Question

2. In a parallelogram ABCD, If ZA = (3x + 12)°, ZB = (2x – 32)º then find the value
of x and then find the measures of ZC and ZD.​

Answer 1

Step-by-step explanation:

ABCD is a parallelogram. [Given] ∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary], ∴ (3x + 12)° + (2x-32)° = 180°

∴ 3x + 12 + 2x – 32 = 180

∴ 5x – 20 = 180

∴ 5x = 180 + 20

∴ 5x = 200

∴ x = 200/5

∴ x = 40

ii. ∠A = (3x + 12)° = [3(40) + 12]° = (120 + 12)° = 132° ∠B = (2x – 32)° = [2(40) – 32]° = (80 – 32)° = 48° ∴ ∠C = ∠A = 132° ∠D = ∠B = 48° [Opposite angles of a parallelogram]

∴ The value of x is 40, and the measures of ∠ZC and ∠ZD are 132° and 48° respectively.

Answer 2

$\Uparrow$See the attachment.

$\huge \bigstar$$\huge\bold{\mathtt{\purple{✍︎A{\pink{N{\green{S{\blue{W{\red{E{\orange{R✍︎}}}}}}}}}}}}}$$\huge \Rightarrow$

$\large\bold \red{Solution:}\Downarrow$

$\Box ABCD$ is a parallelogram.

∴$\angle A \angle B=180°$ ...(Adjacent

angles of parallelogram are supplementary.)

$∴(3x+12)°+(2x-32)°=180°$

$∴ 3x+12+2x-32=180$

$∴5x-20=180$

$∴5x=180+20$

$∴5x=200$

$\large ∴x=\frac{200}{5}$

$∴x =40°$

$\angle C = \angle A$ .(Opposite angles of a parallelogram)

$∴\angle C=3x+12$

$∴\angle C=3(40)+12$

$∴\angle C=120+12$

$∴\angle C=132°$

$∴\angle D=\angle B$ ...(Opposite angles of a parallelogram)

$∴ \angle D=2x-32$

$∴ \angle D= 2(40)-32$

$∴ \angle D= 80-32$

$∴ \angle D=48°$

$\Large\bold \purple{Ans:}\rightarrow$

$x=40°,\angle C = 132°,\angle D=48°.$