Question

the area of quadrilateral in given figure​

Answer 1

Answer:

Step-by-step explanation:

In Triangle DHC

Base = 8cm

Hypotenuse = 17cm

Angle H = 90°

Perpendicular / height = ?

So, by using Pythagoras theorem

H² = P² + B²

(17)² = (P)² + (8)²

289 = P² + 64

289-64 = P²

225 = P²

√225 = P

15 = P
hence height = 15cm
Area = 1/2 ×b ×h
Area = 1/2 × 8 × 15
Area = 60cm²
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In triangle ADH
Base = 12

Hypotenuse = 15

Angle A = 90°

Perpendicular / height = ?
so, by Pythagoras theorem
H² = P² + B²
(15)² = P² + (12)²
225 = P² + 144
225 - 144 = P²
81 = P²
√81 = P
9 = P
Area = 1/2×12×9
Area = 54cm²
___________________
Area of quadrilateral
Area of TriangleDHC + Area of TriangleADH
Area = 60 + 54
Area = 114cm²
___________________
Hope it will help you!
Mark Brainliest.

Answer 2

Answer:

ar(ABCD)= 122 cm²

Step-by-step explanation:

Gn:

CD= 17cm, BC= 8cm, AD= 12cm &∠DBC= ∠DAB= 90°

From the Given Figure and Info, we can understand that

ΔABD & ΔBCD are right triangles.

In ΔBCD, CD is the hypotenuse.

∴CD²= BC²+ BD²

⇒289= 64+ BD²

289- 64= 225= BD²

BD= √225= 15cm

In ΔABD, BD is the hypotenuse and BD= 15cm

∴BD²= AD²+ AB²

⇒225= 144+ AB²

225- 144= AB²

AB²= 81

AB= 9cm.

ar(ABCD)= ar(ABD)+ ar(BCD)

=(1/2×12×9)+ (1/2×8×17)    (As ABD & BCD are right triangle, Area= 1/2× length× breadth)

=(6×9)+ (4×17)

=54+68= 122cm²