Question

ABCD is a parallelogram with side AB = 12 cm.
Its diagonals AC and BD are of lengths 20 cm and
16 cm respectively. Find the area of || gm ABCD.​

Answer 1

Answer:

Let O be the intersecting point of AC and BD

We know,

diagonals of a parallelogram bisect each other

OA =OC = 1/2×AC = 10cm

OB = OD= 1/2×BD = 8cm

in ∆AOB

OA=10cm

OB=8cm

AB=12cm

By using Heron's formula

ar(∆AOB)=√s(s-a)(s-b)(s-c)

s=(a+b+c)/2=(12+8+10)/

2=30/2=15cm

ar(∆AOB)=√15×3×7×5

= 15√7 cm²

we know that the diagonals of a parallelogram divides it into four equal triangles

=>ar(∆AOB)=ar(∆BOC)=ar(∆COD)=ar(∆AOD)= 15√7 cm²

ar(ABCD) = 4*15√7 = 60√7 cm² (Ans.)

$\color{red}{Mark\: as\: Brainlist}$

Answer 2

on:

O को AC और BD का प्रतिच्छेदन बिंदु माना जाता है

हम लोग जान,

एक समांतर चतुर्भुज के विकर्ण एक दूसरे को काटते हैं

OA = OC = 1/2 × AC = 10 सेमी

ओबी = ओडी = 1/2 × बीडी = 8 सेमी

inAOB में

OA = 10 सेमी

ओबी = 8 सेमी

एबी = 12 सेमी

बगुला के सूत्र का उपयोग करके

ar (saAOB) = OBs (sa) (sb) (sc)

s = (a + b + c) / 2 = (12 + 8 + 10) /

2 = 30/2 = 15 सेमी

ar ((AOB) = ∆15 × 3 × 7 × 5

= 15 =7 सेमी√

हम जानते हैं कि एक समांतर चतुर्भुज के विकर्ण इसे चार समान त्रिभुजों में विभाजित करते हैं

=> ar (>AOB) = ar (OCBOC) = ar ((COD) = ar (√AOD) = 15√7 cm∆

ar (ABCD) = 4 * 15√7 = 60²7 cm Ans (Ans।)