Triangles Test

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Triangles Test - 32

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Weekly Quiz Competition

Question 1
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If $$\triangle ABC\sim \triangle PQR,$$ $$ \cfrac{ar(ABC)}{ar(PQR)}=\cfrac{9}{4}$$, $$AB=18$$ $$cm$$ and $$BC=15$$ $$cm$$, then $$QR$$ is equal to:

A
$$10$$ $$cm$$

B
$$12$$ $$cm$$

C
$$\cfrac{20}{3}$$ $$cm$$

D
$$8$$ $$cm$$

Solution

Given, $$\triangle ABC \sim \triangle PQR$$,
Then, $$\dfrac{ar(ABC)}{ar(PQR)} = \dfrac{AB^2}{PQ^2} = \dfrac{BC^2}{QR^2} = \dfrac{AC^2}{PR^2}$$
$$\dfrac{9}{4} = \dfrac{BC^2}{QR^2}$$
$$\dfrac{9}{4} = \dfrac{15^2}{QR^2}$$
$$QR^2 = \dfrac{4 \times 225}{9}$$
$$QR^2 = 100$$
$$QR = 10 \ cm$$
Question 2
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It is given that $$\triangle ABC \sim \triangle PQR$$, $$A(\triangle ABC)=36{cm}^{2}$$ and $$A(\triangle PQR)=25{cm}^{2}$$. If $$QR=6$$cm, find the length of $$BC$$.

A
$$7.2$$cm

B
$$3.6$$cm

C
$$4.8$$cm

D
$$6$$cm

Solution

We know that the area of similar triangles ae proportional to the squares of their corresponding sides
$$\therefore$$ $$\cfrac{A (\triangle ABC)}{A(\triangle PQR)}=\cfrac{{BC}^{2}}{{QR}^{2}}$$
Let $$BC=x$$ cm. Then,
$$\cfrac {36}{25}=\cfrac{{x}^{2}}{{6}^{2}}$$
$${x}^{2}=\cfrac{36\times 36}{25}$$
$$x=(\cfrac{6\times 6}{5})$$
$$x=\cfrac{36}{5}=7.2$$ cm.
Question 3
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$$\triangle ABC$$ and $$\triangle BDE$$ are two equilateral triangles such that $$D$$ is the midpoint of $$BC$$. Ratio of $$A(\triangle ABC)$$ and $$A(\triangle BDE)$$ is:

A
$$2:1$$

B
$$1:2$$

C
$$4:1$$

D
$$1:4$$

Solution

Given: $$\triangle ABC$$ and $$\triangle BDE$$ are equilateral triangles.
D is midpoint of BC.
Since, $$\triangle ABC$$ and $$\triangle BDE$$ are equilateral triangles.
All the angles are $$60^{\circ}$$ and hence they are similar triangles.
Ratio of areas of similar triangles is equal to ratio of squares of their sides:
Now, $$\dfrac{A(\triangle BDE)}{A( \triangle ABC)} = \dfrac{BC^2}{BD^2}$$
$$\dfrac{A (\triangle ABC)}{A(\triangle BDE)} = \dfrac{(2 BD)^2}{BD^2}$$ ....Since $$BC = 2 BD$$
$$\dfrac{A( \triangle ABC)}{A(\triangle BDE)} = 4 : 1$$
Question 4
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Let $$\triangle ABC\sim \triangle DEF$$ and their areas be, respectively $$64\ {cm}^{2}$$ and $$121\ {cm}^{2}$$. If $$EF=15.4\ cm$$, find $$BC$$.

A
$$11.2\ cm$$

B
$$11.6\ cm$$

C
$$11.4\ cm$$

D
$$10.8\ cm$$

Solution

$$\triangle ABC\sim \triangle DEF\quad $$ (Given)
$$\Rightarrow \cfrac { ar(ABC) }{ ar(DEF) } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } $$ (ratio of Areas of Similar triangles are equal to ratio of squares of corresponding sides)
$$\Rightarrow \quad \cfrac { 64 }{ 121 } =\cfrac { { BC }^{ 2 } }{ { EF }^{ 2 } } \quad { \left\{ \cfrac { BC }{ EF } \right\} }^{ 2 }={ \left\{ \cfrac { 8 }{ 11 } \right\} }^{ 2 }$$
$$\Rightarrow \quad \cfrac { BC }{ EF } =\cfrac { 8 }{ 11 } \quad \Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times EF$$
$$\Rightarrow \quad BC=\cfrac { 8 }{ 11 } \times 15.4cm=11.2cm$$
Question 5
1 / -0

In figure, $$DE\parallel BC$$. Find the length of $$EC$$.

A
$$3 cm$$

B
$$4.5 cm$$

C
$$1.5 cm$$

D
$$2 cm$$

Solution

In figure $$DE\parallel BC$$
$$\Rightarrow$$ $$\cfrac{AD}{DB}=\cfrac{AE}{EC}$$ (By basic proportionality theorem)
$$\Rightarrow$$ $$\cfrac{1.5}{3}=\cfrac{1}{EC}$$
($$\because AD=1.5cm, DB=3cm$$ and $$AE=1cm$$)
$$\Rightarrow$$ $$EC=\cfrac{3}{1.5}=2cm$$.
Question 6
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In the diagram given below, $$\displaystyle \angle ABD=\angle CDB=\angle PQD=90^{\circ}.$$ If $$ AB:CD=3:1,$$ the ratio of CD : PQ is

A
$$1 : 0.69$$

B
$$1 : 0.75$$

C
$$1 : 0.72$$

D
None of the above

Solution

$$\triangle s ABD \ ,\ BDC \ ,\ DPQ$$ are similar by AAA, as the corresponding angles are equal.

$$\dfrac{AB}{PQ}=\dfrac{BD}{QD}$$ ..........(1)

$$\dfrac{CD}{PQ}=\dfrac{BD}{BQ}$$ ..........(ii)
Given,
$$\dfrac{AB}{CD}=\dfrac{3}{1}$$ $$\implies AB=3CD$$ ..........(iii)
Substitute for $$BD$$ in (i) from (ii)

$$\dfrac{AB}{PQ}=\dfrac{CD}{PQ}\dfrac{BQ}{QD}$$

$$3CD=\dfrac{CD}{1}\dfrac{BQ}{QD}$$------------(from(iii))

$$\dfrac{BQ}{QD}=\dfrac{3}{1}$$------------(iv)
Consider,
$$\dfrac{CD}{PQ}=\dfrac{BD}{BQ}$$

$$\dfrac{CD}{PQ}=\dfrac{BQ+QD}{BQ}=1+\dfrac{QD}{BQ}$$

$$\dfrac{CD}{PQ}=1+\dfrac{1}{3}=\dfrac{4}{3}$$

$$\dfrac{CD}{PQ}=\dfrac{1}{0.75}$$
$$\therefore CD:PQ=1:0.75$$.
Question 7
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In $$\displaystyle \Delta ABC, PQ\parallel BC$$ and area of quadrilateral $$PBCQ = 42\ sq. cm$$. If $$AP : PB = 2 : 3,$$ then find the area of $$\displaystyle \Delta APQ$$.

A
$$28\ sq. cm$$

B
$$\displaystyle \dfrac{56}{3}\ sq. cm$$

C
$$8\ sq. cm$$

D
$$33.6\ sq. cm$$

Solution

In $$\triangle ABC$$, $$PQ \parallel BC$$

$$\angle BAC = \angle PAQ$$ ...... (Common angle)

$$\angle ABC = \angle APQ$$ ....... (Corresponding angles)

Thus, $$\triangle ABC \sim \triangle APQ$$

$$\Rightarrow \dfrac{Ar(\triangle ABC)}{Ar(\triangle APQ)} = \dfrac{AB^2}{AP^2}$$

$$\Rightarrow \dfrac{Ar(\triangle APQ) + Ar( PBCQ)}{Ar(\triangle APQ)} = \dfrac{25}{4}$$

$$\Rightarrow \dfrac{x + 42}{x} = \dfrac{25}{4}$$

$$\Rightarrow 4x + 168 = 25x$$

$$\Rightarrow 21x = 168$$

$$\Rightarrow x = 8$$

Thus, $$Ar(\triangle APQ) = 8\ sq. cm$$
Question 8
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The corresponding sides of two similar triangles are in the ratio $$2$$ : $$3$$. If the area of the smaller triangle is $$12$$ the area of the larger is

A
$$24$$

B
$$27$$

C
$$18$$

D
$$8$$

Solution

The area of similar triangles is in the ratio of the square of the corresponding sides.

Hence,
$$\dfrac{\text{Area of smaller triangle}}{\text{Area of larger triangle}} = \left(\dfrac{2}{3}\right)^2$$
$$\Rightarrow \dfrac{12}{\text{Area of larger triangle}} = \dfrac{4}{9}$$
$$\Rightarrow \text{Area of larger triangle} = 27$$

Hence, option $$B$$ is correct.
Question 9
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In a $$\displaystyle \Delta ABC$$, $$D$$ is any point on $$AB$$ and $$DE || BC $$ meets $$AC$$ at E if $$AD = 2.5, DB = 6$$ cm and $$AE = 3$$ cm, then $$EC = $$

A
$$9.5$$ cm

B
$$7.2$$ cm

C
$$6.5$$ cm

D
$$12.5$$ cm

Solution

Given: In $$\triangle ABC $$
$$DE \parallel BC$$
Then, $$\dfrac{AD}{DB} = \dfrac{AE}{EC}$$ ...(Basic Proportionality theorem)
$$\dfrac{2.5}{6} = \dfrac{3}{EC}$$
$$EC = \dfrac{18}{2.5}$$.
$$EC = 7.2$$.
Question 10
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If $$\triangle ABC$$ is right angled at $$B$$ and $$M,$$ $$N$$ are the midpoints of $$AB$$ and $$BC$$ respectively, then $$4\left ( AN^{2}+CM^{2} \right )=$$

A
$$4AC^{2}$$

B
$$5AC^{2}$$

C
$$\displaystyle \frac{5}{4}AC^{2}$$

D
$$6AC^{2}$$

Solution

Given, $$\triangle ABC$$, $$M$$ is mid point of $$AB$$ and $$N$$ is mid point of $$BC.$$
In $$\triangle ABN,$$
$$AN^2 = AB^2 + BN^2$$ (Pythagoras Theorem)
$$AN^2 = AB^2 + (\dfrac{BC}{2})^2$$ $$....(1)$$

In $$\triangle BMC,$$
$$MC^2 = BM^2 + BC^2$$ (Pythagoras Theorem)
$$MC^2 = BC^2 + (\dfrac{AB}{2})^2$$ $$....(2)$$
Add $$(1)$$ and $$(2),$$
$$AN^2 + MC^2 = AB^2 + (\dfrac{BC}{2})^2 +BC^2 + (\dfrac{AB}{2})^2$$

$$AN^2 + MC^2 = \dfrac{5}{4} AB^2 + \dfrac{5}{4} BC^2$$

$$4(AN^2 + MC^2) = 5 (AB^2 + BC^2)$$

$$4 (AN^2 + MC^2) = 5 AC^2$$ (Pythagoras Theorem in $$\triangle ABC$$)