Question

solve this..!! asap!​

Answer 1

Answer:

S=AB=10

d= AC=16

AO=d/2=16/2=8

In right ∆ABO

BO=

$\sqrt{ {h}^{2} - {b}^{2} } \\ = \sqrt{ {10}^{2} - {8}^{2} } \\ = \sqrt{100 - 64 } \\ = \sqrt{36} \\ = 6$

BD=BOx2=6x2=12

AREA OF RHOMBUS= AC x BD =16x12=192m²

Answer 2

Step-by-step explanation:

Let’s take Triangle ABO

(I am considering AC = 16 cm)

AB = 10 cm is the hypotenuse

AO = 8 cm (half of AC as diagonals bisect)

hence in the triangle ABO , using Pythagorean theorem ,

AO2+BO2=AB2

BO2+82=102

BO2+64=100

BO2=100−64==>36

√(BO2)=√36

BO = 6 cm

therefore diagonal BD=2∗BO

BD = 12 cm

hence area of rhombus = 1/2 * AC * BD

[1/2∗(16∗12)]cm2

96 cm^2