The Triangle And Its Properties - Class 7 Maths - Extra Questions

#1215097

$$a+b =34cm$$  

$$as\>we \>know$$ $$26^2=a^2+b^2$$  

$$or\>26^2=(34-b)^2+b^2$$  

$$or\>676=34^2+b^2-2. 34. b+b^2$$  

$$or\>676=1156+b^2-68b+b^2$$  

$$or\>2b^2-68b-480=0$$  

$$b^2-34b-240=0$$  

$$b^2-24b-10b-240=0$$  

$$b(b-24)-10(b-24)=0$$  

$$(b-24)(b-10)=0$$  

$$\therefore\>b=10\>or\>b=24$$  

$$then\>a=34-b=34-10=24$$  

$$then\>a=34-b=34-24=10$$

$${\textbf{Step 1: Statement}}$$

                  $${\text{In a right angled triangle}},{\text{the square of the hypotenuse is equal to the sum of the squares }}$$

                   $${\text{of the other two sides}}{\text{.}}$$

                    $${\text{ABC is a triangle in which}}\;\angle ABC = {90^ \circ }$$

                    $$\Rightarrow {\text{A}}{{\text{B}}^{\text{2}}}{\text{ + B}}{{\text{C}}^{\text{2}}}{\text{ = A}}{{\text{C}}^{\text{2}}}$$  

$$\textbf{Hence above statement is required statement}$$

Given:

Lines $$OB$$ and $$OC$$ are the bisectors of $$\angle B$$ and $$\angle C$$ of an isosceles $$\Delta ABC$$ such that $$AB=AC$$ which intersect each other at $$O$$ and $$BO$$ is produced to $$M$$ .

To prove:

$$\angle MOC=\angle ABC$$

Consider the diagram shown below.

Proof:

 In $$\Delta ABC$$ ,

$$AB=AC$$    (given)

$$\angle ACB=\angle ABC$$      (angles opposite to equal sides are equal)

$$\dfrac{1}{2}\angle ACB=\dfrac{1}{2}\angle ABC$$      (dividing both sides by 2)

Therefore,

$$\angle OCB=\angle OBC$$     …… (1)

(Since, $$OB$$ and $$OC$$ are the bisector of $$\angle B$$ and $$\angle C$$ )

Now, from equation (1), we have

$$\angle MOC=\angle OBC+\angle OCB$$         $$\because$$ sum of interior angle = Exterior angle

  $$\angle MOC=2\angle OBC$$

  $$\Rightarrow \angle MOC=2\angle ABC \times\dfrac{1}{2} = \angle ABC$$

(Since, $$OB$$ is the bisector of $$\angle B$$ )

It is given that in $$\Delta PQR$$ ,

$$PQ = QR = PR = 5\;{\rm{cm}}$$

Since all the sides of the given triangle are equal, therefore, the given triangle is an equilateral triangle.

It is given that in $$\Delta XYZ$$ ,

$$m\angle Y = 90^\circ$$ , $$XY = YZ$$

Since, the two sides of the given triangle is equal, therefore, the given triangle is an isosceles triangle.

consider a given right angle triangle

let whose shortest side is base(B) and third side is length(L) and H be a hypotenuse.

According to the given question,

  $$H=2B+6\,\,\,\,......\,\,\,\,\left( 1 \right)$$

  $$L=H-2\,\,\,\,........\,\,\,\left( 2 \right)$$                                    

By equation (1),

$$B=\dfrac{H-6}{2}\,\,\,\,.......\,\,\,\left( 3 \right)$$

We know that,

By Pythagoras theorem,

  $${{H}^{2}}={{B}^{2}}+{{L}^{2}}$$

  $$\Rightarrow {{H}^{2}}={{\left( \dfrac{H-6}{2} \right)}^{2}}+{{\left( H-2 \right)}^{2}}$$

  $$\Rightarrow {{H}^{2}}=\dfrac{{{H}^{2}}+36-12H}{4}+{{H}^{2}}+4-4H$$

  $$\Rightarrow 4{{H}^{2}}={{H}^{2}}+36-12H+4{{H}^{2}}+16-16H$$

  $$\Rightarrow {{H}^{2}}+36-12H+16-16H=0$$

  $$\Rightarrow {{H}^{2}}+52-28H=0$$

  $$\Rightarrow {{H}^{2}}-28H+52=0$$

  $$\Rightarrow {{H}^{2}}-\left( 26+2 \right)H+52=0$$

  $$\Rightarrow {{H}^{2}}-26H-2H+52=0$$

  $$\Rightarrow H\left( H-26 \right)-2\left( H-26 \right)=0$$

  $$\Rightarrow \left( H-26 \right)\left( H-2 \right)=0$$

  $$\Rightarrow H=2,26$$

$$H=2$$ is very small then $$H=26$$ exists.

Put the value of $$H$$ in equation (2) and (3) to, we get

  $$L=H-2\,\,$$

  $$L=26-2=24$$

  $$B=\dfrac{H-6}{2}$$

  $$B=\dfrac{26-6}{2}$$

  $$B=\dfrac{20}{2}$$

  $$B=10$$

Hence, $$B=10 m$$ , $$L=24 m$$    and  $$H=26 m$$

Consider the given figure,

$$\angle ABC={{63}^{0}}$$

Let   $$\angle ABC={{x}^{0}}$$   and   $$\angle ACB={{y}^{0}}$$

Now,

$$\angle ACB+{{125}^{0}}={{180}^{0}}$$

$$\angle ACB={{180}^{0}}-{{125}^{0}}$$

$$\angle ACB={{55}^{0}}$$

In $$\Delta ABC$$

$$\angle ABC+\angle ACB+\angle CAB={{180}^{0}}$$

$${{63}^{0}}+{{55}^{0}}+\angle CAB={{180}^{0}}$$

$$\angle CAB={{180}^{0}}-{{118}^{0}}$$

$$\angle CAB={{62}^{0}}$$

Since, the length of the hypotenuse $$=15$$ unit

We know that, 

Median $$=\dfrac{1}{2}\times$$ the length of the hypotenuse

$$=\dfrac{1}{2}\times 15$$

$$=7.5$$ units

Hence, the value is $$7.5$$ units.

We have.

Equation of two equal sides of isosceles triangle. Are

$$7x-y+3=0\ \ ......\ \ (1)$$

$$x+y+3=0\ \ ......\ \ (2)$$

Solving equation (1) amd (2) and we get.

$$({{x}_{1}},\ {{y}_{1}})=\left( \dfrac{-3}{4},\ \dfrac{-9}{4} \right)$$

And given point

$$(1,\ -10) =({{x}_{2}},\ {{y}_{2}})$$

So,

Equation of third side

$$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$$

$$y+\dfrac{9}{4}=\dfrac{-10+\dfrac{9}{4}}{1+\dfrac{3}{4}}\ \left( x+\dfrac{3}{4} \right)$$

$$y+\dfrac{9}{4}=\dfrac{-\dfrac{-31}{4}}{\dfrac{7}{4}}\ \left( x+\dfrac{3}{4} \right)$$

$$\dfrac{4y+9}{4}=\dfrac{-31}{7}\left( \dfrac{4x+3}{4} \right)$$

$$4y+9=-31(4x+3)$$

$$4y+9=-124x-93$$

$$124x+4y+9+93=0$$

$$124x+4y+{{10}^{2}}=0$$

$$31x+y+23=0$$

Hence, this is the answer.

We have,

$$PQR$$   is an equilateral triangle such that $$QR$$ lies on $$x-axis$$ .

$$O$$ is the midpoint of $$QR$$ .

So,

$$OQ=QR=4\ units.$$

$$\Rightarrow$$ Coordinates of point $$R$$ are $$\left( 4,0 \right)$$ .

Now,

Point $$P$$ lies on the $$y-axis$$ .

Let the coordinate of point $$P$$ be $$\left( 0,y \right).$$

So,

$$PQ=QR$$

$$\Rightarrow \sqrt{{{\left( 0+4 \right)}^{2}}+{{\left( y-0 \right)}^{2}}}=\sqrt{{{\left( 4+4 \right)}^{2}}+{{0}^{2}}}$$

$$\Rightarrow \sqrt{16+{{y}^{2}}}=\sqrt{64}$$

$$\text{On}\,\text{squaring}\,\text{both}\,\text{side}\,\text{and}\,\text{we}\,\text{get,}$$

$$\Rightarrow 16+{{y}^{2}}=64$$

$$\Rightarrow {{y}^{2}}=64-16$$

$${{\Rightarrow }y^{2}}=48$$

$$\Rightarrow y=\pm 4\sqrt{3}$$

Hence, the coordinate of points $$P$$ are $$\left( 0,4\sqrt{3} \right)$$ above $$x-axis.$$

And $$\left( 0,-4\sqrt{3} \right)$$ below $$x-axis$$ .

$${\textbf{Step  - 1: Proving similarity}}$$

                    $${\text{The triangle ABC is right angled at B}}$$

                    $${\text{In }}\vartriangle {\text{ABC and }}\vartriangle {\text{ADB,}}$$

                    $$\angle {\text{ADB  =  }}\angle {\text{ABC  =  90}}^\circ$$

                    $$\angle {\text{BAD  =  }}\angle {\text{BAC (common)}}$$

                    $${\text{AB side common}}$$

                    $$\therefore {\text{ }}\vartriangle {\text{ABC }} \sim {\text{ }}\vartriangle {\text{ADB}}$$

                    $${\text{Similarly, in }}\vartriangle {\text{ABC and }}\vartriangle {\text{BDC}}$$

                    $$\angle {\text{BDC  =  }}\angle {\text{ABC  =  90}}^\circ$$

                    $$\angle {\text{BCD  =  }}\angle {\text{BCA (common)}}$$

                    $${\text{BC side common}}$$

                    $$\therefore {\text{ }}\vartriangle {\text{ABC }} \sim {\text{ }}\vartriangle {\text{BDC}}$$

$${\textbf{Step  - 2: Proving pythagoras theorem}}$$

                  $${\text{Since, }}\vartriangle {\text{ABC }} \sim {\text{ }}\vartriangle {\text{ADB}}$$

                   $$\dfrac{{{\text{AD}}}}{{{\text{AB}}}}{\text{  =  }}\dfrac{{{\text{AB}}}}{{{\text{AC}}}}{\text{ (sides are proportional)}}$$

                   $$\Rightarrow {\text{ AD}} \times {\text{AC  =  A}}{{\text{B}}^{\text{2}}}{\text{ }} \to {\text{ Equation 1}}$$

                   $${\text{Also, }}\vartriangle {\text{ABC }} \sim {\text{ }}\vartriangle {\text{BDC}}$$

                   $$\dfrac{{{\text{CD}}}}{{{\text{BC}}}}{\text{  =  }}\dfrac{{{\text{BC}}}}{{{\text{AC}}}}{\text{ (sides are proportional)}}$$

                   $$\Rightarrow {\text{ CD}} \times {\text{AC  =  B}}{{\text{C}}^{\text{2}}}{\text{ }} \to {\text{ Equation 2}}$$

                   $${\text{Adding Equation 1 and Equation 2,}}$$

                   $$\left( {{\text{AD}} \times {\text{AC}}} \right){\text{  +  }}\left( {{\text{CD}} \times {\text{AC}}} \right){\text{  =   A}}{{\text{B}}^{\text{2}}}{\text{  +  B}}{{\text{C}}^{\text{2}}}$$

                   $$\Rightarrow {\text{ AC}}\left( {{\text{AD  +  CD}}} \right){\text{  =   A}}{{\text{B}}^{\text{2}}}{\text{  +  B}}{{\text{C}}^{\text{2}}}$$

                   $$\Rightarrow {\text{ AC }} \times {\text{ AC  =   A}}{{\text{B}}^{\text{2}}}{\text{  +  B}}{{\text{C}}^{\text{2}}}$$

                   $$\Rightarrow {\text{ A}}{{\text{C}}^{\text{2}}}{\text{  =   A}}{{\text{B}}^{\text{2}}}{\text{  +  B}}{{\text{C}}^{\text{2}}}$$

$${\textbf{Hence, Proved}}$$

$$\angle A=90^{\circ}\\\angle B=45^{\circ}\\\angle BAP=45^{\circ}\\\implies AP=BP\cdots(1)\\\angle ACP=45^{\circ}\\\implies AP=PC\cdots(2)\\\implies AP=BP=CP$$

$$\implies 2AP^2=BP^2+CP^2\\2(AD^2-PD^2)=BP^2+CP^2\\2AD^2=BP^2+PD^2+CP^2+PD^2\\2AD^2=(BP+PD)^2-2BP.PD+(CP-PD)^2+2BP.PD\\2AD^2=BD^2+CD^2$$

• In an isosceles triangle exactly two sides are equal length.