Answer
418.2k+ views
Hint: We draw the line segment AB which represents the perimeter of the triangle XYZ, $XY+YZ+ZX=11$. We construct an angle of ${{30}^{\circ }}$on point A and an angle of ${{60}^{\circ }}$ on the point B which we later bisect to meet at point X. We draw the perpendicular bisector AX which intersects AB on Y and the perpendicular bisector BX which intersects AB on Y. We join XY and XZ to have the required triangle XYZ.\[\]
Complete step-by-step solution
We are given two angles of the triangle XYZ that need to be constructed$\angle Y={{30}^{\circ }}$,$\angle Z={{90}^{\circ }}$ and the sum of lengths of the sides $XY+YZ+ZX=11$cm. \[\]
Step-1: We draw a line segment AB of length $XY+YZ+ZX=11$cm using scale. \[\]
Step-2: We construct angle $\angle LAB={{30}^{\circ }}$ on the point A by first constructing ${{60}^{\circ }}$ by taking an arc with the compass and then bisecting the angle of ${{60}^{\circ }}$. We construct angle $\angle MBA={{90}^{\circ }}$by taking the arc of ${{60}^{\circ }}$twice on the point $B$and then bisecting the second arc. \[\]
Step-3: We bisect angle $\angle LAB={{30}^{\circ }}$and $\angle MBA={{90}^{\circ }}$ by taking width of the arc of ${{30}^{\circ }},{{90}^{\circ }}$ and using the compass. We denote the point of intersection of bisecting rays of $\angle LAB,\angle MBA$ as $X$.\[\]
Step-4: We draw the perpendicular bisector of AX and we denote its point of intersection with AB as Y taking the length of AX as an arc using the compass. We similarly draw the perpendicular bisector of BX and denote its point of intersection with AB as Z taking the length of BX as an arc using the compass. \[\]
Step-5: We join the line segment XY and YZ using the scale. \[\]
XYZ is the required constructed triangle. \[\]
Note: We can justify the above construction first observing the triangle AYX where Y lies on the perpendicular bisector of AX which gives us $AY=YX\Rightarrow \angle YAX=\angle YXA=\dfrac{{{30}^{\circ }}}{2}={{15}^{\circ }}$ which implies$\angle AYX={{180}^{\circ }}-\left( \angle YAX+\angle YXA \right)={{150}^{\circ }}\Rightarrow \angle XYZ={{30}^{\circ }}$. Similarly Z lies on the perpendicular bisector of AX which gives us $ZX=ZB\Rightarrow \angle ZBX=\angle ZXB={{45}^{\circ }}$ which implies$\angle XZY={{90}^{\circ }}$. We also have $XY+YZ+ZX=AY+YZ+ZB=11$cm.
Complete step-by-step solution
We are given two angles of the triangle XYZ that need to be constructed$\angle Y={{30}^{\circ }}$,$\angle Z={{90}^{\circ }}$ and the sum of lengths of the sides $XY+YZ+ZX=11$cm. \[\]
Step-1: We draw a line segment AB of length $XY+YZ+ZX=11$cm using scale. \[\]
![seo images](https://www.vedantu.com/question-sets/6e0b4269-a559-4a4a-bdbf-8234a81813345333883208097400011.png)
Step-2: We construct angle $\angle LAB={{30}^{\circ }}$ on the point A by first constructing ${{60}^{\circ }}$ by taking an arc with the compass and then bisecting the angle of ${{60}^{\circ }}$. We construct angle $\angle MBA={{90}^{\circ }}$by taking the arc of ${{60}^{\circ }}$twice on the point $B$and then bisecting the second arc. \[\]
![seo images](https://www.vedantu.com/question-sets/6b4e3f4a-7dea-4e30-9e2d-d09072c3fb337891841720370934157.png)
Step-3: We bisect angle $\angle LAB={{30}^{\circ }}$and $\angle MBA={{90}^{\circ }}$ by taking width of the arc of ${{30}^{\circ }},{{90}^{\circ }}$ and using the compass. We denote the point of intersection of bisecting rays of $\angle LAB,\angle MBA$ as $X$.\[\]
![seo images](https://www.vedantu.com/question-sets/a3aab308-62cd-4f43-a932-3fdd7fbd583b3978362314615800450.png)
Step-4: We draw the perpendicular bisector of AX and we denote its point of intersection with AB as Y taking the length of AX as an arc using the compass. We similarly draw the perpendicular bisector of BX and denote its point of intersection with AB as Z taking the length of BX as an arc using the compass. \[\]
![seo images](https://www.vedantu.com/question-sets/159eb242-5663-4c8c-9d72-94f9dd6df3051080466653680001435.png)
Step-5: We join the line segment XY and YZ using the scale. \[\]
![seo images](https://www.vedantu.com/question-sets/dd41e0b2-e322-45fc-9ac1-2b86cfa8d8f3205141188190815949.png)
XYZ is the required constructed triangle. \[\]
Note: We can justify the above construction first observing the triangle AYX where Y lies on the perpendicular bisector of AX which gives us $AY=YX\Rightarrow \angle YAX=\angle YXA=\dfrac{{{30}^{\circ }}}{2}={{15}^{\circ }}$ which implies$\angle AYX={{180}^{\circ }}-\left( \angle YAX+\angle YXA \right)={{150}^{\circ }}\Rightarrow \angle XYZ={{30}^{\circ }}$. Similarly Z lies on the perpendicular bisector of AX which gives us $ZX=ZB\Rightarrow \angle ZBX=\angle ZXB={{45}^{\circ }}$ which implies$\angle XZY={{90}^{\circ }}$. We also have $XY+YZ+ZX=AY+YZ+ZB=11$cm.
Recently Updated Pages
Mark and label the given geoinformation on the outline class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
When people say No pun intended what does that mea class 8 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Name the states which share their boundary with Indias class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give an account of the Northern Plains of India class 9 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Advantages and disadvantages of science
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
10 examples of evaporation in daily life with explanations
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
How do you graph the function fx 4x class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)