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given triangle str, find the measure of a,b, and h​

Given Triangle Str Find The Measure Of Ab And H class=

Sagot :

FierceFireflieeidn

Answer:-

See the attachment first

If we triangle like this then we can solve the above problem easily.

Solution:-

According to the attachment and basic proportionality theorem.

[tex]\boxed{\sf SV^2=RV\times VT}[/tex]

[tex]\\ \sf\longmapsto h^2=20\times 4[/tex]

[tex]\\ \sf\longmapsto h^2=80[/tex]

[tex]\\ \sf\longmapsto h=\sqrt{80}[/tex]

[tex]\\ \sf\longmapsto h=8.4[/tex]

[tex]\\ \sf\longmapsto SV=8.4cm[/tex]

____________________________

Now

∆SVR is a right angle triangle

  • Perpendicular=p=8.4
  • Base=b=20
  • Hypontuse=h=?

Using Pythagoras Therom

[tex]\\ \sf\longmapsto h=\sqrt{p^2+b^2}[/tex]

[tex]\\ \sf\longmapsto h=\sqrt{(20)^2+(8.4)^2}[/tex]

[tex]\\ \sf\longmapsto h=\sqrt{400+70.5}[/tex]

[tex]\\ \sf\longmapsto h=\sqrt{470.5}[/tex]

[tex]\\ \sf\longmapsto h=21.6[/tex]

[tex]\therefore\boxed{\sf a=21.6cm}[/tex]

___________________________________

Again

∆SVT is a right angle triangle

  • P=8.4
  • b=4
  • h=?

[tex]\\ \sf\longmapsto h^2=p^2+b^2[/tex]

[tex]\\ \sf\longmapsto h^2=(8.4)^2+(4)^2[/tex]

[tex]\\ \sf\longmapsto h^2=70.5+16[/tex]

[tex]\\ \sf\longmapsto h^2=86.5[/tex]

[tex]\\ \sf\longmapsto h=\sqrt{86.5}[/tex]

[tex]\\ \sf\longmapsto h=9.5[/tex]

[tex]\therefore\boxed{\sf b=9.5cm}[/tex]

View image FierceFirefliee
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