Question

heron's formula find the area of trapezium whose two parallel sides are 60 cm and 77 cm and the other side the 25 cm and 26 CM​

Answer 1

$\large{\underline{\red{\rm{AnswEr:}}}}$

Given

• The 2 parallel sides of trapezium are
• PS=77cm and QR=60cm
• The other sides are
• PQ=25cm and RS=26cm

Here, QU and RT is joining

∴QR=UT=60cm

Now,

• In right angle ∆RTS
• RT²=RS²-TS²

⟹RT²=26²-(17-x)²

⟹here , RT=QU

⟹25²-x²=26²-(17²-33x+x²)

⟹625-x²=676-189+34x-x²

⟹625=387+34x

⟹625-387=34x

⟹238=34x

⟹x=7cm

Thus

⟹UQ=√25²-x²

⟹UQ=√625-49

⟹UQ=√576

⟹UQ=24cm

Hence,

⟹Area of trapezium=1/2*(sum of parall sides)* height

⟹Area of trapezium=1/2*(77+60)*24

⟹Area of trapezium=1/2*137*24

⟹Area of trapezium=137*12

⟹Area of trapezium=1644cm²

$\bold\purple{\boxed{ANSWER=>1644\:cm^2}}$

Answer 2

Answer:-

Draw a line BC parallel to AD 

Draw a perpendicular line BE on DF

ABCD is a parallelogram.

BC=AD=25cm

CD=AB=60cm

CF=77-CD=17cm

For Δ BCF

perimeter of triangle=(25+26+17)/2

=68/2=34

By Heron's Formula of triangle=√s(s-a)(s-b)(s-c)

=√34(34-25)(34-26)(34-17)

=204cm²

Now area of ΔBCF=1/2xbase x height

=1/2 BExCF204cm²

=1/2x BEx17BE

=408/17

=24cm

Area of Trapezium =1/2(AB+DF)xBE

=1/2(60+77)x24

=1644cm²

Hence the area of trapezium is 1644cm².