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]
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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]
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Again
∆SVT is a right angle triangle
[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]
