1 Which of the following
equations has two distinct real
roots.
2X2-32x+9/4=0
2 x+x-5=0
x² +3x+2^2=0
5x2-3x+1=0
Answers
{\bold{\pink{\underline{\green{G}\purple{iv}\orange{en}\red{:-}}}}}
Given:−
\bf\:perimeter=42\:cmperimeter=42cm
\bf\:First\:side\:(a)=18\:cmFirstside(a)=18cm
\bf\:second\:side\:(b)=10\:cmsecondside(b)=10cm
\bf\:Third\:side\:(c)=perimeter-(a+b)Thirdside(c)=perimeter−(a+b)
\bf\:Third\:side=42-(18+10)\:cmThirdside=42−(18+10)cm
\bf\:Third\:side=42-28\:cmThirdside=42−28cm
\bf\:Third\:side=14\:cmThirdside=14cm
{\bold{\blue{\underline{\red{To}\:\pink{Fi}\green{nd}\purple{:-}}}}}
ToFind:−
\bf\green{{Area\:of\:triangle=?}}Areaoftriangle=?
{\bold{\pink{\underline{\red{So}\purple{lut}\green{ion}\orange{:-}}}}}
Solution:−
Now we have,
a = 18 cm ,b = 10 cm and c = 14 cm
\bf\:Semi\:perimeter=\dfrac{a+b+c}{2}Semiperimeter=
2
a+b+c
\bf\:Semi\:perimeter=\dfrac{18+10+14}{2}Semiperimeter=
2
18+10+14
\bf\:Semi\:perimeter=\dfrac{42}{2}Semiperimeter=
2
42
\bf\red{{Semi\:perimeter=21\:cm}}Semiperimeter=21cm
\bf\boxed\star\pink{\underline{\underline{{Using\:heron's\:formula:-}}}}
⋆
Usingheron
′
sformula:−
\bf\:Area=\sqrt{s(s-a)(s-b)(s-c)}Area=
s(s−a)(s−b)(s−c)
\bf\:Area=\sqrt{21(21-18)(21-10)(21-14)}Area=
21(21−18)(21−10)(21−14)
\bf\:Area=\sqrt{21\times\:3\times\:11\times\:7}Area=
21×3×11×7
\bf\:Area=\sqrt{3\times\:7\times\:3\times\:11\times\:7}Area=
3×7×3×11×7
\bf\:Area=3\times\:7\sqrt{11}\:c{m}^{2}Area=3×7
11
cm
2
\bf\orange{{Area=21\sqrt{11}\:c{m}^{2}}}Area=21
11
cm
2
Hence,area of triangle will be 21√11 cm²