### Theorems related to triangles - class-X

 Description: theorems related to triangles Number of Questions: 37 Created by: Sara Dalvi Tags: geometry maths pythagoras theorem
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Find the side of the square whose diagonal is $16 \sqrt 2$ cm.

1. $4$ cm

2. $16$ cm

3. $8$ cm

4. $16\sqrt 2$ cm

Correct Option: B
Explanation:

We know that,

1) All angles of a square are congruent. i.e $90^o$
2) Diagonal of a square bisects each of its angles.
Therefore, the square gets divided into $2$ triangles of degrees $45^o-45^o-90^o$
$\therefore \sin 45^o = \cfrac {\text {side}}{\text {hyp}}$
$\therefore \cfrac {1}{\sqrt 2} = \cfrac {\text {side}}{16 \sqrt 2}$
$\therefore$ side of the square $= 16$ cm.

The sides of triangle are in A.P. and the greatest angle exceeds the least by 90. The sides are in the ratio _____________.

1. $1 : 2 : \sqrt { 2 }$

2. $1 : \sqrt { 3 } : 2$

3. $\sqrt { 7 } + 1 : \sqrt { 7 } : \sqrt { 7 } - 1$

4. $\sqrt { 3 } + 1 : 1 : \sqrt { 3 } - 1$

Correct Option: C

If  H is orthocenter of triangle PQR then PH + QH + RH is

1. QR cot P + PR cot Q + PQ cot R

2. (pq + QR + RP) (cot P + cot Q + copt R)

3. $\dfrac{1}{2r}(cot P + cotQ + cot R)$

4. none of these

Correct Option: A

ABC is a triangle right angle at B. D is a point on AC such that $\angle ABD = 45^0$. If AC =$6$ and AD =$2$ , then AB is

1. $\dfrac{6}{\sqrt{5}}$

2. ${3}{\sqrt{2}}$

3. $\dfrac{12}{\sqrt{5}}$

4. $2$

Correct Option: A

consider a triangle PQR in which the relation $QR^2+PR^2=5*PQ^2$ holds. let G be the point of intersection of the medians PM and QN . then angle QGM is always

1. less then 45 degree

2. obtuse

3. a right angle

4. acute and larger than 45 degree

Correct Option: A

If in a $\Delta ABC,\sin A=\sin^{2} B$ and $2\cos^{2}A=3\cos^{2}B$, then the $\Delta ABC$ is

1. Right angled

2. Obtuse angled

3. Isosceles

4. Equilateral

Correct Option: A

Which of the following  can be the sides of a right-angled triangle?

1. $0.5cm, 1.2 cm, 1.3cm$

2. $2.4cm, 3.2 cm, 7.9cm$

3. $5.0cm, 5.25 cm, 7.25cm$

4. $1.6cm, 3.0 cm, 3.4cm$

Correct Option: A

Let $\Delta _1$ denotes the area of the triangle formed by the vertices $(a^3m^3 _1, am _1), (a^3m^3 _2am _2), (a^3m^3 _3, am _3)$ and $\Delta _2$ denotes the area of the triangle formed by the vertices $(2am _1m _2, a^2(m^2 _1+m^2 _2))$, $(2am _2m _3, a^2(m^2 _2+m^2 _3))$ and $(2am _3m _1, a^2(m^2 _3+m^2 _1))$. Then $\dfrac{\Delta _1}{\Delta _2}(a > 0)$ equals?

1. $\dfrac{a}{2}$

2. $2a$

3. $\dfrac{a^3}{8}$

4. $8a^3$

Correct Option: A

The sides of $\Delta ABC$ are 5, 7, 8 units then $AG^2 + BG^2 + CG^2$ where G is centroid of $\Delta ABC$ is

1. $48.2$

2. $49.2$

3. $92.2$

4. $69.2$

Correct Option: A

For a right-angled triangle, two small sides are of $6$cm and $8$cm length. Length of third side will be

1. $14$cm

2. $9$cm

3. $10$cm

4. none of these

Correct Option: A

In $\Delta ABC, \, m \angle B = 90$ and $\overline{BM}$ is an altitude. If AB = 2 AM, then AC = ......

1. 2 AM

2. 4 AM

3. 6 AM

4. 8 AM

Correct Option: A

P, Q, R are the points of intersection of a line 1 with sides BC, CA, AB of a $\Delta$ ABC
respectively, then $\dfrac{BP}{PC} \dfrac{CQ}{QA} \dfrac{AR}{RB}$

1. 1

2. 2

3. -1

4. -2

Correct Option: A

Sides of triangle are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

1. 7 cm, 24 cm, 25 cmj

2. 3 cm, 8 cm, 6 cm

3. 50 cm, 80 cm, 100 cm

4. 13 cm, 12 cm, 5 cm

Correct Option: A

The hypotenuse and the semi-perimeter of right triangle are 20 cm and 24 cm, respectively. The other two sides of the triangle are :

1. 16 cm, 15 cm

2. 14 cm, 16 cm

3. 20 cm, 16 cm

4. None of these

Correct Option: A

In $\Delta PQR,\angle P$ is right angle and $\bar{PM}$ is an altitude. $PQ=\sqrt{20}$ and QM=4 then RM=____.

1. 8

2. 5

3. 9

4. 1

Correct Option: A

The lengths of the medians through acute angles of a right-angled triangle are 3 and 4. Find the area of the triangle:

1. $\displaystyle \frac{4}{3}\sqrt{11}$

2. $\displaystyle \frac{2}{3}\sqrt{11}$

3. $\displaystyle \frac{1}{3}\sqrt{11}$

4. None of these

Correct Option: A
Explanation:

Given, $AD=3,CE=4$
Using Appaloneaus theorem for median $AD$
We have $\displaystyle{ c }^{ 2 }+{ b }^{ 2 }=2\left( \frac { { a }^{ 2 } }{ 4 } +9 \right)$   ...(1)
Using Appaloneaus theorem for median $CE$
We have $\displaystyle{ b }^{ 2 }+{ a }^{ 2 }=2\left( \frac { { c }^{ 2 } }{ 4 } +10 \right)$   ...(2)
Also, ${ a }^{ 2 }+{ c }^{ 2 }={ b }^{ 2 }$
Adding (1) and (2)
$\displaystyle 3{ b }^{ 2 }=2\left( \frac { { b }^{ 2 } }{ 4 } +25 \right) \Rightarrow { b }^{ 2 }=20$
Solving (1) and (2) we get,
$\displaystyle c=\frac { 4 }{ \sqrt { 3 } }$ and $\displaystyle a=2\frac { 4 }{ \sqrt { 3 } }$
Hence, area of triangle
$\displaystyle = \frac{1}{2}\left ( \frac{4}{\sqrt{3}} \right )\left ( 2\sqrt{\frac{11}{3}} \right )= \frac{4}{3}\sqrt{11}$.

The length of the hypotenuse of an isosceles right triangle whose one side is $4\surd {2}\ cm$  is___________$cm$

1. $12$

2. $8$

3. $8\surd {2}$

4. $12\surd {2}$

Correct Option: A

Let $ABC$ be a fixed triangle and $P$ be variable point in the plane of a triangle $ABC$. Suppose $a, b, c$ are lengths of sides $BC, CA,AB$ opposite to angles $A, B, C$ respectively. If $a(PA)^{2} + b(PB)^{2} + c(PC)^{2}$ is minimum, then the point $P$ with respect to $\triangle{ABC}$ is

1. Centroid

2. Circumcenter

3. Orthocenter

4. Incenter

Correct Option: A

If $AD,BE$ and $CF$ are the medians of a $\Delta ABC,$ then evaluate  $\displaystyle \left ( AD^{2}+BE^{2}+CF^{2} \right ):\left ( BC^{2}+CA^{2}+AB^{2} \right )=$

1. $3:4$

2. $4:3$

3. $5:3$

4. $4:1$

Correct Option: A
Explanation:

Given, $AD,BE$ and $CF$ are the medians of a $\Delta ABC$.
$\Rightarrow AB^2+AC^2=2(AD^2+BD^2)$
$\Rightarrow AB^2+AC^2=2AD^2+\displaystyle\frac{BC^2}{2}$
$\Rightarrow 2AD^2=AB^2+AC^2-\displaystyle\frac{BC^2}{2}$ -----(1)
Similarly,
$2BE^2=BC^2+BA^2-\displaystyle\frac{AC^2}{2}$ -----(2)
$2CF^2=CA^2+CB^2-\displaystyle\frac{AB^2}{2}$ -----(3)
Adding equation 1,2 and 3, we get
$2(AD^2+BE^2+CF^2)=\displaystyle\frac{3}{2}(AB^2+BC^2+CA^2)$
$\therefore (AD^2+BE^2+CF^2):(AB^2+BC^2+CA^2)=3:4$

The distances of the circumcentre of the acute-angled $\Delta \mathrm{ABC}$ from the sides $\mathrm{BC},$ CA and AB are in the ratio

1. asinA: bsinB:csinC

2. $\cos A : \cos B : \cos C$

3. $\operatorname{acot} A : \operatorname{bcot} B : \operatorname{ccot} C$

4. none of these

Correct Option: A

Let ABC be a triangle having its centroid at G. If S is any point in the plane of the triangle, then $S\vec { A } +S\vec { B } +S\vec { C } =$

1. $S\vec { G }$

2. $2S\vec { G }$

3. $3S\vec { G }$

4. $\vec { 0 }$

Correct Option: A

Mark the correct alternative of the following.
In a right triangle, one of the acute angles is four times the other. Its measure is?

1. $68^o$

2. $84^o$

3. $80^o$

4. $72^o$

Correct Option: D
Explanation:

In the right-angled triangle the sum of the other two angles is $90^o$.

Let one acute angle is $x$, then the other angle is $4x$. [ Given]
Then we get,
$4x+x=90^o$
or, $5x=90^o$
or, $x=18^o$.
So the measure of that angle is $4\times 18^o=72^o$.

Find the perimeter of an isosceles right triangle with each of its congruent as 7cm.

1. $7\sqrt 2$ cm

2. $14$ cm

3. $(2+ \sqrt 2)$ cm

4. $7(2+ \sqrt 2)$ cm

Correct Option: D
Explanation:

Let the other side of triangle is x cm

Then in isosceles right angle triangle two congruent sides are 7 cm
$x^{2}=(7)^{2}+(7)^{2}$
$\Rightarrow x^{2}=49+49$
$\Rightarrow x^{2}=198$
$\Rightarrow x=7\sqrt{2}$
Then perimeter of right angle isosceles triangle =$7+7+7\sqrt{2}=14+7\sqrt{2}=7(2+\sqrt{2})$

So, option D is correct.

If the sides of a triangle are in the ratio $1\, :\, \sqrt2\, :\, 1$, then the triangle is:

1. an equilateral triangle

2. an isosceles triangle

3. a right angled triangle

4. a right angled isosceles triangle

Correct Option: D
Explanation:

Given ratio of sides of the triangle, $1 : \sqrt{2} : 1$
Let the triangle be $ABC$ and sides be
$AB = x$
$BC = x$
$AC = \sqrt{2}x$

Clearly, $AC^2 = BC^2 + AB^2$
Hence, by converse of Pythagoras theorem, $ABC$ is a right-angled isosceles triangle.

In $\triangle ABC$, AP is the median. If $AP=7$ and $AB^2+AC^2=260$, then find BC.

1. $14$ cm

2. $18$ cm

3. $15$ cm

4. $12$ cm

Correct Option: B
Explanation:

By Apollonius Theorem (states that "the sum of the squares of any two sides of any triangle equals twice the square on half the third side, together with twice the square on the median bisecting the third side".) , we have

$BC^2=AB^2+AC^2+2AP^2$

$BC^2=260+2(7)^2$

$BC^2=260+98=358$

$BC=18.92$

Find the length of median. If the sides of triangle are:
$a = 5, b = 6, c = 8$. and $m = 3, n = 2$.

1. $\sqrt{\dfrac{206}{5}}$

2. $\sqrt{206}$

3. $\dfrac{\sqrt{206}}{5}$

4. None

Correct Option: A
Explanation:
We have from Appollonius theorem,

$a(mn+p^2)=b^2m+c^2n$

$5(3\times2+p^2)=6^2\times3+8^2\times2$

$5(6+p^2)=36\times3+64\times2$

$30+5p^2=108+128$

$5p^2=236−30 \ \implies 5p^2=206$

$p^2=\dfrac{206}{5}$

$p=\sqrt{\dfrac{206}{5}}$

Option A.

In a $\Delta$ $ABC, AD = 3, BC = 2, AB = 1$, find the value of $AC$. (Use Apollonius theorem).

1. $2.35$

2. $3.42$

3. $4.35$

4. $5.61$

Correct Option: C
Explanation:

According to the Apollonius theorem,
$AB^{2}+AC^{2}= 2[AD^{2}+\dfrac {BC}{2}^{2}]$
$1^{2}+AC^{2}= 2[3^{2}+\dfrac{2}{2}^{2}]$
$1+AC^{2}= 2[9 + 1]$
$AC^{2}=20 -1$
$AC^{2}= 19$
$AC = \sqrt{19}$
$AC = 4.35$

In a $\Delta$ $ABC, AC = 6, BC = 2, AB = 4$, find the value of $AD$. (Use Apollonius theorem).

1. 5

2. 4

3. 3

4. 2

Correct Option: A
Explanation:

According to the Apollonius theorem,
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$4^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$16+36= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

In a $\Delta$ $ABC, AC = 8, BC = 2, AB = 6$, find the value of $AD$. (Use Apollonius theorem).

1. 3

2. 5

3. 7

4. 9

Correct Option: C
Explanation:

According to the Apollonius theorem,
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$6^{2}+8^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$36+64= 2[AD^{2} + 1]$
$2AD^{2}=100 -2$
$2AD^{2}= 98$
$AD^{2} = \dfrac{98}{2}$
$AD^{2} = 49$
$AD = 7$

In a $\Delta$ $ABC, AC = 4, BC = 2, AB = 6$, find the value of $AD$. (Use Apollonius theorem).

1. 2

2. 3

3. 4

4. 5

Correct Option: D
Explanation:

According to the Apollonius theorem,
$AB^{2}+AC^{2}= 2\left [AD^{2}+\dfrac{BC}{2}^{2}\right]$
$6^{2}+4^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$36+16= 2[AD^{2} + 1]$
$2AD^{2}=52 -2$
$2AD^{2}= 50$
$AD^{2} = \dfrac{50}{2}$
$AD^{2} = 25$
$AD = 5$

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side is called ______ theorem.

1. Pythagoras

2. Apollinius

3. Stewart

4. Ceva's

Correct Option: B
Explanation:

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side is called Apollinius theorem.

Option $B$ is correct.

In a $\Delta$ $ABC, AC = 6, BC = 2, AB = 8$, find the value of $AD$. (Use Apollonius theorem).

1. 5

2. 6

3. 7

4. 8

Correct Option: C
Explanation:

According to the Apollonius theorem,
$AB^{2}+AC^{2}= 2\left[AD^{2}+\dfrac{BC}{2}^{2}\right]$
$8^{2}+6^{2}= 2\left [AD^{2}+\dfrac{2}{2}^{2}\right]$
$64+36= 2[AD^{2} + 1]$
$2AD^{2}=100 -2$
$2AD^{2}= 98$
$AD^{2} = \dfrac{98}{2}$
$AD^{2} = 49$
$AD = 7$

Which one of the following formula is used to find apollinius theorem for isosceles triangle?

1. $a^{2}+b^{2}=2m^{2}+\dfrac{c}{2}^{2}$

2. $b^{2}=m^{2}+\dfrac{c}{4}^{2}$

3. $b^{2}+b^{2}=2m^{2}+\dfrac{c}{2}^{2}$

4. $a^{2}+b^{2}=2m^{2}+\dfrac{b}{2}^{2}$

Correct Option: B
Explanation:

In any triangle, the sum of the squares on any two sides is equal to twice the square on half the third side together with twice the square on the median which bisects the third side.
Apollinius theorem formula,
$a^{2}+b^{2}= 2[m^{2}+\dfrac{c}{2}^{2}]$
When the given triangle is isosceles, than $b = a$.
So, $b^{2}=m^{2}+\dfrac{c}{4}^{2}$ is the formula used for isosceles triangle.

In a $\triangle ABC$, $AB= 4$ cm and $AC = 8$ cm. If M is the midpoint of BC and $AM = 3$ cm, then the length of $BC$ in cm is:

1. ${2\sqrt{26}}$

2. ${2\sqrt{31}}$

3. ${\sqrt{31}}$

4. ${\sqrt{26}}$

Correct Option: B
Explanation:

Given : In $\triangle ABC$, $AB=4$ cm and $AC=8$ cm.

M is the midpoint of BC and $AM=3$ cm
Using Apollonius theorem,
$AB^2+AC^2=2(AM^2+BM^2)$
$\implies$ $4^2+8^2=2(3^2+BM^2)$
$\implies$ $16+64=2(9+BM^2)$
$\implies$ $BM^2=31$
$\implies$ $BM=\sqrt{31}$.
$\because$ $BC=2BM$
$\therefore$ $BC=2\sqrt{31}$.

In $\triangle PQR$, $\angle P=30^o$, $\angle Q=60^0$, $\angle R= 90^o$ and $PQ=10$ units.

Find $PR$ and $QR$.

1. $PR =$ $10$ units, $QR =$ $5\sqrt 3$ units

2. $PR =$ $5$ units, $QR =$ $5\sqrt 3$ units

3. $PR =$ $5\sqrt 3$ units, $QR =$ $5$ units

4. $PR =$ $5$ units, $QR =$ $10\sqrt 3$ units

Correct Option: C
Explanation:

$\triangle PQR$ is a $30^o-60^o-90^o$ triangle        ....given

Since, $\angle R = 90^o$, side $PQ$ is hypotenuse.

$\Rightarrow$ By $30^o-60^o-90^o$ theorem,

$\Rightarrow PR = \dfrac {\sqrt 3}{2} \times PQ$ and $QR = \dfrac 12 \times PQ$

$\Rightarrow PR = \dfrac {\sqrt 3}{2} \times 10$ and $QR = \dfrac 12 \times 10$

$\Rightarrow PR = 5 \sqrt 3$ units and $QR = 5$ units
So, option C is correct.

According to Apolloneous Theorem, if $\overline AD$ is a median of $\triangle ABC$, then $AB^{2}+AC^{2}=$

1. $AD+BD$

2. $AD-BD$

3. $2(AD^{2}+BD^{2})$

4. $2(AD^{2}-BD^{2})$

Correct Option: C
Explanation:

$\begin{array}{l} Then,\, \, A{ B^{ 2 } }+A{ C^{ 2 } }=? & \ AD\, \, is\, \, median & \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2\left| { \frac { { B{ C^{ 2 } } } }{ 4 } } \right| +2A{ D^{ 2 } } & \left[ \begin{array}{l} BC=2BD \ or\, \, BC=DC \end{array} \right] \ \Rightarrow A{ B^{ 2 } }+A{ C^{ 2 } }=2{ \left| { \frac { { B{ C^{ } } } }{ 2 } } \right| ^{ 2 } }+2A{ D^{ 2 } } & \ A{ B^{ 2 } }+A{ C^{ 2 } }=2B{ D^{ 2 } }+2A{ D^{ 2 } }=2\left( { A{ D^{ 2 } }+B{ D^{ 2 } } } \right) & \end{array}$

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