similarity theorem" ,we get that$\frac{\text{ar}(\triangle\text{AMN})}{\text{ar}(\triangle\text{ABC})}=\frac{\text{AM}^2}{\text{AB}^2}=\frac{1}{3^2}=\frac{1}{9}$
Hence, the required ratio is 1 : 9.
Hence, the required ratio is 1 : 9.
Question 51 Mark
A ladder 10m long reaches a window 8m above the ground. The distance of the foot of the ladder from the base of the wall is _________ m.
Answer
View full question & answer→A ladder 10m long reaches a window 8m above the ground. The distance of the foot of the ladder from the base of the wall is 6m.
Solution:
Image
Let AB be A ladder and B is the window at 8m above the ground C. Now, In right triangle ABC
By using Pythagoras theorem, we have
$AB^2 = BC^2 + CA^2$
$\Rightarrow 10^2 = 8^2 + CA^2$
$\Rightarrow CA^2 = 100 - 64$
$\Rightarrow CA^2 = 36$
$\Rightarrow CA = 6m.$
Hence, the distance of the foot of the ladder from the base of the wall is 6m.
Solution:
Image
Let AB be A ladder and B is the window at 8m above the ground C. Now, In right triangle ABC
By using Pythagoras theorem, we have
$AB^2 = BC^2 + CA^2$
$\Rightarrow 10^2 = 8^2 + CA^2$
$\Rightarrow CA^2 = 100 - 64$
$\Rightarrow CA^2 = 36$
$\Rightarrow CA = 6m.$
Hence, the distance of the foot of the ladder from the base of the wall is 6m.
Question 61 Mark
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is _____________.
Answer
View full question & answer→ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is 4 : 1. Solution: 
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are both equilateral triangles.
$\therefore\triangle\text{ABC}\sim\text{BDE}$
[Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2}$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2$
$\Rightarrow[\therefore\text{AB}=\text{BC}=\text{CA}]$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{2\text{BD}}{\text{BD}}\Big)^2$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac41$
Hence, 4 : 1 is the correct answer.
$\triangle\text{ABC}$ and $\triangle\text{BDE}$ are both equilateral triangles.$\therefore\triangle\text{ABC}\sim\text{BDE}$
[Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2}$
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2$
$\Rightarrow[\therefore\text{AB}=\text{BC}=\text{CA}]$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{2\text{BD}}{\text{BD}}\Big)^2$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac41$
Hence, 4 : 1 is the correct answer.
Question 71 Mark
Given $\triangle\text{ABC}\sim\triangle\text{PQR},$ if $\frac{\text{AB}}{\text{PQ}}=\frac13,$ then $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR)}}=$ _________.
Answer
View full question & answer→Given $\triangle\text{ABC}\sim\triangle\text{PQR},$ if $\frac{\text{AB}}{\text{PQ}}=\frac13,$ then $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR)}}=$ $\frac19.$ Solution:$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{PQR)}}=\frac{\text{(AB)}^2}{(\text{PQ})^2}(\frac13)^2=\frac19.$
Question 81 Mark
ABC is an equilateral triangle of side 2a, then length of one of its altitude is _______.
Answer
View full question & answer→ABC is an equilateral triangle of side 2a, then length of one of its altitude is $\sqrt3\text{a}.$ Solution: ABC is an equilateral triangle side = 2a AB = AC = BC = 2a
In $\triangle\text{ACM}$$\sin60^\circ=\frac{\text{CM}}{\text{AC}}$
$\frac{\sqrt3}{2}=\frac{\text{CM}}{2\text{a}}$
$\text{CM}=2\text{a}\times\frac{\sqrt3}{2}$
Altitude $\text{CM}=\sqrt{3}\text{a}$
In $\triangle\text{ACM}$$\sin60^\circ=\frac{\text{CM}}{\text{AC}}$
$\frac{\sqrt3}{2}=\frac{\text{CM}}{2\text{a}}$
$\text{CM}=2\text{a}\times\frac{\sqrt3}{2}$
Altitude $\text{CM}=\sqrt{3}\text{a}$
Question 91 Mark
Two triangles are similar, if their corresponding angles are .......... (proportional, equal).
Answer
View full question & answer→Two triangles are similar, if their corresponding angles are equal.
Question 101 Mark
Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are ........... (equal, proportional).
Answer
View full question & answer→Two polygons of the same number of sides are similar, if (a) their corresponding angles are and (b) their corresponding sides are proportional.
Question 111 Mark
All circles are ________ (congruent, similar).
Answer
View full question & answer→All circles are similar.
Question 121 Mark
All squares are ________. (similar, congruent)
Answer
View full question & answer→All squares are similar.
Question 131 Mark
All circles are ......... (congruent, similar).
Answer
View full question & answer→All circles are similar.
Question 141 Mark
All squares are ........ (similar, congruent).
Answer
View full question & answer→All squares are similar.
Question 151 Mark
All ________ triangles are similar. (isosceles, equilateral)
Answer
View full question & answer→All equilateral triangles are similar.
Question 161 Mark
All .......... triangles are similar (isosceles, equilateral).
Answer
View full question & answer→All equilateral triangles are similar.
Question 171 Mark
Two triangles are similar, if their corresponding sides are ........... (proportional, equal).
Answer
View full question & answer→Two triangles are similar, if their corresponding sides are proportional.
Question 181 Mark
Two polygons of the same number of sides are similar, if (a) their corresponding angles are ________ and (b) their corresponding sides are ________ . (equal, proportional)
Answer
View full question & answer→Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.