From the adjoining figure, find(i) $x$ (ii) $∠$ $D A B$
(iii) $∠$ $A D B$

Open in App

Solution

(i) From the given figure, $A B C D$ is a quadrilateral
$\begin{matrix} ∠ A + ∠ B + ∠ C + ∠ D = 360^{\circ} (3 x + 4) + (50 + x) + (5 x + 8) + (3 x + 10) = 360^{\circ} 3 x + 4 + 50 + x + 5 x + 8 + 3 x + 20 = 360^{\circ} 12 x + 72 = 360^{\circ} \end{matrix}$
By transposing we get,
$\begin{matrix} 12 x = 360^{\circ} - 72 12 x = 288 x = 288 / 12 x = 24 \end{matrix}$
(ii) $∠ D A B = (3 x + 4)$
$= ((3 \times 24) + 4)$
$\begin{matrix} = 72 + 4 = 76^{\circ} \end{matrix}$
Therefore, $∠ D A B = 76^{\circ}$
(iii) Consider the triangle ABD, We know that, sum of interior angles of triangle is equal to $180^{\circ}$ ,
$\begin{matrix} ∠ D A B + ∠ A B D + ∠ A D B = 180^{\circ} 76^{\circ} + 50^{\circ} + ∠ A D B = 180^{\circ} ∠ A D B + 126^{\circ} = 180^{\circ} ∠ A D B = 180^{\circ} - 126^{\circ} \end{matrix}$
Therefore, $∠ A D B = 54^{\circ}$

Suggest Corrections

Similar questions

Q. From the adjoining figure, find
i) the area of

△ A B C

ii) length of

B C

iii) the length of altitude from

A

B C

In the adjoining figure, ABCD is a parallelogram in which $∠ D A B = 80^{\circ} a n d ∠ D B C = 60^{\circ}$ . Calculate $∠ C D B a n d ∠ A D B$ .

Q. In the adjoining figure,

A B C D

is a parallelogram and diagonals intersect at

O

. Find
(i)

∠ C A D

(ii)

∠ A C D

(iii)

∠ A D C

Q. In the adjoining figure,

A B C D

is a rhombus and

E D C

is an equilateral triangle. If

∠ D A B = 48^{o}

, find
i)

∠ B E C

ii)

∠ D E B

iii)

∠ B F C

In the adjoining figure, $A B = A C$ and $B D = D C$ . Prove that $Δ A D B ≅ Δ A D C$ and hence show that
$(i) ∠ A D B = ∠ A D C = 90^{\circ}$
$(i i) ∠ B A D = ∠ C A D$

Join BYJU'S Learning Program

From the adjoining figure, find(i) x(ii)∠ DAB(iii) ∠ ADB

From the adjoining figure, find(i) $x$ (ii) $∠$ $D A B$
(iii) $∠$ $A D B$