If - - -cm- - - - cm- - - -cm- - - - cm- - - - cm and - - - - - - - Find the perimeter of quadrilateral - -

AD is the angle bisector of angle 𝑪𝑨𝑩, so by the angle bisector theorem, 𝑪𝑫: 𝑩𝑫 = 𝑪𝑨: 𝑩𝑨. 3/4 = CA/6 4×CA = 18 CA = 4.5 cm AF is the angle bisector of angle ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Internal Angle Bisector Theorem

If 𝑪𝑫 = 𝟑c...

Question

If 𝑪𝑫 = 𝟑cm, 𝑫𝑩 = 𝟒 cm, 𝑩𝑭 = 𝟒cm, 𝑭𝑬 = 𝟓 cm, 𝑨𝑩 = 𝟔 cm and ∠𝑪𝑨𝑫 ≅ ∠𝑫𝑨𝑩 ≅ ∠𝑩𝑨𝑭 ≅ ∠𝑭𝑨𝑬, Find the perimeter of quadrilateral 𝑨𝑬𝑩𝑪 .

Open in App

Solution

AD is the angle bisector of angle 𝑪𝑨𝑩, so by the angle bisector theorem, 𝑪𝑫: 𝑩𝑫 = 𝑪𝑨: 𝑩𝑨.

$\frac{3}{4} = \frac{C A}{6}$

4 $\times$ CA = 18

CA = 4.5 cm

AF is the angle bisector of angle 𝑩𝑨𝑬, so by the angle bisector theorem, 𝑩𝑭:𝑬𝑭 = 𝑩𝑨:𝑬𝑨.

$\frac{4}{5} = \frac{6}{E A}$

4 $\times$ EA = 30

EA = 7.5 cm

The perimeter of quadrilateral
𝑨𝑬𝑩𝑪 = 𝑪𝑫 + 𝑫𝑩 + 𝑩𝑭 + 𝑭𝑬 + 𝑬𝑨 + 𝑨𝑪
𝑨𝑬𝑩𝑪 = 𝟑 + 𝟒 + 𝟒 + 𝟓 + 𝟕. 𝟓 + 𝟒. 𝟓
𝑨𝑬𝑩𝑪 = 28 cm

Suggest Corrections

Similar questions

Q. Find the area of quadrilateral whose diagonal length is

10

cm, height

h_{1} = 5

cm and height

h_{2}

7

cm.

Q. Battery acid is 4.22 M aqueous

H_{2} S O_{4}

solution, and has density of 1.21 g cm

^{- 3}

. What is the molality of

H_{2} S O_{4}

[H = 1, S = 32, O = 16]

Q. A catalyst has a total surface area of

10^{3}

^{2}

and adsorption sites available on it are

6 \times

^{15}

sites per cm

^{2}

H_{2}

gas is adsorbed on this surface occupying only

5 %

of the available sites of total surface area. On heating, the catalyst releases

H_{2}

back and whole of the

H_{2}

adsorbed is measured and found to occupy

2.46

mL at

0.03

atm and

300

K. The number of

H_{2}

molecules adsorbed on each site is:

The correct statement with regard to $H_{2}^{+}$ & $H_{2}^{-}$ is

Q. Find the area of the given figure, if AC = 10 cm, AD = 12 cm, and

h_{1}, h_{2}, h_{3}

are 3, 4, 5 in cm respectively.

Join BYJU'S Learning Program