please this time answer and send pic i only want pic
these two...!!!
Answers
Step-by-step explanation:
Solutions:-
C)
ABCD is a rectangle
AB = CD = 14 cm
AD = BC = 10 cm
Method-1:-
Area of a rectangle = length × breadth
area of the rectangle ABCD=>AB×BC
area of the rectangle ABCD=>14×10
area of the rectangle ABCD=140 sq.cm
Base of the un shaded triangle = AB = 14 cm
Height of the un shaded triangle = BC = 10 cm
Area of a triangle = base×height/2 sq.units
Area of the un shaded triangle =
=>AB×BC/2
=>14×10/2 sq.cm
=>7×10
=>70 sq.cm
Area of un shaded triangle = 70 sq.cm
Area of the shaded region =
Area of the rectangle - Area of un shaded triangle
=>140-70
=>70 sq.cm
method-2:-
Area of a triangle of base 9cm and height is 10 cm
Area of a right angled triangle = 9×10/2
=>9×5
=>45 sq.cm
Area of the triangle whose base 5 cm and height is 10 cm
=>5×10/2
=>5×5
=>25 sq.cm
Area of two shaded triangles = 45+25 = 70 sq.cm
D)
See the above attachment
Step -1
Draw the rough digram for the above trinagle
Step -2:-
Construction
See the attached
Step-3:-
Steps of Construction:-
1.Draw a linesegment QR with 9.5 cm
2. Construct angel 60° at Q and extend it.
3.Draw an arc with 4.3 cm from Q such that it intersect PQ at P.
4.Join P and R
5.∆ PQR is the required triangle .
