Question

Plz solve Q 6,7,8 PLZ​

Answer 1

Step-by-step explanation:

Q.6.) Given : ΔABC is isosceles triangle

a) In ΔADB and ΔADC

AB = AC ....... Given

∠ABD = ∠ACD ......... Angles opposite to equal side are equal

AD = AD ....... Common side

∴ ΔADB ≅ ΔADC ......... S. A .S

∴ BD = DC.......C.P.CT.C

Hence, AD bisects BC

b) Consequently,

∠BAD = ∠CAD......C.P.C.T.C

Hence AD bisects ∠A

Q.7.) y² - 5y + 6

y² - 3y - 2y + 6 ....... Splitting middle term

(y² - 3y) - (2y + 6)

y(y - 3) - 2(y - 3)

(y - 3)(y - 2)

Q.8) $\frac{5 + \sqrt{3} }{5 - \sqrt{3} }$

Rationalising factor of 5 - √3 = 5 + √3

$\frac{5+\sqrt{3} }{5-\sqrt{3} } * \frac{5+\sqrt{3} }{5+\sqrt[]{3} }$

$\frac{(5+\sqrt{3}) ^{2} }{(5)^{2} - (\sqrt{3}) ^{2} }$

$\f5^{2} + (\sqrt{3})^{2}+ 2*5*\sqrt{3}$ = 25 + 3 + 10√3 = 28 + 10√3

(5)² - (√3)² = 25 - 3 = 22

$\frac{28+10\sqrt{3} }{22}$

Fraction is $\frac{14 + 5\sqrt{3} }{11}$