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Basic proportionality theorem - class-X

Description: basic proportionality theorem
Number of Questions: 22
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Tags: triangles similar triangles similarity similarity in geometrical shapes geometry maths
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Basic proportionality theorem  is also known as

  1. Basic theorem

  2. Thales Theorem

  3. Potential theorem

  4. Unknown


Correct Option: B
Explanation:

Basic proportionality theorem is also known as Thales Theorem. Thales was a famous Greek mathematician who gave an important truth relating two equiangular triangles.
Therefore, C is the correct answer.

In $\triangle ABC,A-P-B, A-Q-C$ and $\overline {PQ} \parallel \overline {BC}$. If $PQ=5, AP=4$ and $PB=8$, then $BC=$.....

  1. $6$

  2. $10$

  3. $12.5$

  4. $15$


Correct Option: A

ABC is a triangle with AB = $13$ cm, BC =$14$ cm and CA=$15$ cm. AD and BE are the altitudes from A to B to BC and AC respectively. H is the point of intersection of the AD and BE. Then the ratio of $\frac { HD }{ HB } =$ 

  1. $\dfrac { 3 }{ 5 } $

  2. $\dfrac { 12 }{ 13 } $

  3. $\dfrac { 4 }{ 5 } $

  4. $\dfrac { 5 }{ 9 } $


Correct Option: A
Explanation:
According to the question,
Triangle $BEC$ and triangle $BDH$ are similar, because they have the same angles this means that the sides  of these two triangles are in the same ratio.

So,

$\dfrac{{HD}}{{BD}} = \dfrac{{CE}}{{BC}}$

Note,However that $\displaystyle \frac{{CE}}{{BC}}$=$cosC$, 

Hence$\displaystyle \frac{{HD}}{{BD}}$=$cosC$, so we proceed to find $cosC$ using the cosine rule,

${c^2} = {a^2} + {b^2} - 2ab\cos C$

${13^2} = {14^2} + {15^2} - 2(14)(15)cosC$

$\cos C = \dfrac{{{{13}^2} - {{14}^2} - {{15}^2}}}{{ - 2 \times 14 \times 15}} = \dfrac{3}{5}$

$so\, \, \dfrac{{HD}}{{HB}} = \dfrac{3}{5}$












In a triangle ABC, D and E are the point on the line segment BC and AC respectively, such that 2 BD = DC and 3 AE = 2 EC. The lines AD and BE meet at P,the line CP and AB F, then :

  1. AP:PD = 2:1

  2. BP : PE =4:

  3. BP:PE =5:4

  4. CP:PF = 7:2


Correct Option: A

Let  $ABC$  be a triangle and  $D$  and  $E$  be two points on side  $AB$  such that  $AD = BE$.  If  $D P | B C$  and  $E Q | A C,$ then $P Q | A C.$

  1. True

  2. False


Correct Option: B

If the sides a, b, c, of a triangle are such that a: b: c: :1:$\sqrt{3}$: 2, then the A:B:C is -

  1. 3 : 2 : 1

  2. 3 : 1 : 2

  3. 1 : 3 : 2

  4. 1 : 2 : 3


Correct Option: A

In any $\Delta$ABC , $4\Delta(cotA+cotB+cotC)$ is equal to 

  1. $3(a^2+b^2+c^2)$

  2. $2(a^2+b^2+c^2)$

  3. $(a^2+b^2+c^2)$

  4. none of these


Correct Option: A

$ABCD$ is a rectangl $P$ and $Q$ are poits on $AB$ and $BC$ respectively such that the area of triangle $APD=5$ area of triangle $PBQ=4$ and area of triangle $QCD=3$, all area in square units. THen the area of the triangle $DPQ$ in square units is

  1. $12$

  2. $\dfrac {20}{3}$

  3. $2\sqrt {21}$

  4. $\sqrt {21}$


Correct Option: A

If G is the centroid of $\Delta ABC$ and if area of $\Delta AGB$ is 5 sq. nits then the area of $\Delta ABC$ is 

  1. 20 sq. unit

  2. 10 sq.unit

  3. 15 sq. unit

  4. 25 sq. unit


Correct Option: A

The areas of two similar triangle are $18\ cm^{2}$ and $32\ cm^{2}$ respectively. What is the ratio of their corresponding sides?

  1. $3:4$

  2. $4:3$

  3. $9:16$

  4. $16:9$


Correct Option: A

In a triangle PQR, S and T are points on QR and PR respectively, such that QS = 3SR and PT = 4TR Let M be the point of intersection of PS and QT.  FInd the ration QM : MT 

  1. 15: 8

  2. 16 : 5

  3. 15 : 4

  4. 5 : n2


Correct Option: A

In a triangle PQR, S and T are points on QR and PR respectively, such that QS = 3SR and PT = 4TR Let M be the point of intersection of PS and QT.  FInd the ration QM : MT 

  1. 15 : 8

  2. 16 : 5

  3. 15 : 4

  4. 5 :2


Correct Option: A

If the areas of two similar triangles are equal,  then they are congruent.

  1. True

  2. False


Correct Option: A

In a $\Delta ABC$, let $M$ be the mid-point of segment $AB$ and let $D$ be the foot of the bisector of $\angle C$. Then the ratio $\dfrac{Area\Delta CDM}{Area \Delta ABC}$ is $\left(A>B\right)$

  1. $\dfracc{1}{4}\dfrac{a-b}{a+b}$

  2. $\dfracc{1}{2}\dfrac{a-b}{a+b}$

  3. $\dfracc{1}{2}\tan\dfrac{A-B}{2}\cot\dfrac{A+B}{2}$

  4. $\dfracc{1}{4}\cot\dfrac{A-B}{2}\tan\dfrac{A+B}{2}$


Correct Option: A

If $\triangle ABC \cong \triangle QPR$ and $\dfrac {ar(\triangle ABC)}{ar(\triangle PQR)}=\dfrac {9}{4}$, $AB=18\ cm$ and $BC=15\ cm$, then $PR$ is equal to________ $cm$

  1. $10$

  2. $12$

  3. $20/3$

  4. $8$


Correct Option: A

The sides of a triangle are $3x+4y,\,4x+3y$ and $5x+5y$ units, where $x,y>0$.The triangle is ______________.

  1. right  angled 

  2. equilateral

  3. obtuse angled

  4. none of these


Correct Option: C
Explanation:
Let $a=3x+4y,\,b=4x+3y$ and $c=5x+5y$ be the largest side
$\Rightarrow \cos{C}=\dfrac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}$
$=\dfrac{{\left(3x+4y\right)}^{2}+{\left(4x+3y\right)}^{2}-{\left(5x+5y\right)}^{2}}{2\left(3x+4y\right)\left(4x+3y\right)}$
$\Rightarrow \cos{C}=\dfrac{9{x}^{2}+16{y}^{2}+24xy+16{x}^{2}+9{y}^{2}+24xy-25{x}^{2}-25{y}^{2}-50xy}{2\left(3x+4y\right)\left(4x+3y\right)}<0,\,\,\,x,y>0$
$\Rightarrow \cos{C}=\dfrac{-2xy}{2\left(3x+4y\right)\left(4x+3y\right)}<0,\,\,x,y>0$
$\Rightarrow \theta>{90}^{\circ}$
$\therefore,\, $ the triangle is obtuse angled.

D and E are respectively the points on the sides AB and AC of a $\displaystyle \Delta ABC$ such that $AB = 12 cm$, $AD = 8 cm$, $AE = 12 cm$ and $AC = 18 cm$, then

  1. DE $\parallel$ BD is true

  2. DE $\parallel$ BC is true

  3. AD $\parallel$ BD is true

  4. AD $\parallel$ CD is true


Correct Option: B
Explanation:

We have,
AB = 12 cm, AC = 18 cm, AD = 8 cm and AE = 12 cm.
$\displaystyle \therefore \quad BD=AB-AD=\left( 12-8 \right) cm=4cm$
$\displaystyle CE=AC-AE=\left( 18-12 \right) cm=6cm$
Now, $\displaystyle \frac { AD }{ BD } =\frac { 8 }{ 4 } =\frac { 2 }{ 1 } $
And, $\displaystyle \frac { AE }{ CE } =\frac { 12 }{ 6 } =\frac { 2 }{ 1 } $
$\displaystyle \Rightarrow \quad \frac { AD }{ BD } =\frac { AE }{ CE } $
Thus, DE divides sides AB and AC of $\displaystyle \Delta ABC$ in the same ratio. Therefore, by the converse of basic proportionality theorem, we have
$\displaystyle DE\parallel BC$.

Match the column.

1. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR},\angle A=\angle P$ (a) AA similarity criterion 
2. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \angle A=\angle P,\angle B=\angle Q$ (b) SAS similarity criterion 
3. In $\displaystyle \Delta ABC$ and $\displaystyle \Delta PQR$,$\displaystyle \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$$\angle A=\angle P$ (c) SSS similarity criterion 
4. In $\displaystyle \Delta ACB,DE
  1. $\displaystyle 1\rightarrow a,2\rightarrow b,3\rightarrow c,4\rightarrow d$

  2. $\displaystyle a\rightarrow d,2\rightarrow a,3\rightarrow c,4\rightarrow b$

  3. $\displaystyle 1\rightarrow b,2\rightarrow a,3\rightarrow c,4\rightarrow d$

  4. $\displaystyle 1\rightarrow c,2\rightarrow b,3\rightarrow d,4\rightarrow a$


Correct Option: C
Explanation:

In $\triangle ABC$ and $\triangle PQR$

Option A:

If $\angle A = \angle P$      ....Given

And, $\dfrac {AB}{PQ} = \dfrac {AC}{PR}$    ...Given

$\triangle ABC \sim \triangle PQR$        ...SAS test of similarity


Option B:

If $\angle A = \angle P$      ....Given

And $\angle B = \angle Q$      ....Given

$\triangle ABC \sim \triangle PQR$        ...AA test of similarity


Option C:

If $\angle A = \angle P$      ....Given

And $\dfrac {AB}{PQ} = \dfrac {AC}{PR} = \dfrac {BC}{QR}$      ....Given

$\triangle ABC \sim \triangle PQR$        ...SS S test of similarity


Option D:

In $\triangle ACB, DE \parallel BC$

$\dfrac {AD}{BD} = \dfrac {AE}{CE} $      ....Given

This is known as basic proportionality theorem.

In an isosceles $\Delta A B C$ the base $A B$ is produced both the ways to $P$ and $Q$ such that $A P \times BO = A C ^ { 2 }$ then $\Delta A P C \sim \Delta B C Q$

  1. True

  2. False


Correct Option: A

In the sides $BC,CA,AB$ of a triangle $ABC$, three points $D,E,F$ are taken such that each of $BD,CE,AE$ is equal to one-third of the corresponding side, then
$\triangle DEF=\dfrac {1}{2}\triangle ABC$.

  1. True

  2. False


Correct Option: B

In any triangle, medians meet at a point and divide each other as the ratio of $2:3$

  1. True

  2. False


Correct Option: B

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$, respectivetly where $\triangle ABC \sim \triangle PQR$, then  $\dfrac {AB}{PR}=\dfrac {AC}{PM}$.

  1. True

  2. False


Correct Option: B
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