Triangles Test

Result

Triangles Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The areas of two similar triangles are $$16\ cm^{2}$$ and $$36\ cm^{2}$$ respectively. If the altitude of the first triangle is $$3\ cm$$, then the corresponding altitude of the other triangle is

A
$$6.5\ cm$$

B
$$6\ cm$$

C
$$4\ cm$$

D
$$4.5\ cm$$

Solution

Given: Area of $$two$$ similar triangle $$16{cm}^{2}$$ and $$36{cm}^{2}$$
Altitude of first triangle $$=3cm$$
For similar triangle,
$${\text{Ratio of altitudes}}^{2}=\text {Ratio of their Area}$$
$$\therefore$$ $$\cfrac { { 3 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 16 }{ 36 } $$
$$\cfrac { 9}{ { x }^{ 2 }\ } =\cfrac {16}{ 36 } $$
$$16{x}^{2}=9\times 36$$
$${x}^{2}=\cfrac{9\times 36}{16}$$
$${x}^{2}=\cfrac{324}{16}=20.25$$
$$x=4.5cm$$
Question 2
1 / -0

In the figure, $$DE\parallel BC$$ and $$\dfrac {AD}{BD} = \dfrac {3}{5}$$, calculate the value of $$\dfrac {area\ of\ \triangle ADE}{area\ of\ \triangle ABC}$$,

A
$$\dfrac {9}{64}$$

B
$$\dfrac {19}{64}$$

C
$$\dfrac {9}{34}$$

D
None of these

Solution
Question 3
1 / -0

Triangles $$ABC$$ and $$DEF$$ are similar. If their areas are $$100\ cm^{2}$$ and $$49\ cm^{2}$$ respectively and $$BC$$ is $$8.2\ cm$$ then $$EF =$$

A
$$5.47\ cm$$

B
$$5.74\ cm$$

C
$$6.47\ cm$$

D
$$6.74\ cm$$

Solution

Given: Area of similar triangles $$100{cm}^{2}$$ and $$49{cm}^{2}$$
$$BC=8.2cm$$
Let ,$$EF=x$$
For similar triangle,
$${\text{Ratio of corresponding sides}}^{2}=\text {Ratio of their Area}$$
$$\therefore$$ $$\cfrac { { 8.2 }^{ 2 } }{ { x }^{ 2 } } =\cfrac { 100 }{ 49 } $$
$$\cfrac { 67.24}{ { x }^{ 2 }\ } =\cfrac { 100 }{ 49 } $$
$$100{x}^{2}=67.24\times 49$$
$${x}^{2}=\cfrac{67.24\times 49}{100}$$
$${x}^{2}=32.9476$$
$$x=5.74cm$$
$$\Rightarrow EF=5.74cm$$
Question 4
1 / -0

If the sides of two similar triangles are in the ratio $$2:3$$, then their areas are in the ratio

A
$$9:4$$

B
$$4:9$$

C
$$2:3$$

D
$$3:2$$

Solution

Let $$ABC$$ and $$DEF$$ be two similar triangles.

Then, $$\dfrac {AB}{DE}=\dfrac {BC}{EF}=\dfrac {AC}{DF}=\dfrac {2}{3}$$ [Given]

The ratio of areas of these triangles will be equal to the square of the ratio of the corresponding sides

$$\dfrac {Area (ABC)}{Area (DEF)}=\left (\dfrac {2}{3}\right)^2$$
$$=\dfrac {4}{9}$$

Hence, their areas are in the ratio of $$4:9$$.

Therefore, option $$B$$ is correct.
Question 5
1 / -0

In the given figure, $$XY\parallel BC$$, then $$\cfrac { AX }{ BX } =$$

A
$$\cfrac { AY }{ AC } $$

B
$$\cfrac { YC }{ AY } $$

C
$$\cfrac { AX }{ AB } $$

D
$$\cfrac { AY }{ CY } $$

Solution

Given : $$XY \parallel BC$$.
Then the ratio two parts of one side is equal to the ratio of two parts of other side.(Basic Proportionality)
Hence $$\dfrac{AX}{XB}=\dfrac{AY}{YC}$$.
Question 6
1 / -0

__________ triangles become always similar.

A
Acute-angled

B
Equilateral

C
Obtuse-angled

D
None of the above

Solution

If both triangles are equilateral, so all angles measure $$60^o$$ each.
$$\Delta ABC \sim \Delta PQR$$ ....A.A.A test of similarity

Therefore, all equilateral triangles are always similar.
Question 7
1 / -0

In the figure, $$\overline{XY} || BC$$
AX = 1 cm
XB = 3 cm
BC = 6 cm
then XY = ................

A
2 cm

B
1.5 cm

C
1 cm

D
3 cm

Solution

Given $$\overline{XY} || \overline{BC}$$
In $$\triangle AXY$$ and $$\triangle ABC$$,
$$\angle XAY=\angle BAC$$ ....... [Same vertex]
$$\angle AXY=\angle ABC$$ and $$\angle AYX=\angle ACB$$ ...... $$[\because XY\parallel BC]$$
$$\therefore \triangle AXY \sim \triangle ABC$$ ........ [By AAA criterion]
$$\therefore \dfrac{AX}{AB} = \dfrac{XY}{BC}$$ .........$$(i)$$ (Corresponding sides of similar triangle are proportional)
Now, $$AB = AX + XB = 1 + 3=4$$
$$\implies AB=4$$
$$\therefore \dfrac{1}{4} = \dfrac{XY}{6}$$ ........ From $$(i)$$

$$\therefore XY = \dfrac{6}{4}$$
$$\therefore XY = 1.5 cm$$
Hence, the answer is $$1.5$$ cm.
Question 8
1 / -0

D, E and F are respectively mid points of sides $$\overline{AB}, \overline{BC}$$ and $$\overline{CA}$$ of $$\Delta ABC$$. Which of the following correspondence between $$\Delta ABC$$ and $$\Delta DEF$$ of similarity.

A
$$ABC \sim EFD$$

B
$$ABC \sim FED$$

C
$$ABC \sim DEF$$

D
$$ABC \sim EDF$$

Solution

D, E and F are the mid-points of the sides AB, BC and CA of $$\Delta$$ ABC.
FD || BC
$$FD = \dfrac{1}{2} BC \Rightarrow \dfrac{FD}{BC} = \dfrac{1}{2}$$
DE || AC
$$DE = \dfrac{1}{2} AC \Rightarrow \dfrac{DE}{AC} = \dfrac{1}{2}$$
EF || AB
$$EF = \dfrac{1}{2} AB \Rightarrow \dfrac{EF}{AB} = \dfrac{1}{2}$$

Then, quadrilaterals $$ADEF, \,BEFD, \,CFDE$$ will be $$||gm$$
$$\Rightarrow \angle A=\angle E\,and\,\angle B=\angle F\,and\,\angle C=\angle D$$
$$\therefore ABC\sim EFD$$
Question 9
1 / -0

In $$\triangle ABC, \angle B=90^{o}$$. $$\square PQRS$$ is a square of side $$5$$ units and $$\square WJYZ$$ is a square of side $$10$$ units. $$MS$$ and $$SY$$ are the length and breadth respectively of rectangle $$MNSY$$. What is the length of $$MS$$ ?

A
$$10$$ units

B
$$12$$ units

C
$$15$$ units

D
$$20$$ units

Solution

Lets observe few points.
$$\angle MPQ =\angle BMN$$
$$\angle PMQ = \angle BNM$$ (because sum has to be $$180^{\circ}$$)
Also,
$$\angle BNM =\angle NJW$$
$$\angle BMN=\angle WNJ$$
Thus we can say,
$$\angle MPQ =\angle WNJ$$
$$\angle PMQ =\angle NJW$$

Finally, since all $$3$$ angles of $$\triangle PQM= \triangle NWJ$$
They are similar by AAA condition.
Let length $$NW=x$$,
then $$MQ=x+5$$
By ratio of sides of similar triangles,
$$\dfrac{5}{x+5}$$ = $$\dfrac{x}{10}$$
$$x^{2}+5x-50=0$$
$$(x-5)(x+10)=0$$
$$x=5$$ (neglect $$x= -10$$ as length can't be negative)
$$MS=x+5+5=15$$
Question 10
1 / -0

A tree is broken at a height of $$5\ m$$ from the ground and its top touches the ground at a distance of $$12\ m$$ from the base of the tree. Find the original height of the tree.

A
$$20\ m$$

B
$$36\ m$$

C
$$18\ m$$

D
$$25\ m$$

Solution

Let $$KB$$ is original height of the tree
In $$ \triangle ABC$$, we have
$$AC^2=AB^2+BC^2$$
$$\Rightarrow 5^2+12^2=169$$
Therefore, $$ AC=\sqrt { 169 } =13$$ m
$$KB=KA+AB=AC+AB$$ ....Since $$\left( KA=AC \right) =(13+5)$$ m $$=18$$ m
Thus original height of the tree is $$18$$ m.