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