Question

A, what is the measure of <Bad?
B, what is the length of BD?
C, what is the measure of <ACD?
D, what is the length of DC? ​

Answer 1

$\huge \mathtt{Question:}$

In the adjoining figure,

A, what is the measure of <BAD?

B, what is the length of BD?

C, what is the measure of <ACD?

D, what is the length of DC?

$\large \mathtt{Question \:A:}$

Given,

<ABD = 45°

<ADB = 90°

Solution:

<ABD + <ADB + <BAD = 180° (Angle sum property)

45° + 90° + <BAD = 180°

135° + <BAD = 180°

<BAD = 180° - 135°

= 45°

• Note:- Since two angles of ∆ABD are equal, it is an isosceles triangle.

$\large \mathtt{Question \:B:}$

Given,

AD = 8cm

Solution:

BD = AD = 8cm (Sides opposite to equal angles of an isosceles triangle)

$\large \mathtt{Question \:C:}$

Given,

<DAC = 60°

Solution:

<ADC = 90° (Linear pair)

<ACD + 60° + 90° = 180° (Angle sum property)

<ACD + 150° = 180°

<ACD = 30°

$\large \mathtt{Question \:D:}$

Given,

AD = 8 cm

<DAC = 60°

<ADC = 90°

<ACD = 30°

Solution:

tan(60°) = √3

√3 = $\huge \frac{DC}{8}$

DC = 8 × √3

= 13.8564064606cm

$\large \mathtt{Question \:E:}$

Given,

DC = 13.8564064606cm

BD = 8cm

AD = 8cm

Solution:

Area of triangle = $\huge \frac{1}{2}$× base × height

Base = DC + BD = 21.8564064606cm

Height = 8cm

Area of triangle = $\huge \frac{1}{2}$× base × height

Area = $\huge \frac{1}{2}$× 21.8564064606 × 8

= 87.4258418.