Question

10th question pls answer with explanation,
Will mark as brainiest

Answer 1

1.

A= ½× b ×h

294= ½×3×4x=

294 = 6x

x=294/6

x=49

so b = 3(49) h = 4(49)

b = 147cm h = 196cm

2.

Assuming length QU=RA

The figure can be split into trapezium ERQS and a rectangle QRUA

Area of a trapezium:

A = ½ (a+b)× h

Area = ½ (8+16) (30-22)

Area = ½ (24) (8)

Area =96cm²

Find are of a rectangle

area of a rectangle = lxb

Area = 22×16= 352cm²

Total area = 96+352 = 448cm²

Answer is 448cm²

3.

360cm² is number 3

4.

for number 4 I attached in the photo

5.

If the parallel sides of a trapezium are in the ratio 3:5, its height is 16 cm, and its area is 768 cm^2, what are the lengths of the parallel sides?

GIVEN: A trapezium ABCD , DC // AB

DC:AB = 3:5 , So, If DC = 3x AB = 5x

Height CG of the trapezium = 16 cm

Area ( trapezium ABCD) = 768cm²

TO FIND: DC & AB

CONSTRUCTION: Cut AE = DC. Since AE // DC

So, AECD is a parellelogram ( as one pair of opposite sides are equal & parallel)

= AE = 3x, EB = 2x

Area( parallelogram AECD) = side * corresponding height

=> 3x * 16 = 48x cm ² ………(1)

Area ( triangle CEB) = 1/2 * side * corresponding height.

Area( triangle CEB) = 1/2 * 2x * 16 = 16x cm²…(2)

Now by adding (1) +(2)

48x + 16x = 64x = 768

=> x = 768/64 = 12

So, 3x = 36cm

& 5x = 60cm

(I DID HALF IM TO LAZY TO DO THE REMAINING BUT I WILL TELL YOU NUMBER 6 INCASE)

6.

Area of a trapezium is A=(B+b)×h/2 where

B is the large paralel line

b is the small paralel line

h is altitude

P is perimeter and c the nonparalel line

P=a+b+2c from this a+b=P-2c a+b=52-20

a+b=32

A=(32×8)/2=128 area A=128

7.

is 21

8.

h=20

DO NUMBER 9 AND 10 YOURSELF IM TIRED