Wellcome!

Aman Senapati

Given,

1. Area of trapezium ABCE = 475 sq.cm
2. Height = 19 cm.
3. one side is 4 cm greater than the other.

Area of trapezium = 1/2 ( sum of // sides ) * H

let the // sides of trapezium are x and x + 4.

475 sq.cm = 1/2 ( x + x + 4) * 19cm.

475/19  = 2x + 42

25 * 2  = 2x + 4

50 - 4 = 2x

46 cm = 2x

46/2 = x

23=x

x + 4 = 23 + 4 = 27.

∴ // sides of trapezium are 23 cm and 27 cm.

-Answered by Aman Senapati On 25 May 2020:10:02:27 AM

Question 2. Asked on :25 May 2020:08:00:22 AM

Q-1)Find the area of trapezium PQRS with height PQ where SP=12cm,SR=13cm,QR=7cm?

-Added by Aman Senapati Mathematics » Herons Formula

Aman Senapati

Given,

• PS=12m
• SR=13m
• QR=7m

PQRS is a trapezium.

let SUR is a triangle and PQUR is a rectangle.

then SU=PS-QR

SU=12m-7m

SU=5m.

By Pythagorous theorm:-

(H sq.)=(P sq.) + (B sq.)

(13m sq.)=(P sq.) + (5m sq.)

169 sq. m=(P sq.) + 25 sq.m

169-25=(P sq.)

144=(P sq.)

√144=P

√12*12=P

12m=P

UR = 12m

According to the property of rectangle ( Opposite side's are equal )

UR=PQ

∴ PQ=12m.

-Answered by Aman Senapati On 25 May 2020:08:46:56 AM

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