Question

(6)
In the given figure, ABCD is a rectangle, whose diagonals intersect at 'O'.
Diagonal AC is produced to E and ZDCE = 145º.
Find:
D
(0) <CAB
(ii) ZAOB
(iii)
ZACB
he leitude and area of an isosceles triangle whose perimeter is 64 C​

Answer 1

please give the image...

Answer 2

(6)  In the given figure, ABCD is a rectangle, whose diagonals intersect at 'O'.

Diagonal AC is produced to E and ZDCE = 145º.

Consider the attached figure while going through the following steps.

Given,

ABCD is a rectangle

Diagonal intersect at O

Diagonal AC is produced to E.

∠ DCE = 145°

Now, consider

∠ ACD = 180° - 145°       (ACE is a straight line)

∴ ∠ ACD = 35°

∠ COD = ∠ DOA = ∠ AOB = ∠ BOC (diagonal of rectangle intersect at 90°)

∴ ∠ AOB = 90°

∠ ACD + ∠ ACB = 90°  (angle between sides of a rectangle equals 90°)

35° + ∠ ACB = 90°

∠ ACB = 90° - 35°

∴ ∠ ACB = 55°

In Δ ACB,

∠ CBA = 90° (angle between sides of a rectangle equals 90°)

∠ ACB + ∠ CBA + ∠ CAB = 180°

55° + 90° + ∠ CAB = 180°

145° + ∠ CAB = 180°

∠ CAB = 180° - 145°

∴ ∠ CAB = 35°

The area of isosceles triangle given perimeter is,

A = 1/4 b √ [P (P - 2b) ]

where, b = altitude

            P = perimeter of triangle

            A = area of triangle

A = 1/4 b √ [64 (64 - 2b) ]

= 1/4 b √ [64 × 64 - 64 × 2b) ]

= 1/4 b √ [64² - 128b ]

= b √ [64²/16 - 128b/16 ]

∴ A = b √ (256 - 8b)