Triangles Test

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Triangles Test - 39

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Weekly Quiz Competition

Question 1
1 / -0

Ram goes $$\sqrt{12.5}km$$ towards west from a certain point. Then after turning to his right he again goes same distance. In the end he goes $$25km$$ towards South-East. How far is he now from his starting point?

A
$$10km$$

B
$$20km$$

C
$$15km$$

D
$$5km$$

Solution

As shown in figure, Ram starts from C and goes to B and consecutively to A, and then travels 25 km. He will cross point C because ABC is right angled isosceles triangle, so by Pythagoras theorem
$${ AC }^{ 2 }={ AB }^{ 2 }+{ BC }^{ 2 }$$
$${ AC }^{ 2 }={12.5}+{12.5}$$
$$AC = 5$$
So, further $$20 km$$ more will be walked.
Question 2
1 / -0

ABCD is parallelogram and P is the mid point of the side AD. The line BP meets the diagonal AC in Q. Find the ratio AQ $$:$$ QC.

A
$$1:2$$

B
$$2:1$$

C
$$1:3$$

D
$$3:1$$

Solution

Join $$AC$$ and $$BP$$.
$$\angle AQP = \angle CQB$$ [Vertically opposite angles]
$$ \angle APQ = \angle CBQ $$ [Alternate interior angles between two parallel lines $$AD$$ and $$BC$$]

Hence, by $$AAA$$ similarity criterion,
$$\Delta APQ \sim \Delta CBQ $$

Hence, $$ \dfrac{AP}{BC} = \dfrac{AQ}{QC}$$
Since, $$AD = BC,\ \dfrac{AP}{BC} = \dfrac{AP}{AD}$$

Now, $$P$$ is the midpoint of $$AD$$
Hence, $$\dfrac{AP}{BC} = \dfrac{AP}{AD} = \dfrac{1}{2} $$

$$\Rightarrow \dfrac{AQ}{QC} = \dfrac{1}{2}$$

Hence, $$AQ:QC=1:2$$.
Question 3
1 / -0

In a $$\triangle ABC, P$$ and $$Q$$ are the points on sides $$AB$$ and $$BC$$ respectively, such that $$PQ\parallel BC$$. Also, $$AP=2\ cm, PB=4\ cm, AQ=3\ cm$$ and $$QC=6\ cm$$ then find $$BC$$

A
$$PQ$$

B
$$2PQ$$

C
$$3PQ$$

D
$$4PQ$$

Solution

Let $$ABC$$ be a triangle. Let $$P,Q$$ are the points on sides $$AB,BC$$ respectively such that $$PQ\: || \:BC$$.

Given $$AP=2\:cm, PB=4\: cm, AQ=3\:cm, QC=6\:cm$$.

We have to find $$BC$$.

The pictorial representation of the problem is shown below

Since we have $$PQ\: || \:BC$$ we get $$\Delta APQ \sim \Delta ABC$$ by AAA

$$\Rightarrow \dfrac{AP}{AB}=\dfrac{AQ}{AC}=\dfrac{PQ}{BC}$$

Substituting the values we get,

$$\Rightarrow \dfrac{2}{6}=\dfrac{3}{9}=\dfrac{PQ}{BC}$$

$$\Rightarrow \dfrac{1}{3}=\dfrac{1}{3}=\dfrac{PQ}{BC}$$

$$\Rightarrow \dfrac{1}{3}=\dfrac{PQ}{BC}$$

$$\Rightarrow BC=3PQ$$
Question 4
1 / -0

Triangles ABC and DEF are similar. If their areas are 64 $$cm^2$$ and 49 $$cm^2$$ and if AB is 7 cm, then find the value of DE.

A
8 cm

B
$$\dfrac{49}{8}$$ cm

C
$$\dfrac{8}{49}$$ cm

D
$$\dfrac{64}{7}$$cm

Solution

$$\Delta ABC \Delta DEF$$
$$\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF}$$.
We know that,
$$\dfrac{Area of \Delta ABC}{ Area of \Delta DEF} = $$ $$\Rightarrow \dfrac{64}{49} = \left ( \dfrac{7}{DE} \right )^2$$
$$\Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \frac{7}{DE} \right )^2 \Rightarrow \left ( \dfrac{8}{7} \right )^2 = \left ( \dfrac{7}{DE} \right )^2 = \dfrac{49}{8}$$cm
Question 5
1 / -0

In $$\triangle DEF$$ &$$ \triangle PQR$$, $$\angle R = \angle F$$, then both triangles are similar if

A
$$\dfrac{DF}{PR}=\dfrac{FE}{RQ}$$

B
$$\dfrac{DE}{PQ}=\dfrac{DF}{PR}$$

C
$$\dfrac{DE}{PR}=\dfrac{DF}{RQ}$$

D
$$\dfrac{DE}{QR}=\dfrac{EF}{PR}$$

Solution

In$$\triangle DEF$$ and $$\triangle PQR$$
if $$\angle R=\angle F$$
then the ratios of the sides including these angles must be equal.
So, that SAS criteria of similarity can be applied.
i.e $$\dfrac{DF}{PR}=\dfrac{FE}{RQ}$$
Question 6
1 / -0

Check if the triangles are similar. If similar, write the similarity in symbolic form. Mention the similarity condition used.

A
$$ASA$$

B
$$SAS$$

C
$$AAS$$

D
NONE

Solution

$$AB=XY$$
$$ =>\cfrac { AB }{ XY } =1\quad \quad -(i)$$
$$ BC=XZ$$
$$ =>\cfrac { BC }{ XZ } =1\quad \quad -(ii)$$
$$ \angle ABC=\angle YXZ$$
$$ \therefore \cfrac { AB }{ XY } =\cfrac { BC }{ XZ } $$
$$ and\quad included\quad angles\quad equal$$
$$ =>\angle ABC=\angle YXZ$$
$$ \therefore \triangle ABC\sim \triangle XYZ\quad by\quad S.A.S\quad similarity$$
Question 7
1 / -0

$$ABC$$ and $$BDE$$ are two equilateral triangles such that $$D$$ is the mid point of $$BC$$. Ratio of the areas of triangle $$ABC$$ and $$BDE$$ is

A
$$2:1$$

B
$$1:2$$

C
$$4:1$$

D
$$1:4$$

Solution

$$\triangle ABC \sim \triangle BDE$$ (both are equilateral triangles)

According to the given condition, $$BC=\dfrac{BD}{2}$$

$$\Rightarrow \triangle ABC : \triangle BDE = AB^2 : BD^2$$

$$= AB^2 : (\dfrac{1}{2} BC)^{2} $$

$$ = AB^2 : \dfrac{1}{4} BC^2 $$

$$= 4 : 1$$ $$(\because AB = BC)$$
Question 8
1 / -0

Given ΔABC is similar to ΔPQR. Where AB=1cm and PQ =3 cm .Find the ratio of their areas.

A
$$\dfrac{1}{9}$$

B
$$\dfrac{1}{8}$$

C
$$\dfrac{8}{9}$$

D
$$\dfrac{9}{1}$$

Solution

$$\dfrac{AB}{PQ}=\dfrac{1}{3}$$
$$\dfrac{ar\Delta ABC}{ar\Delta PQR}=\left(\dfrac{AB}{PQ}\right)^2=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{9}$$.
Question 9
1 / -0

Triangle $$DEF$$ is similar to Triangle $$ABC$$, $$DE:AB=2:3$$ and area of triangle $$DEF =44 sq. cm.$$. Area of triangle $$ABC =$$

A
$$99$$

B
$$33$$

C
$$11$$

D
$$66$$
Question 10
1 / -0

Given, $$\triangle ABC$$ $$\sim \triangle PQR.$$ If $$\dfrac{ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac{9}{4}$$ and $$AB=18$$ cm, then find the length of $$PQ.$$

A
$$19$$

B
$$12$$

C
$$32$$

D
$$44$$

Solution