Question

Please Solve the question 7 8 9

Answer 1

7) Draw BM such that AD || BM and draw AM such that AM || BC...
In quad. ABMD ..
two sides are parallel therefore it is a parallelogram...
Now AD = BM (opposite sides of ||gm)
Similarly in quad. ABCM..
AM = BC (opposite sides of ||gm)

Now ar(∆AMD) ...
s = 13+13+10/2 = 36/2 = 18cm
by Herons Formula ,
area = √s(s-a)(s-b)(s-c)
= √18(18-13)(18-13)(18-10)
= √9×2×5×5×2×4
= 3×2×5×2
= 60cm²

Now , All three triangles have equal sides and are isosceles triangle with equal length 13cm and base 10cm..so,

= 3×ar(∆AMD)
= 3×60
= 180cm²

8)
ar(trapezium) = 384cm²
let the side be x , now
3:5 = 3x + 5x
h = 12cm
now,
384 = 1/2(8x)12
768 = 8x×12
8x = 768/12
x = 64/8 = 8
A = 3x = 3×8 = 24cm
B = 5x = 5×8 = 40cm

9) a)

ar(ABCD)...
a = 24cm
b = 15cm
h = 12cm

area = 1/2×24+15×12
= 12+15×15
= 24×15 = 360cm²

ar(DCFE)
h = 10cm
a = 15cm
b = 21cm

area = 1/2×15+21×10
= 36×5
= 180cm²

area of the figure = 360+180
= 540cm²

(b)
ar(EDCF)
h = 24cm
a = 8cm
b = 15cm

area = 1/2×8+15×24
= 19×24
= 456cm²

figure FCAB is a rectangle as opposite sides are equal..

ar(FCAB)..
l = 15cm
b = 20cm

area = lb
= 15×20
= 300cm²

Area of the figure = 456+300
= 756cm²