Question

evaluate x and y in each of the figures given below.​

Answer 1

Answer:

$x = 5 \\ y = 6$

may it help u

may it help u

may it help u

may it help u

Answer 2

$\huge \red \star \large \green \star \tiny \orange \star \huge \underline{ \boxed{ \red{ \frak{s}} \frak{ \orange{o} \blue{lu} \pink{t} \green{io} \red{n}}}} \tiny \orange \star \large \green \star \huge \red \star$

✪ Given :

• AB = 17cm
• DO = 19cm
• AO = OC
• ∠DOC = ∠AOB
• ∠OAB = ∠OCD

✪ To Find :

• (3x + 2)cm
• (2y + 7)cm

✪ Solving :

In △AOB and △DOC :

• ∠DOC = ∠AOB (Given)
• AO = OC (Given)
• ∠OAB = ∠OCD (Given)

$\large \sf\mapsto \pink{ \: therefore \: \: △AOB \cong △DOC \: (a.s.a)}$

A.S.A - Angle,Side,Angle congruence condition

• AB = DC (c.p.c.t) ____(1)
• DO = OB (c.p.c.t)_____(2)

➜ From (1) :

• AB = DC = 17cm
• AB = 3x + 2

$\large \red \implies \sf \: (3x + 2) = 17cm$

⍟ On Transposing The Terms :

$\large \red \implies \sf \: 3x = 17 - 2cm \\ \\ \large \red \implies \sf \: x = \frac{15}{3} cm$

$\\ \large \therefore \red{ \underline{ \boxed{ \tt{ \green{x = 5cm}}}}}$

✪ Check :

(3x + 2 )cm= 17cm

⇒ 3(5) + 2cm = 17cm

⇒ 15 + 2cm = 17cm

17cm = 17cm

L.H.S = R.H.S

Hence Proved

➜ From (2) :

• DO = OB = 19cm
• OB = (2y + 7)cm

$\large \red \implies \sf \: (2y + 7) = 19cm \\ \large \red \implies \sf \: 2y = 19 - 7cm \\ \\ \large \red \implies \sf \: y = \frac{12cm}{2}$

$\\ \large \therefore \red{ \underline{ \boxed{ \tt{ \green{y = 6cm}}}}}$

✪ Check :

(2y + 7)cm = 19cm

⇒ 2(6) + 7cm = 19cm

⇒ 12 + 7cm = 19cm

19cm = 19cm

L.H.S = R.H.S

Hence Proved